inputs
stringlengths 50
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stringlengths 4
655k
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|---|---|
Polycarp has $n$ different binary words. A word called binary if it contains only characters '0' and '1'. For example, these words are binary: "0001", "11", "0" and "0011100".
Polycarp wants to offer his set of $n$ binary words to play a game "words". In this game, players name words and each next word (starting from the second) must start with the last character of the previous word. The first word can be any. For example, these sequence of words can be named during the game: "0101", "1", "10", "00", "00001".
Word reversal is the operation of reversing the order of the characters. For example, the word "0111" after the reversal becomes "1110", the word "11010" after the reversal becomes "01011".
Probably, Polycarp has such a set of words that there is no way to put them in the order correspondent to the game rules. In this situation, he wants to reverse some words from his set so that: the final set of $n$ words still contains different words (i.e. all words are unique); there is a way to put all words of the final set of words in the order so that the final sequence of $n$ words is consistent with the game rules.
Polycarp wants to reverse minimal number of words. Please, help him.
-----Input-----
The first line of the input contains one integer $t$ ($1 \le t \le 10^4$) — the number of test cases in the input. Then $t$ test cases follow.
The first line of a test case contains one integer $n$ ($1 \le n \le 2\cdot10^5$) — the number of words in the Polycarp's set. Next $n$ lines contain these words. All of $n$ words aren't empty and contains only characters '0' and '1'. The sum of word lengths doesn't exceed $4\cdot10^6$. All words are different.
Guaranteed, that the sum of $n$ for all test cases in the input doesn't exceed $2\cdot10^5$. Also, guaranteed that the sum of word lengths for all test cases in the input doesn't exceed $4\cdot10^6$.
-----Output-----
Print answer for all of $t$ test cases in the order they appear.
If there is no answer for the test case, print -1. Otherwise, the first line of the output should contain $k$ ($0 \le k \le n$) — the minimal number of words in the set which should be reversed. The second line of the output should contain $k$ distinct integers — the indexes of the words in the set which should be reversed. Words are numerated from $1$ to $n$ in the order they appear. If $k=0$ you can skip this line (or you can print an empty line). If there are many answers you can print any of them.
-----Example-----
Input
4
4
0001
1000
0011
0111
3
010
101
0
2
00000
00001
4
01
001
0001
00001
Output
1
3
-1
0
2
1 2
|
for _ in range(int(input())):
n = int(input())
mass = []
zo = 0
oz = 0
zz = 0
oo = 0
ozs = []
zos = []
ozss = set()
zoss = set()
for j in range(n):
k = input()
mass.append(k)
if k[0] == '0' and k[-1] == '1':
zoss.add(k)
zos.append(j + 1)
zo += 1
elif k[0] == '1' and k[-1] == '0':
ozss.add(k)
ozs.append(j + 1)
oz += 1
elif k[0] == '0' and k[-1] == '0':
zz += 1
else:
oo += 1
if zz and oo and not oz and not zo:
print(-1)
continue
else:
if zo > oz:
print((zo - oz) // 2)
ans = []
need = (zo - oz) // 2
i = 0
while need:
zzz = mass[zos[i] - 1][len(mass[zos[i] - 1]) - 1:: -1]
if zzz not in ozss:
ans.append(zos[i])
need -= 1
i += 1
print(*ans)
else:
print((oz - zo) // 2)
ans = []
need = (oz - zo) // 2
i = 0
while need:
zzz = mass[ozs[i] - 1][len(mass[ozs[i] - 1]) - 1:: -1]
if zzz not in zoss:
ans.append(ozs[i])
need -= 1
i += 1
print(*ans)
|
Polycarp has $n$ different binary words. A word called binary if it contains only characters '0' and '1'. For example, these words are binary: "0001", "11", "0" and "0011100".
Polycarp wants to offer his set of $n$ binary words to play a game "words". In this game, players name words and each next word (starting from the second) must start with the last character of the previous word. The first word can be any. For example, these sequence of words can be named during the game: "0101", "1", "10", "00", "00001".
Word reversal is the operation of reversing the order of the characters. For example, the word "0111" after the reversal becomes "1110", the word "11010" after the reversal becomes "01011".
Probably, Polycarp has such a set of words that there is no way to put them in the order correspondent to the game rules. In this situation, he wants to reverse some words from his set so that: the final set of $n$ words still contains different words (i.e. all words are unique); there is a way to put all words of the final set of words in the order so that the final sequence of $n$ words is consistent with the game rules.
Polycarp wants to reverse minimal number of words. Please, help him.
-----Input-----
The first line of the input contains one integer $t$ ($1 \le t \le 10^4$) — the number of test cases in the input. Then $t$ test cases follow.
The first line of a test case contains one integer $n$ ($1 \le n \le 2\cdot10^5$) — the number of words in the Polycarp's set. Next $n$ lines contain these words. All of $n$ words aren't empty and contains only characters '0' and '1'. The sum of word lengths doesn't exceed $4\cdot10^6$. All words are different.
Guaranteed, that the sum of $n$ for all test cases in the input doesn't exceed $2\cdot10^5$. Also, guaranteed that the sum of word lengths for all test cases in the input doesn't exceed $4\cdot10^6$.
-----Output-----
Print answer for all of $t$ test cases in the order they appear.
If there is no answer for the test case, print -1. Otherwise, the first line of the output should contain $k$ ($0 \le k \le n$) — the minimal number of words in the set which should be reversed. The second line of the output should contain $k$ distinct integers — the indexes of the words in the set which should be reversed. Words are numerated from $1$ to $n$ in the order they appear. If $k=0$ you can skip this line (or you can print an empty line). If there are many answers you can print any of them.
-----Example-----
Input
4
4
0001
1000
0011
0111
3
010
101
0
2
00000
00001
4
01
001
0001
00001
Output
1
3
-1
0
2
1 2
|
k = int(input())
for i in range(k):
is_t = set()
a = dict()
a['00'] = []
a['11'] = []
a['01'] = []
a['10'] = []
n = int(input())
s = []
for i in range(n):
b = input()
a[b[0] + b[-1]].append(i)
s.append(b)
is_t.add(b)
c = len(a['10'])
d = len(a['01'])
if c + d == 0:
if len(a['00']) == 0 or len(a['11']) == 0:
print(0)
else:
print(-1)
elif c > d:
ans = []
i = 0
m = (d + c) // 2
while d != m and i < len(a['10']):
s1 = s[a['10'][i]]
if s1[::-1] not in is_t:
d += 1
ans.append(a['10'][i] + 1)
i += 1
if d != m:
print(-1)
else:
print(len(ans))
print(*ans)
else:
ans = []
i = 0
m = (d + c) // 2
while c != m and i < len(a['01']):
s1 = s[a['01'][i]]
if s1[::-1] not in is_t:
c += 1
ans.append(a['01'][i] + 1)
i += 1
if c != m:
print(-1)
else:
print(len(ans))
print(*ans)
|
Polycarp has $n$ different binary words. A word called binary if it contains only characters '0' and '1'. For example, these words are binary: "0001", "11", "0" and "0011100".
Polycarp wants to offer his set of $n$ binary words to play a game "words". In this game, players name words and each next word (starting from the second) must start with the last character of the previous word. The first word can be any. For example, these sequence of words can be named during the game: "0101", "1", "10", "00", "00001".
Word reversal is the operation of reversing the order of the characters. For example, the word "0111" after the reversal becomes "1110", the word "11010" after the reversal becomes "01011".
Probably, Polycarp has such a set of words that there is no way to put them in the order correspondent to the game rules. In this situation, he wants to reverse some words from his set so that: the final set of $n$ words still contains different words (i.e. all words are unique); there is a way to put all words of the final set of words in the order so that the final sequence of $n$ words is consistent with the game rules.
Polycarp wants to reverse minimal number of words. Please, help him.
-----Input-----
The first line of the input contains one integer $t$ ($1 \le t \le 10^4$) — the number of test cases in the input. Then $t$ test cases follow.
The first line of a test case contains one integer $n$ ($1 \le n \le 2\cdot10^5$) — the number of words in the Polycarp's set. Next $n$ lines contain these words. All of $n$ words aren't empty and contains only characters '0' and '1'. The sum of word lengths doesn't exceed $4\cdot10^6$. All words are different.
Guaranteed, that the sum of $n$ for all test cases in the input doesn't exceed $2\cdot10^5$. Also, guaranteed that the sum of word lengths for all test cases in the input doesn't exceed $4\cdot10^6$.
-----Output-----
Print answer for all of $t$ test cases in the order they appear.
If there is no answer for the test case, print -1. Otherwise, the first line of the output should contain $k$ ($0 \le k \le n$) — the minimal number of words in the set which should be reversed. The second line of the output should contain $k$ distinct integers — the indexes of the words in the set which should be reversed. Words are numerated from $1$ to $n$ in the order they appear. If $k=0$ you can skip this line (or you can print an empty line). If there are many answers you can print any of them.
-----Example-----
Input
4
4
0001
1000
0011
0111
3
010
101
0
2
00000
00001
4
01
001
0001
00001
Output
1
3
-1
0
2
1 2
|
N = int(input())
def ceildiv(x, y):
if x % y == 0:
return x // y
else:
return x // y + 1
for _ in range(N):
doms = []
oc, zc = 0, 0
n = int(input())
used = set()
fulls = dict()
for i in range(n):
d = input()
used.add(d)
if d[0] != d[-1]:
fulls[i] = d
doms.append((i, (d[0], d[-1])))
else:
if d[0] == '0':
zc = 1
else:
oc = 1
if len(doms) == 0:
if zc == 1 and oc == 1:
print(-1)
else:
print(0)
else:
# print(doms)
_01 = 0
_10 = 0
_01_indexes = []
_10_indexes = []
for dom in doms:
if dom[1] == ('0', '1'):
_01 += 1
_01_indexes.append(dom[0])
else:
_10 += 1
_10_indexes.append(dom[0])
if _10 < _01:
_01, _10 = _10, _01
_01_indexes, _10_indexes = _10_indexes, _01_indexes
_10_indexes = [x for x in _10_indexes if fulls[x][::-1] not in used]
need = ceildiv(_10-_01-1, 2)
if len(_10_indexes) >= need:
print(need)
print( ' '.join(list([str(x+1) for x in _10_indexes[:need]])) )
else:
print(-1)
# print("===")
# print(ceil(abs(doms.count(('0', '1')) - doms.count(('1', '0'))) - 1, 2))
|
Polycarp has $n$ different binary words. A word called binary if it contains only characters '0' and '1'. For example, these words are binary: "0001", "11", "0" and "0011100".
Polycarp wants to offer his set of $n$ binary words to play a game "words". In this game, players name words and each next word (starting from the second) must start with the last character of the previous word. The first word can be any. For example, these sequence of words can be named during the game: "0101", "1", "10", "00", "00001".
Word reversal is the operation of reversing the order of the characters. For example, the word "0111" after the reversal becomes "1110", the word "11010" after the reversal becomes "01011".
Probably, Polycarp has such a set of words that there is no way to put them in the order correspondent to the game rules. In this situation, he wants to reverse some words from his set so that: the final set of $n$ words still contains different words (i.e. all words are unique); there is a way to put all words of the final set of words in the order so that the final sequence of $n$ words is consistent with the game rules.
Polycarp wants to reverse minimal number of words. Please, help him.
-----Input-----
The first line of the input contains one integer $t$ ($1 \le t \le 10^4$) — the number of test cases in the input. Then $t$ test cases follow.
The first line of a test case contains one integer $n$ ($1 \le n \le 2\cdot10^5$) — the number of words in the Polycarp's set. Next $n$ lines contain these words. All of $n$ words aren't empty and contains only characters '0' and '1'. The sum of word lengths doesn't exceed $4\cdot10^6$. All words are different.
Guaranteed, that the sum of $n$ for all test cases in the input doesn't exceed $2\cdot10^5$. Also, guaranteed that the sum of word lengths for all test cases in the input doesn't exceed $4\cdot10^6$.
-----Output-----
Print answer for all of $t$ test cases in the order they appear.
If there is no answer for the test case, print -1. Otherwise, the first line of the output should contain $k$ ($0 \le k \le n$) — the minimal number of words in the set which should be reversed. The second line of the output should contain $k$ distinct integers — the indexes of the words in the set which should be reversed. Words are numerated from $1$ to $n$ in the order they appear. If $k=0$ you can skip this line (or you can print an empty line). If there are many answers you can print any of them.
-----Example-----
Input
4
4
0001
1000
0011
0111
3
010
101
0
2
00000
00001
4
01
001
0001
00001
Output
1
3
-1
0
2
1 2
|
t=int(input())
for _ in range(t):
n=int(input())
k={"01":0,"00":0,"11":0,"10":0}
ab=[]
ba=[]
a=[]
ra=set()
rb=set()
for i in range(n):
s=input()
ts=s[0]+s[-1]
k[ts]+=1
if ts=="01":
ab.append([str(i+1),s])
ra.add(s)
if ts=="10":
ba.append([str(i+1),s])
rb.add(s)
if k["01"]==0 and k["10"]==0 and k["00"]>0 and k["11"]>0:
ans=-1
else:
if k["01"]==k["10"] or k["01"]==k["10"]+1 or k["01"]==k["10"]-1:
ans=0
else:
m=(k["01"]+k["10"])//2 if (k["01"]+k["10"])%2==0 else (k["01"]+k["10"])//2+1
if k["01"]>m:
ans=k["01"]-m
for i in range(len(ab)):
psp=ab[i][1]
nn=list(psp)
nn.reverse()
psp="".join(nn)
c1=len(rb)
rb.add(psp)
c2=len(rb)
if c1!=c2:
a.append(ab[i][0])
if len(a)>=ans:
a=a[:ans]
else:
ans=-1
else:
ans=k["10"]-m
for i in range(len(ba)):
psp=ba[i][1]
nn=list(psp)
nn.reverse()
psp="".join(nn)
c1=len(ra)
ra.add(psp)
c2=len(ra)
if c1!=c2:
a.append(ba[i][0])
if len(a)>=ans:
a=a[:ans]
else:
ans=-1
print(ans)
if ans>0:
print(" ".join(a))
|
Polycarp has $n$ different binary words. A word called binary if it contains only characters '0' and '1'. For example, these words are binary: "0001", "11", "0" and "0011100".
Polycarp wants to offer his set of $n$ binary words to play a game "words". In this game, players name words and each next word (starting from the second) must start with the last character of the previous word. The first word can be any. For example, these sequence of words can be named during the game: "0101", "1", "10", "00", "00001".
Word reversal is the operation of reversing the order of the characters. For example, the word "0111" after the reversal becomes "1110", the word "11010" after the reversal becomes "01011".
Probably, Polycarp has such a set of words that there is no way to put them in the order correspondent to the game rules. In this situation, he wants to reverse some words from his set so that: the final set of $n$ words still contains different words (i.e. all words are unique); there is a way to put all words of the final set of words in the order so that the final sequence of $n$ words is consistent with the game rules.
Polycarp wants to reverse minimal number of words. Please, help him.
-----Input-----
The first line of the input contains one integer $t$ ($1 \le t \le 10^4$) — the number of test cases in the input. Then $t$ test cases follow.
The first line of a test case contains one integer $n$ ($1 \le n \le 2\cdot10^5$) — the number of words in the Polycarp's set. Next $n$ lines contain these words. All of $n$ words aren't empty and contains only characters '0' and '1'. The sum of word lengths doesn't exceed $4\cdot10^6$. All words are different.
Guaranteed, that the sum of $n$ for all test cases in the input doesn't exceed $2\cdot10^5$. Also, guaranteed that the sum of word lengths for all test cases in the input doesn't exceed $4\cdot10^6$.
-----Output-----
Print answer for all of $t$ test cases in the order they appear.
If there is no answer for the test case, print -1. Otherwise, the first line of the output should contain $k$ ($0 \le k \le n$) — the minimal number of words in the set which should be reversed. The second line of the output should contain $k$ distinct integers — the indexes of the words in the set which should be reversed. Words are numerated from $1$ to $n$ in the order they appear. If $k=0$ you can skip this line (or you can print an empty line). If there are many answers you can print any of them.
-----Example-----
Input
4
4
0001
1000
0011
0111
3
010
101
0
2
00000
00001
4
01
001
0001
00001
Output
1
3
-1
0
2
1 2
|
t=int(input())
for i in range(t):
n=int(input())
i0,i1=[],[]
l0,l1=[],[]
h0,h1=False,False
for i in range(n):
t=input()
if t[0]=='0' and t[-1]=='1':
i0.append(i)
l0.append(t)
elif t[0]=='1' and t[-1]=='0':
i1.append(i)
l1.append(t)
elif t[0]==t[-1]=='1':
h1=True
elif t[0]==t[-1]=='0':
h0=True
c0,c1=len(l0),len(l1)
req,sl=0,[]
s0=set(l0)
s1=set(l1)
if c0>0 or c1>0:
if c0-c1>1:
req=(c0-c1)//2
sel=0
sl=[]
for tt in range(len(l0)):
t=l0[tt]
if not t[::-1] in s1:
req-=1
sl.append(i0[tt]+1)
if req==0:
break
elif c1-c0>1:
req=(c1-c0)//2
sel=0
sl=[]
for tt in range(len(l1)):
t=l1[tt]
if not t[::-1] in s0:
req-=1
sl.append(i1[tt]+1)
if req==0:
break
if req>0:
print(-1)
else:
print(len(sl))
print(*sl)
else:
if h0 and h1:
print(-1)
else:
print(0)
print(*[])
|
Mikhail walks on a Cartesian plane. He starts at the point $(0, 0)$, and in one move he can go to any of eight adjacent points. For example, if Mikhail is currently at the point $(0, 0)$, he can go to any of the following points in one move: $(1, 0)$; $(1, 1)$; $(0, 1)$; $(-1, 1)$; $(-1, 0)$; $(-1, -1)$; $(0, -1)$; $(1, -1)$.
If Mikhail goes from the point $(x1, y1)$ to the point $(x2, y2)$ in one move, and $x1 \ne x2$ and $y1 \ne y2$, then such a move is called a diagonal move.
Mikhail has $q$ queries. For the $i$-th query Mikhail's target is to go to the point $(n_i, m_i)$ from the point $(0, 0)$ in exactly $k_i$ moves. Among all possible movements he want to choose one with the maximum number of diagonal moves. Your task is to find the maximum number of diagonal moves or find that it is impossible to go from the point $(0, 0)$ to the point $(n_i, m_i)$ in $k_i$ moves.
Note that Mikhail can visit any point any number of times (even the destination point!).
-----Input-----
The first line of the input contains one integer $q$ ($1 \le q \le 10^4$) — the number of queries.
Then $q$ lines follow. The $i$-th of these $q$ lines contains three integers $n_i$, $m_i$ and $k_i$ ($1 \le n_i, m_i, k_i \le 10^{18}$) — $x$-coordinate of the destination point of the query, $y$-coordinate of the destination point of the query and the number of moves in the query, correspondingly.
-----Output-----
Print $q$ integers. The $i$-th integer should be equal to -1 if Mikhail cannot go from the point $(0, 0)$ to the point $(n_i, m_i)$ in exactly $k_i$ moves described above. Otherwise the $i$-th integer should be equal to the the maximum number of diagonal moves among all possible movements.
-----Example-----
Input
3
2 2 3
4 3 7
10 1 9
Output
1
6
-1
-----Note-----
One of the possible answers to the first test case: $(0, 0) \to (1, 0) \to (1, 1) \to (2, 2)$.
One of the possible answers to the second test case: $(0, 0) \to (0, 1) \to (1, 2) \to (0, 3) \to (1, 4) \to (2, 3) \to (3, 2) \to (4, 3)$.
In the third test case Mikhail cannot reach the point $(10, 1)$ in 9 moves.
|
q=int(input())
for e in range(q):
x,y,k=list(map(int,input().split()))
x,y=abs(x),abs(y)
x,y=max(x,y),min(x,y)
if(x%2!=k%2):
k-=1
y-=1
if(x>k):
print(-1)
continue
if((x-y)%2):
k-=1
x-=1
print(k)
|
Mikhail walks on a Cartesian plane. He starts at the point $(0, 0)$, and in one move he can go to any of eight adjacent points. For example, if Mikhail is currently at the point $(0, 0)$, he can go to any of the following points in one move: $(1, 0)$; $(1, 1)$; $(0, 1)$; $(-1, 1)$; $(-1, 0)$; $(-1, -1)$; $(0, -1)$; $(1, -1)$.
If Mikhail goes from the point $(x1, y1)$ to the point $(x2, y2)$ in one move, and $x1 \ne x2$ and $y1 \ne y2$, then such a move is called a diagonal move.
Mikhail has $q$ queries. For the $i$-th query Mikhail's target is to go to the point $(n_i, m_i)$ from the point $(0, 0)$ in exactly $k_i$ moves. Among all possible movements he want to choose one with the maximum number of diagonal moves. Your task is to find the maximum number of diagonal moves or find that it is impossible to go from the point $(0, 0)$ to the point $(n_i, m_i)$ in $k_i$ moves.
Note that Mikhail can visit any point any number of times (even the destination point!).
-----Input-----
The first line of the input contains one integer $q$ ($1 \le q \le 10^4$) — the number of queries.
Then $q$ lines follow. The $i$-th of these $q$ lines contains three integers $n_i$, $m_i$ and $k_i$ ($1 \le n_i, m_i, k_i \le 10^{18}$) — $x$-coordinate of the destination point of the query, $y$-coordinate of the destination point of the query and the number of moves in the query, correspondingly.
-----Output-----
Print $q$ integers. The $i$-th integer should be equal to -1 if Mikhail cannot go from the point $(0, 0)$ to the point $(n_i, m_i)$ in exactly $k_i$ moves described above. Otherwise the $i$-th integer should be equal to the the maximum number of diagonal moves among all possible movements.
-----Example-----
Input
3
2 2 3
4 3 7
10 1 9
Output
1
6
-1
-----Note-----
One of the possible answers to the first test case: $(0, 0) \to (1, 0) \to (1, 1) \to (2, 2)$.
One of the possible answers to the second test case: $(0, 0) \to (0, 1) \to (1, 2) \to (0, 3) \to (1, 4) \to (2, 3) \to (3, 2) \to (4, 3)$.
In the third test case Mikhail cannot reach the point $(10, 1)$ in 9 moves.
|
#
import collections, atexit, math, sys, bisect
sys.setrecursionlimit(1000000)
def getIntList():
return list(map(int, input().split()))
try :
#raise ModuleNotFoundError
import numpy
def dprint(*args, **kwargs):
print(*args, **kwargs, file=sys.stderr)
dprint('debug mode')
except ModuleNotFoundError:
def dprint(*args, **kwargs):
pass
inId = 0
outId = 0
if inId>0:
dprint('use input', inId)
sys.stdin = open('input'+ str(inId) + '.txt', 'r') #标准输出重定向至文件
if outId>0:
dprint('use output', outId)
sys.stdout = open('stdout'+ str(outId) + '.txt', 'w') #标准输出重定向至文件
atexit.register(lambda :sys.stdout.close()) #idle 中不会执行 atexit
Q, = getIntList()
for _ in range(Q):
N, M, K = getIntList()
if max(N,M) >K:
print(-1)
continue
r = K
if N%2!= K%2:
r-=1
if M%2!= K%2:
r-=1
print(r)
|
Mikhail walks on a Cartesian plane. He starts at the point $(0, 0)$, and in one move he can go to any of eight adjacent points. For example, if Mikhail is currently at the point $(0, 0)$, he can go to any of the following points in one move: $(1, 0)$; $(1, 1)$; $(0, 1)$; $(-1, 1)$; $(-1, 0)$; $(-1, -1)$; $(0, -1)$; $(1, -1)$.
If Mikhail goes from the point $(x1, y1)$ to the point $(x2, y2)$ in one move, and $x1 \ne x2$ and $y1 \ne y2$, then such a move is called a diagonal move.
Mikhail has $q$ queries. For the $i$-th query Mikhail's target is to go to the point $(n_i, m_i)$ from the point $(0, 0)$ in exactly $k_i$ moves. Among all possible movements he want to choose one with the maximum number of diagonal moves. Your task is to find the maximum number of diagonal moves or find that it is impossible to go from the point $(0, 0)$ to the point $(n_i, m_i)$ in $k_i$ moves.
Note that Mikhail can visit any point any number of times (even the destination point!).
-----Input-----
The first line of the input contains one integer $q$ ($1 \le q \le 10^4$) — the number of queries.
Then $q$ lines follow. The $i$-th of these $q$ lines contains three integers $n_i$, $m_i$ and $k_i$ ($1 \le n_i, m_i, k_i \le 10^{18}$) — $x$-coordinate of the destination point of the query, $y$-coordinate of the destination point of the query and the number of moves in the query, correspondingly.
-----Output-----
Print $q$ integers. The $i$-th integer should be equal to -1 if Mikhail cannot go from the point $(0, 0)$ to the point $(n_i, m_i)$ in exactly $k_i$ moves described above. Otherwise the $i$-th integer should be equal to the the maximum number of diagonal moves among all possible movements.
-----Example-----
Input
3
2 2 3
4 3 7
10 1 9
Output
1
6
-1
-----Note-----
One of the possible answers to the first test case: $(0, 0) \to (1, 0) \to (1, 1) \to (2, 2)$.
One of the possible answers to the second test case: $(0, 0) \to (0, 1) \to (1, 2) \to (0, 3) \to (1, 4) \to (2, 3) \to (3, 2) \to (4, 3)$.
In the third test case Mikhail cannot reach the point $(10, 1)$ in 9 moves.
|
q = int(input())
for i in range(q):
x, y, k = list(map(int, input().split()))
if x > y: x, y = y, x
m = y
d = y
if (y - x) % 2 == 1:
d -= 1
if k < m:
print(-1)
continue
r = k - m
if r % 2 != 0:
r -= 1
if d != m:
d += 1
else:
d -= 1
d += r
print(d)
|
Mikhail walks on a Cartesian plane. He starts at the point $(0, 0)$, and in one move he can go to any of eight adjacent points. For example, if Mikhail is currently at the point $(0, 0)$, he can go to any of the following points in one move: $(1, 0)$; $(1, 1)$; $(0, 1)$; $(-1, 1)$; $(-1, 0)$; $(-1, -1)$; $(0, -1)$; $(1, -1)$.
If Mikhail goes from the point $(x1, y1)$ to the point $(x2, y2)$ in one move, and $x1 \ne x2$ and $y1 \ne y2$, then such a move is called a diagonal move.
Mikhail has $q$ queries. For the $i$-th query Mikhail's target is to go to the point $(n_i, m_i)$ from the point $(0, 0)$ in exactly $k_i$ moves. Among all possible movements he want to choose one with the maximum number of diagonal moves. Your task is to find the maximum number of diagonal moves or find that it is impossible to go from the point $(0, 0)$ to the point $(n_i, m_i)$ in $k_i$ moves.
Note that Mikhail can visit any point any number of times (even the destination point!).
-----Input-----
The first line of the input contains one integer $q$ ($1 \le q \le 10^4$) — the number of queries.
Then $q$ lines follow. The $i$-th of these $q$ lines contains three integers $n_i$, $m_i$ and $k_i$ ($1 \le n_i, m_i, k_i \le 10^{18}$) — $x$-coordinate of the destination point of the query, $y$-coordinate of the destination point of the query and the number of moves in the query, correspondingly.
-----Output-----
Print $q$ integers. The $i$-th integer should be equal to -1 if Mikhail cannot go from the point $(0, 0)$ to the point $(n_i, m_i)$ in exactly $k_i$ moves described above. Otherwise the $i$-th integer should be equal to the the maximum number of diagonal moves among all possible movements.
-----Example-----
Input
3
2 2 3
4 3 7
10 1 9
Output
1
6
-1
-----Note-----
One of the possible answers to the first test case: $(0, 0) \to (1, 0) \to (1, 1) \to (2, 2)$.
One of the possible answers to the second test case: $(0, 0) \to (0, 1) \to (1, 2) \to (0, 3) \to (1, 4) \to (2, 3) \to (3, 2) \to (4, 3)$.
In the third test case Mikhail cannot reach the point $(10, 1)$ in 9 moves.
|
q = int(input())
otvet = []
for i in range(q):
g = input().split()
n = int(g[0])
m = int(g[1])
k = int(g[2])
if n < 0:
n = -n
if m < 0:
m = -m
if m > k or n > k:
otvet.append(-1)
elif m % 2 == k % 2 and n % 2 == k % 2:
otvet.append(k)
elif m % 2 == k % 2 or n % 2 == k % 2:
otvet.append(k - 1)
else:
otvet.append(k - 2)
for i in otvet:
print(i)
|
Mikhail walks on a Cartesian plane. He starts at the point $(0, 0)$, and in one move he can go to any of eight adjacent points. For example, if Mikhail is currently at the point $(0, 0)$, he can go to any of the following points in one move: $(1, 0)$; $(1, 1)$; $(0, 1)$; $(-1, 1)$; $(-1, 0)$; $(-1, -1)$; $(0, -1)$; $(1, -1)$.
If Mikhail goes from the point $(x1, y1)$ to the point $(x2, y2)$ in one move, and $x1 \ne x2$ and $y1 \ne y2$, then such a move is called a diagonal move.
Mikhail has $q$ queries. For the $i$-th query Mikhail's target is to go to the point $(n_i, m_i)$ from the point $(0, 0)$ in exactly $k_i$ moves. Among all possible movements he want to choose one with the maximum number of diagonal moves. Your task is to find the maximum number of diagonal moves or find that it is impossible to go from the point $(0, 0)$ to the point $(n_i, m_i)$ in $k_i$ moves.
Note that Mikhail can visit any point any number of times (even the destination point!).
-----Input-----
The first line of the input contains one integer $q$ ($1 \le q \le 10^4$) — the number of queries.
Then $q$ lines follow. The $i$-th of these $q$ lines contains three integers $n_i$, $m_i$ and $k_i$ ($1 \le n_i, m_i, k_i \le 10^{18}$) — $x$-coordinate of the destination point of the query, $y$-coordinate of the destination point of the query and the number of moves in the query, correspondingly.
-----Output-----
Print $q$ integers. The $i$-th integer should be equal to -1 if Mikhail cannot go from the point $(0, 0)$ to the point $(n_i, m_i)$ in exactly $k_i$ moves described above. Otherwise the $i$-th integer should be equal to the the maximum number of diagonal moves among all possible movements.
-----Example-----
Input
3
2 2 3
4 3 7
10 1 9
Output
1
6
-1
-----Note-----
One of the possible answers to the first test case: $(0, 0) \to (1, 0) \to (1, 1) \to (2, 2)$.
One of the possible answers to the second test case: $(0, 0) \to (0, 1) \to (1, 2) \to (0, 3) \to (1, 4) \to (2, 3) \to (3, 2) \to (4, 3)$.
In the third test case Mikhail cannot reach the point $(10, 1)$ in 9 moves.
|
q = int(input())
for i in range(q):
a, b, k = list(map(int, input().split()))
if a < b:
a, b, = b, a
if a > k:
print(-1)
elif a % 2 == b % 2 != k % 2:
print(k - 2)
elif (a + b) % 2 != 0:
print(k - 1)
else:
print(k)
|
Mikhail walks on a Cartesian plane. He starts at the point $(0, 0)$, and in one move he can go to any of eight adjacent points. For example, if Mikhail is currently at the point $(0, 0)$, he can go to any of the following points in one move: $(1, 0)$; $(1, 1)$; $(0, 1)$; $(-1, 1)$; $(-1, 0)$; $(-1, -1)$; $(0, -1)$; $(1, -1)$.
If Mikhail goes from the point $(x1, y1)$ to the point $(x2, y2)$ in one move, and $x1 \ne x2$ and $y1 \ne y2$, then such a move is called a diagonal move.
Mikhail has $q$ queries. For the $i$-th query Mikhail's target is to go to the point $(n_i, m_i)$ from the point $(0, 0)$ in exactly $k_i$ moves. Among all possible movements he want to choose one with the maximum number of diagonal moves. Your task is to find the maximum number of diagonal moves or find that it is impossible to go from the point $(0, 0)$ to the point $(n_i, m_i)$ in $k_i$ moves.
Note that Mikhail can visit any point any number of times (even the destination point!).
-----Input-----
The first line of the input contains one integer $q$ ($1 \le q \le 10^4$) — the number of queries.
Then $q$ lines follow. The $i$-th of these $q$ lines contains three integers $n_i$, $m_i$ and $k_i$ ($1 \le n_i, m_i, k_i \le 10^{18}$) — $x$-coordinate of the destination point of the query, $y$-coordinate of the destination point of the query and the number of moves in the query, correspondingly.
-----Output-----
Print $q$ integers. The $i$-th integer should be equal to -1 if Mikhail cannot go from the point $(0, 0)$ to the point $(n_i, m_i)$ in exactly $k_i$ moves described above. Otherwise the $i$-th integer should be equal to the the maximum number of diagonal moves among all possible movements.
-----Example-----
Input
3
2 2 3
4 3 7
10 1 9
Output
1
6
-1
-----Note-----
One of the possible answers to the first test case: $(0, 0) \to (1, 0) \to (1, 1) \to (2, 2)$.
One of the possible answers to the second test case: $(0, 0) \to (0, 1) \to (1, 2) \to (0, 3) \to (1, 4) \to (2, 3) \to (3, 2) \to (4, 3)$.
In the third test case Mikhail cannot reach the point $(10, 1)$ in 9 moves.
|
q = int(input())
for i in range(q):
n, m, k = list(map(int, input().split()))
m, n = abs(m), abs(n)
mx = max(m, n)
remaining = k - mx
if remaining < 0:
print(-1)
elif m == n == 0:
if k == 1:
print(-1)
elif k % 2:
print(k - 1)
else:
print(k)
elif abs(m - n) % 2 == 0:
if remaining % 2 == 0:
print(k)
else:
print(k - 2)
else:
if not remaining:
print(k - 1)
elif remaining % 2 == 0:
print(k - 1)
else:
print(k - 1)
|
Mikhail walks on a Cartesian plane. He starts at the point $(0, 0)$, and in one move he can go to any of eight adjacent points. For example, if Mikhail is currently at the point $(0, 0)$, he can go to any of the following points in one move: $(1, 0)$; $(1, 1)$; $(0, 1)$; $(-1, 1)$; $(-1, 0)$; $(-1, -1)$; $(0, -1)$; $(1, -1)$.
If Mikhail goes from the point $(x1, y1)$ to the point $(x2, y2)$ in one move, and $x1 \ne x2$ and $y1 \ne y2$, then such a move is called a diagonal move.
Mikhail has $q$ queries. For the $i$-th query Mikhail's target is to go to the point $(n_i, m_i)$ from the point $(0, 0)$ in exactly $k_i$ moves. Among all possible movements he want to choose one with the maximum number of diagonal moves. Your task is to find the maximum number of diagonal moves or find that it is impossible to go from the point $(0, 0)$ to the point $(n_i, m_i)$ in $k_i$ moves.
Note that Mikhail can visit any point any number of times (even the destination point!).
-----Input-----
The first line of the input contains one integer $q$ ($1 \le q \le 10^4$) — the number of queries.
Then $q$ lines follow. The $i$-th of these $q$ lines contains three integers $n_i$, $m_i$ and $k_i$ ($1 \le n_i, m_i, k_i \le 10^{18}$) — $x$-coordinate of the destination point of the query, $y$-coordinate of the destination point of the query and the number of moves in the query, correspondingly.
-----Output-----
Print $q$ integers. The $i$-th integer should be equal to -1 if Mikhail cannot go from the point $(0, 0)$ to the point $(n_i, m_i)$ in exactly $k_i$ moves described above. Otherwise the $i$-th integer should be equal to the the maximum number of diagonal moves among all possible movements.
-----Example-----
Input
3
2 2 3
4 3 7
10 1 9
Output
1
6
-1
-----Note-----
One of the possible answers to the first test case: $(0, 0) \to (1, 0) \to (1, 1) \to (2, 2)$.
One of the possible answers to the second test case: $(0, 0) \to (0, 1) \to (1, 2) \to (0, 3) \to (1, 4) \to (2, 3) \to (3, 2) \to (4, 3)$.
In the third test case Mikhail cannot reach the point $(10, 1)$ in 9 moves.
|
from collections import deque
from sys import stdin
lines = deque(line.strip() for line in stdin.readlines())
def nextline():
return lines.popleft()
def types(cast, sep=None):
return tuple(cast(x) for x in strs(sep=sep))
def ints(sep=None):
return types(int, sep=sep)
def strs(sep=None):
return tuple(nextline()) if sep == '' else tuple(nextline().split(sep=sep))
def main():
# lines will now contain all of the input's lines in a list
T = int(nextline())
for testCase in range(1, T + 1):
n, m, k = ints()
min_k = max(n, m)
if min_k > k:
print(-1)
continue
if (n - m) % 2 == 0:
if k % 2 == n % 2:
print(k)
continue
print(k - 2)
continue
print(k - 1)
def __starting_point():
main()
__starting_point()
|
Mikhail walks on a Cartesian plane. He starts at the point $(0, 0)$, and in one move he can go to any of eight adjacent points. For example, if Mikhail is currently at the point $(0, 0)$, he can go to any of the following points in one move: $(1, 0)$; $(1, 1)$; $(0, 1)$; $(-1, 1)$; $(-1, 0)$; $(-1, -1)$; $(0, -1)$; $(1, -1)$.
If Mikhail goes from the point $(x1, y1)$ to the point $(x2, y2)$ in one move, and $x1 \ne x2$ and $y1 \ne y2$, then such a move is called a diagonal move.
Mikhail has $q$ queries. For the $i$-th query Mikhail's target is to go to the point $(n_i, m_i)$ from the point $(0, 0)$ in exactly $k_i$ moves. Among all possible movements he want to choose one with the maximum number of diagonal moves. Your task is to find the maximum number of diagonal moves or find that it is impossible to go from the point $(0, 0)$ to the point $(n_i, m_i)$ in $k_i$ moves.
Note that Mikhail can visit any point any number of times (even the destination point!).
-----Input-----
The first line of the input contains one integer $q$ ($1 \le q \le 10^4$) — the number of queries.
Then $q$ lines follow. The $i$-th of these $q$ lines contains three integers $n_i$, $m_i$ and $k_i$ ($1 \le n_i, m_i, k_i \le 10^{18}$) — $x$-coordinate of the destination point of the query, $y$-coordinate of the destination point of the query and the number of moves in the query, correspondingly.
-----Output-----
Print $q$ integers. The $i$-th integer should be equal to -1 if Mikhail cannot go from the point $(0, 0)$ to the point $(n_i, m_i)$ in exactly $k_i$ moves described above. Otherwise the $i$-th integer should be equal to the the maximum number of diagonal moves among all possible movements.
-----Example-----
Input
3
2 2 3
4 3 7
10 1 9
Output
1
6
-1
-----Note-----
One of the possible answers to the first test case: $(0, 0) \to (1, 0) \to (1, 1) \to (2, 2)$.
One of the possible answers to the second test case: $(0, 0) \to (0, 1) \to (1, 2) \to (0, 3) \to (1, 4) \to (2, 3) \to (3, 2) \to (4, 3)$.
In the third test case Mikhail cannot reach the point $(10, 1)$ in 9 moves.
|
q = int(input())
for _ in range(q):
n, m, k = list(map(int, input().split()))
if max([n, m]) > k:
print(-1)
else:
if (n + m) % 2 == 0:
if max([n, m]) % 2 != k % 2:
print(k - 2)
else:
print(k)
else:
print((k - 1));
|
Mikhail walks on a Cartesian plane. He starts at the point $(0, 0)$, and in one move he can go to any of eight adjacent points. For example, if Mikhail is currently at the point $(0, 0)$, he can go to any of the following points in one move: $(1, 0)$; $(1, 1)$; $(0, 1)$; $(-1, 1)$; $(-1, 0)$; $(-1, -1)$; $(0, -1)$; $(1, -1)$.
If Mikhail goes from the point $(x1, y1)$ to the point $(x2, y2)$ in one move, and $x1 \ne x2$ and $y1 \ne y2$, then such a move is called a diagonal move.
Mikhail has $q$ queries. For the $i$-th query Mikhail's target is to go to the point $(n_i, m_i)$ from the point $(0, 0)$ in exactly $k_i$ moves. Among all possible movements he want to choose one with the maximum number of diagonal moves. Your task is to find the maximum number of diagonal moves or find that it is impossible to go from the point $(0, 0)$ to the point $(n_i, m_i)$ in $k_i$ moves.
Note that Mikhail can visit any point any number of times (even the destination point!).
-----Input-----
The first line of the input contains one integer $q$ ($1 \le q \le 10^4$) — the number of queries.
Then $q$ lines follow. The $i$-th of these $q$ lines contains three integers $n_i$, $m_i$ and $k_i$ ($1 \le n_i, m_i, k_i \le 10^{18}$) — $x$-coordinate of the destination point of the query, $y$-coordinate of the destination point of the query and the number of moves in the query, correspondingly.
-----Output-----
Print $q$ integers. The $i$-th integer should be equal to -1 if Mikhail cannot go from the point $(0, 0)$ to the point $(n_i, m_i)$ in exactly $k_i$ moves described above. Otherwise the $i$-th integer should be equal to the the maximum number of diagonal moves among all possible movements.
-----Example-----
Input
3
2 2 3
4 3 7
10 1 9
Output
1
6
-1
-----Note-----
One of the possible answers to the first test case: $(0, 0) \to (1, 0) \to (1, 1) \to (2, 2)$.
One of the possible answers to the second test case: $(0, 0) \to (0, 1) \to (1, 2) \to (0, 3) \to (1, 4) \to (2, 3) \to (3, 2) \to (4, 3)$.
In the third test case Mikhail cannot reach the point $(10, 1)$ in 9 moves.
|
import math
q = int(input())
for i in range(q):
x, y, k = map(int, input().split())
if x > k or y > k:
print(-1)
else:
if (x+y)%2 == 0:
if (k-max(x,y)) % 2 == 0:
print(k)
else:
print(k - 2)
else:
if (k-max(x,y)) % 2 == 0:
print(k-1)
else:
print(k-1)
|
Mikhail walks on a Cartesian plane. He starts at the point $(0, 0)$, and in one move he can go to any of eight adjacent points. For example, if Mikhail is currently at the point $(0, 0)$, he can go to any of the following points in one move: $(1, 0)$; $(1, 1)$; $(0, 1)$; $(-1, 1)$; $(-1, 0)$; $(-1, -1)$; $(0, -1)$; $(1, -1)$.
If Mikhail goes from the point $(x1, y1)$ to the point $(x2, y2)$ in one move, and $x1 \ne x2$ and $y1 \ne y2$, then such a move is called a diagonal move.
Mikhail has $q$ queries. For the $i$-th query Mikhail's target is to go to the point $(n_i, m_i)$ from the point $(0, 0)$ in exactly $k_i$ moves. Among all possible movements he want to choose one with the maximum number of diagonal moves. Your task is to find the maximum number of diagonal moves or find that it is impossible to go from the point $(0, 0)$ to the point $(n_i, m_i)$ in $k_i$ moves.
Note that Mikhail can visit any point any number of times (even the destination point!).
-----Input-----
The first line of the input contains one integer $q$ ($1 \le q \le 10^4$) — the number of queries.
Then $q$ lines follow. The $i$-th of these $q$ lines contains three integers $n_i$, $m_i$ and $k_i$ ($1 \le n_i, m_i, k_i \le 10^{18}$) — $x$-coordinate of the destination point of the query, $y$-coordinate of the destination point of the query and the number of moves in the query, correspondingly.
-----Output-----
Print $q$ integers. The $i$-th integer should be equal to -1 if Mikhail cannot go from the point $(0, 0)$ to the point $(n_i, m_i)$ in exactly $k_i$ moves described above. Otherwise the $i$-th integer should be equal to the the maximum number of diagonal moves among all possible movements.
-----Example-----
Input
3
2 2 3
4 3 7
10 1 9
Output
1
6
-1
-----Note-----
One of the possible answers to the first test case: $(0, 0) \to (1, 0) \to (1, 1) \to (2, 2)$.
One of the possible answers to the second test case: $(0, 0) \to (0, 1) \to (1, 2) \to (0, 3) \to (1, 4) \to (2, 3) \to (3, 2) \to (4, 3)$.
In the third test case Mikhail cannot reach the point $(10, 1)$ in 9 moves.
|
q = int(input())
for _ in range(q):
n, m, k = list(map(int, input().split()))
if k == 0:
if n == 0 and m == 0:
print(0)
else:
print(-1)
elif k == 1:
if max(abs(n), abs(m)) != 1:
print(-1)
elif abs(n) == abs(m) == 1:
print(1)
else:
print(0)
else:
if max(abs(n), abs(m)) > k:
print(-1)
elif abs(n) == abs(m):
if (k - abs(n)) % 2 == 0:
print(k)
else:
print(k - 2)
elif (max(abs(n), abs(m)) - min(abs(n), abs(m))) % 2 == 0:
if (k - max(abs(n), abs(m))) % 2 == 0:
print(k)
else:
print(k - 2)
else:
print(k - 1)
|
Mikhail walks on a Cartesian plane. He starts at the point $(0, 0)$, and in one move he can go to any of eight adjacent points. For example, if Mikhail is currently at the point $(0, 0)$, he can go to any of the following points in one move: $(1, 0)$; $(1, 1)$; $(0, 1)$; $(-1, 1)$; $(-1, 0)$; $(-1, -1)$; $(0, -1)$; $(1, -1)$.
If Mikhail goes from the point $(x1, y1)$ to the point $(x2, y2)$ in one move, and $x1 \ne x2$ and $y1 \ne y2$, then such a move is called a diagonal move.
Mikhail has $q$ queries. For the $i$-th query Mikhail's target is to go to the point $(n_i, m_i)$ from the point $(0, 0)$ in exactly $k_i$ moves. Among all possible movements he want to choose one with the maximum number of diagonal moves. Your task is to find the maximum number of diagonal moves or find that it is impossible to go from the point $(0, 0)$ to the point $(n_i, m_i)$ in $k_i$ moves.
Note that Mikhail can visit any point any number of times (even the destination point!).
-----Input-----
The first line of the input contains one integer $q$ ($1 \le q \le 10^4$) — the number of queries.
Then $q$ lines follow. The $i$-th of these $q$ lines contains three integers $n_i$, $m_i$ and $k_i$ ($1 \le n_i, m_i, k_i \le 10^{18}$) — $x$-coordinate of the destination point of the query, $y$-coordinate of the destination point of the query and the number of moves in the query, correspondingly.
-----Output-----
Print $q$ integers. The $i$-th integer should be equal to -1 if Mikhail cannot go from the point $(0, 0)$ to the point $(n_i, m_i)$ in exactly $k_i$ moves described above. Otherwise the $i$-th integer should be equal to the the maximum number of diagonal moves among all possible movements.
-----Example-----
Input
3
2 2 3
4 3 7
10 1 9
Output
1
6
-1
-----Note-----
One of the possible answers to the first test case: $(0, 0) \to (1, 0) \to (1, 1) \to (2, 2)$.
One of the possible answers to the second test case: $(0, 0) \to (0, 1) \to (1, 2) \to (0, 3) \to (1, 4) \to (2, 3) \to (3, 2) \to (4, 3)$.
In the third test case Mikhail cannot reach the point $(10, 1)$ in 9 moves.
|
import sys
#sys.stdin=open("data.txt")
input=sys.stdin.readline
for _ in range(int(input())):
n,m,k=list(map(int,input().split()))
n=abs(n)
m=abs(m)
if max(n,m)>k:
print("-1")
else:
# you can't 0 0 1 me :D
bad1=((n+k)%2==1)
bad2=((m+k)%2==1)
print(k-bad1-bad2)
|
Mikhail walks on a Cartesian plane. He starts at the point $(0, 0)$, and in one move he can go to any of eight adjacent points. For example, if Mikhail is currently at the point $(0, 0)$, he can go to any of the following points in one move: $(1, 0)$; $(1, 1)$; $(0, 1)$; $(-1, 1)$; $(-1, 0)$; $(-1, -1)$; $(0, -1)$; $(1, -1)$.
If Mikhail goes from the point $(x1, y1)$ to the point $(x2, y2)$ in one move, and $x1 \ne x2$ and $y1 \ne y2$, then such a move is called a diagonal move.
Mikhail has $q$ queries. For the $i$-th query Mikhail's target is to go to the point $(n_i, m_i)$ from the point $(0, 0)$ in exactly $k_i$ moves. Among all possible movements he want to choose one with the maximum number of diagonal moves. Your task is to find the maximum number of diagonal moves or find that it is impossible to go from the point $(0, 0)$ to the point $(n_i, m_i)$ in $k_i$ moves.
Note that Mikhail can visit any point any number of times (even the destination point!).
-----Input-----
The first line of the input contains one integer $q$ ($1 \le q \le 10^4$) — the number of queries.
Then $q$ lines follow. The $i$-th of these $q$ lines contains three integers $n_i$, $m_i$ and $k_i$ ($1 \le n_i, m_i, k_i \le 10^{18}$) — $x$-coordinate of the destination point of the query, $y$-coordinate of the destination point of the query and the number of moves in the query, correspondingly.
-----Output-----
Print $q$ integers. The $i$-th integer should be equal to -1 if Mikhail cannot go from the point $(0, 0)$ to the point $(n_i, m_i)$ in exactly $k_i$ moves described above. Otherwise the $i$-th integer should be equal to the the maximum number of diagonal moves among all possible movements.
-----Example-----
Input
3
2 2 3
4 3 7
10 1 9
Output
1
6
-1
-----Note-----
One of the possible answers to the first test case: $(0, 0) \to (1, 0) \to (1, 1) \to (2, 2)$.
One of the possible answers to the second test case: $(0, 0) \to (0, 1) \to (1, 2) \to (0, 3) \to (1, 4) \to (2, 3) \to (3, 2) \to (4, 3)$.
In the third test case Mikhail cannot reach the point $(10, 1)$ in 9 moves.
|
USE_STDIO = False
if not USE_STDIO:
try: import mypc
except: pass
def main():
q, = list(map(int, input().split(' ')))
for _ in range(q):
n, m, k = list(map(int, input().split(' ')))
if n > k or m > k:
print(-1)
elif (n - m) % 2:
print(k - 1)
elif (n - k) % 2:
print(k - 2)
else:
print(k)
def __starting_point():
main()
__starting_point()
|
Mikhail walks on a Cartesian plane. He starts at the point $(0, 0)$, and in one move he can go to any of eight adjacent points. For example, if Mikhail is currently at the point $(0, 0)$, he can go to any of the following points in one move: $(1, 0)$; $(1, 1)$; $(0, 1)$; $(-1, 1)$; $(-1, 0)$; $(-1, -1)$; $(0, -1)$; $(1, -1)$.
If Mikhail goes from the point $(x1, y1)$ to the point $(x2, y2)$ in one move, and $x1 \ne x2$ and $y1 \ne y2$, then such a move is called a diagonal move.
Mikhail has $q$ queries. For the $i$-th query Mikhail's target is to go to the point $(n_i, m_i)$ from the point $(0, 0)$ in exactly $k_i$ moves. Among all possible movements he want to choose one with the maximum number of diagonal moves. Your task is to find the maximum number of diagonal moves or find that it is impossible to go from the point $(0, 0)$ to the point $(n_i, m_i)$ in $k_i$ moves.
Note that Mikhail can visit any point any number of times (even the destination point!).
-----Input-----
The first line of the input contains one integer $q$ ($1 \le q \le 10^4$) — the number of queries.
Then $q$ lines follow. The $i$-th of these $q$ lines contains three integers $n_i$, $m_i$ and $k_i$ ($1 \le n_i, m_i, k_i \le 10^{18}$) — $x$-coordinate of the destination point of the query, $y$-coordinate of the destination point of the query and the number of moves in the query, correspondingly.
-----Output-----
Print $q$ integers. The $i$-th integer should be equal to -1 if Mikhail cannot go from the point $(0, 0)$ to the point $(n_i, m_i)$ in exactly $k_i$ moves described above. Otherwise the $i$-th integer should be equal to the the maximum number of diagonal moves among all possible movements.
-----Example-----
Input
3
2 2 3
4 3 7
10 1 9
Output
1
6
-1
-----Note-----
One of the possible answers to the first test case: $(0, 0) \to (1, 0) \to (1, 1) \to (2, 2)$.
One of the possible answers to the second test case: $(0, 0) \to (0, 1) \to (1, 2) \to (0, 3) \to (1, 4) \to (2, 3) \to (3, 2) \to (4, 3)$.
In the third test case Mikhail cannot reach the point $(10, 1)$ in 9 moves.
|
q=int(input())
Q=[list(map(int,input().split())) for i in range(q)]
for n,m,k in Q:
if n>k or m>k:
print(-1)
continue
x=max(n,m)-min(n,m)
y=k-max(n,m)
if x%2==0 and y%2==0:
print(k)
elif x%2==0 and y%2==1:
print(k-2)
elif x%2==1 and y%2==0:
print(k-1)
elif x%2==1 and y%2==1:
print(k-1)
|
Mikhail walks on a Cartesian plane. He starts at the point $(0, 0)$, and in one move he can go to any of eight adjacent points. For example, if Mikhail is currently at the point $(0, 0)$, he can go to any of the following points in one move: $(1, 0)$; $(1, 1)$; $(0, 1)$; $(-1, 1)$; $(-1, 0)$; $(-1, -1)$; $(0, -1)$; $(1, -1)$.
If Mikhail goes from the point $(x1, y1)$ to the point $(x2, y2)$ in one move, and $x1 \ne x2$ and $y1 \ne y2$, then such a move is called a diagonal move.
Mikhail has $q$ queries. For the $i$-th query Mikhail's target is to go to the point $(n_i, m_i)$ from the point $(0, 0)$ in exactly $k_i$ moves. Among all possible movements he want to choose one with the maximum number of diagonal moves. Your task is to find the maximum number of diagonal moves or find that it is impossible to go from the point $(0, 0)$ to the point $(n_i, m_i)$ in $k_i$ moves.
Note that Mikhail can visit any point any number of times (even the destination point!).
-----Input-----
The first line of the input contains one integer $q$ ($1 \le q \le 10^4$) — the number of queries.
Then $q$ lines follow. The $i$-th of these $q$ lines contains three integers $n_i$, $m_i$ and $k_i$ ($1 \le n_i, m_i, k_i \le 10^{18}$) — $x$-coordinate of the destination point of the query, $y$-coordinate of the destination point of the query and the number of moves in the query, correspondingly.
-----Output-----
Print $q$ integers. The $i$-th integer should be equal to -1 if Mikhail cannot go from the point $(0, 0)$ to the point $(n_i, m_i)$ in exactly $k_i$ moves described above. Otherwise the $i$-th integer should be equal to the the maximum number of diagonal moves among all possible movements.
-----Example-----
Input
3
2 2 3
4 3 7
10 1 9
Output
1
6
-1
-----Note-----
One of the possible answers to the first test case: $(0, 0) \to (1, 0) \to (1, 1) \to (2, 2)$.
One of the possible answers to the second test case: $(0, 0) \to (0, 1) \to (1, 2) \to (0, 3) \to (1, 4) \to (2, 3) \to (3, 2) \to (4, 3)$.
In the third test case Mikhail cannot reach the point $(10, 1)$ in 9 moves.
|
n = int(input())
for i in range(n):
a, b, c = [int(el) for el in input().split()]
if ( a > c or b > c):
print(-1)
else:
if (a% 2 + b % 2 == 1):
print(c - 1)
elif (a%2 == b%2 == c%2):
print(c)
else:
print(c - 2)
|
Mikhail walks on a Cartesian plane. He starts at the point $(0, 0)$, and in one move he can go to any of eight adjacent points. For example, if Mikhail is currently at the point $(0, 0)$, he can go to any of the following points in one move: $(1, 0)$; $(1, 1)$; $(0, 1)$; $(-1, 1)$; $(-1, 0)$; $(-1, -1)$; $(0, -1)$; $(1, -1)$.
If Mikhail goes from the point $(x1, y1)$ to the point $(x2, y2)$ in one move, and $x1 \ne x2$ and $y1 \ne y2$, then such a move is called a diagonal move.
Mikhail has $q$ queries. For the $i$-th query Mikhail's target is to go to the point $(n_i, m_i)$ from the point $(0, 0)$ in exactly $k_i$ moves. Among all possible movements he want to choose one with the maximum number of diagonal moves. Your task is to find the maximum number of diagonal moves or find that it is impossible to go from the point $(0, 0)$ to the point $(n_i, m_i)$ in $k_i$ moves.
Note that Mikhail can visit any point any number of times (even the destination point!).
-----Input-----
The first line of the input contains one integer $q$ ($1 \le q \le 10^4$) — the number of queries.
Then $q$ lines follow. The $i$-th of these $q$ lines contains three integers $n_i$, $m_i$ and $k_i$ ($1 \le n_i, m_i, k_i \le 10^{18}$) — $x$-coordinate of the destination point of the query, $y$-coordinate of the destination point of the query and the number of moves in the query, correspondingly.
-----Output-----
Print $q$ integers. The $i$-th integer should be equal to -1 if Mikhail cannot go from the point $(0, 0)$ to the point $(n_i, m_i)$ in exactly $k_i$ moves described above. Otherwise the $i$-th integer should be equal to the the maximum number of diagonal moves among all possible movements.
-----Example-----
Input
3
2 2 3
4 3 7
10 1 9
Output
1
6
-1
-----Note-----
One of the possible answers to the first test case: $(0, 0) \to (1, 0) \to (1, 1) \to (2, 2)$.
One of the possible answers to the second test case: $(0, 0) \to (0, 1) \to (1, 2) \to (0, 3) \to (1, 4) \to (2, 3) \to (3, 2) \to (4, 3)$.
In the third test case Mikhail cannot reach the point $(10, 1)$ in 9 moves.
|
Q = int(input())
src = [tuple(map(int,input().split())) for i in range(Q)]
ans = []
for x,y,k in src:
d = max(x,y)
if (x+y)%2:
ans.append(-1 if d > k else k-1)
else:
if d > k:
ans.append(-1)
else:
ans.append(k-2 if (d+k)%2 else k)
print(*ans,sep='\n')
|
Mikhail walks on a Cartesian plane. He starts at the point $(0, 0)$, and in one move he can go to any of eight adjacent points. For example, if Mikhail is currently at the point $(0, 0)$, he can go to any of the following points in one move: $(1, 0)$; $(1, 1)$; $(0, 1)$; $(-1, 1)$; $(-1, 0)$; $(-1, -1)$; $(0, -1)$; $(1, -1)$.
If Mikhail goes from the point $(x1, y1)$ to the point $(x2, y2)$ in one move, and $x1 \ne x2$ and $y1 \ne y2$, then such a move is called a diagonal move.
Mikhail has $q$ queries. For the $i$-th query Mikhail's target is to go to the point $(n_i, m_i)$ from the point $(0, 0)$ in exactly $k_i$ moves. Among all possible movements he want to choose one with the maximum number of diagonal moves. Your task is to find the maximum number of diagonal moves or find that it is impossible to go from the point $(0, 0)$ to the point $(n_i, m_i)$ in $k_i$ moves.
Note that Mikhail can visit any point any number of times (even the destination point!).
-----Input-----
The first line of the input contains one integer $q$ ($1 \le q \le 10^4$) — the number of queries.
Then $q$ lines follow. The $i$-th of these $q$ lines contains three integers $n_i$, $m_i$ and $k_i$ ($1 \le n_i, m_i, k_i \le 10^{18}$) — $x$-coordinate of the destination point of the query, $y$-coordinate of the destination point of the query and the number of moves in the query, correspondingly.
-----Output-----
Print $q$ integers. The $i$-th integer should be equal to -1 if Mikhail cannot go from the point $(0, 0)$ to the point $(n_i, m_i)$ in exactly $k_i$ moves described above. Otherwise the $i$-th integer should be equal to the the maximum number of diagonal moves among all possible movements.
-----Example-----
Input
3
2 2 3
4 3 7
10 1 9
Output
1
6
-1
-----Note-----
One of the possible answers to the first test case: $(0, 0) \to (1, 0) \to (1, 1) \to (2, 2)$.
One of the possible answers to the second test case: $(0, 0) \to (0, 1) \to (1, 2) \to (0, 3) \to (1, 4) \to (2, 3) \to (3, 2) \to (4, 3)$.
In the third test case Mikhail cannot reach the point $(10, 1)$ in 9 moves.
|
def m():
[x, y, k] = [int(i) for i in input().split()]
d=min(x, y)
x-=d
y-=d
k-=d
if k-x-y<0:
print(-1)
else:
x+=y
if x%2 > 0 and k%2>0:
print(d+k-1)
elif x%2 >0:
print(d+k-1)
elif k%2>0:
print(d+k-2)
else:
print(d+k)
n=int(input())
for i in range(n):
m()
|
Mikhail walks on a Cartesian plane. He starts at the point $(0, 0)$, and in one move he can go to any of eight adjacent points. For example, if Mikhail is currently at the point $(0, 0)$, he can go to any of the following points in one move: $(1, 0)$; $(1, 1)$; $(0, 1)$; $(-1, 1)$; $(-1, 0)$; $(-1, -1)$; $(0, -1)$; $(1, -1)$.
If Mikhail goes from the point $(x1, y1)$ to the point $(x2, y2)$ in one move, and $x1 \ne x2$ and $y1 \ne y2$, then such a move is called a diagonal move.
Mikhail has $q$ queries. For the $i$-th query Mikhail's target is to go to the point $(n_i, m_i)$ from the point $(0, 0)$ in exactly $k_i$ moves. Among all possible movements he want to choose one with the maximum number of diagonal moves. Your task is to find the maximum number of diagonal moves or find that it is impossible to go from the point $(0, 0)$ to the point $(n_i, m_i)$ in $k_i$ moves.
Note that Mikhail can visit any point any number of times (even the destination point!).
-----Input-----
The first line of the input contains one integer $q$ ($1 \le q \le 10^4$) — the number of queries.
Then $q$ lines follow. The $i$-th of these $q$ lines contains three integers $n_i$, $m_i$ and $k_i$ ($1 \le n_i, m_i, k_i \le 10^{18}$) — $x$-coordinate of the destination point of the query, $y$-coordinate of the destination point of the query and the number of moves in the query, correspondingly.
-----Output-----
Print $q$ integers. The $i$-th integer should be equal to -1 if Mikhail cannot go from the point $(0, 0)$ to the point $(n_i, m_i)$ in exactly $k_i$ moves described above. Otherwise the $i$-th integer should be equal to the the maximum number of diagonal moves among all possible movements.
-----Example-----
Input
3
2 2 3
4 3 7
10 1 9
Output
1
6
-1
-----Note-----
One of the possible answers to the first test case: $(0, 0) \to (1, 0) \to (1, 1) \to (2, 2)$.
One of the possible answers to the second test case: $(0, 0) \to (0, 1) \to (1, 2) \to (0, 3) \to (1, 4) \to (2, 3) \to (3, 2) \to (4, 3)$.
In the third test case Mikhail cannot reach the point $(10, 1)$ in 9 moves.
|
q = int(input())
for i in range(q):
(x, y, k) = list(map(int, input().split()))
if max(x, y) > k:
print(-1)
elif x == y and k == x + 1:
print(k - 2)
continue
elif x % 2 == 1 and y % 2 == 1 and k % 2 == 0:
print(k - 2)
continue
elif x % 2 == 0 and y % 2 == 0 and k % 2 == 1:
print(k - 2)
continue
elif (x + y) % 2 == 0:
print(k)
else:
print(k - 1)
|
Mikhail walks on a Cartesian plane. He starts at the point $(0, 0)$, and in one move he can go to any of eight adjacent points. For example, if Mikhail is currently at the point $(0, 0)$, he can go to any of the following points in one move: $(1, 0)$; $(1, 1)$; $(0, 1)$; $(-1, 1)$; $(-1, 0)$; $(-1, -1)$; $(0, -1)$; $(1, -1)$.
If Mikhail goes from the point $(x1, y1)$ to the point $(x2, y2)$ in one move, and $x1 \ne x2$ and $y1 \ne y2$, then such a move is called a diagonal move.
Mikhail has $q$ queries. For the $i$-th query Mikhail's target is to go to the point $(n_i, m_i)$ from the point $(0, 0)$ in exactly $k_i$ moves. Among all possible movements he want to choose one with the maximum number of diagonal moves. Your task is to find the maximum number of diagonal moves or find that it is impossible to go from the point $(0, 0)$ to the point $(n_i, m_i)$ in $k_i$ moves.
Note that Mikhail can visit any point any number of times (even the destination point!).
-----Input-----
The first line of the input contains one integer $q$ ($1 \le q \le 10^4$) — the number of queries.
Then $q$ lines follow. The $i$-th of these $q$ lines contains three integers $n_i$, $m_i$ and $k_i$ ($1 \le n_i, m_i, k_i \le 10^{18}$) — $x$-coordinate of the destination point of the query, $y$-coordinate of the destination point of the query and the number of moves in the query, correspondingly.
-----Output-----
Print $q$ integers. The $i$-th integer should be equal to -1 if Mikhail cannot go from the point $(0, 0)$ to the point $(n_i, m_i)$ in exactly $k_i$ moves described above. Otherwise the $i$-th integer should be equal to the the maximum number of diagonal moves among all possible movements.
-----Example-----
Input
3
2 2 3
4 3 7
10 1 9
Output
1
6
-1
-----Note-----
One of the possible answers to the first test case: $(0, 0) \to (1, 0) \to (1, 1) \to (2, 2)$.
One of the possible answers to the second test case: $(0, 0) \to (0, 1) \to (1, 2) \to (0, 3) \to (1, 4) \to (2, 3) \to (3, 2) \to (4, 3)$.
In the third test case Mikhail cannot reach the point $(10, 1)$ in 9 moves.
|
n = int(input())
for q in range(n):
x, y, k = list(map(int, input().split()))
if max(x, y) > k:
print(-1)
else:
if 0 == (x + y) % 2:
if k % 2 == max(x, y) % 2:
print(k)
else:
print(k - 2)
else:
print(k - 1)
|
Mikhail walks on a Cartesian plane. He starts at the point $(0, 0)$, and in one move he can go to any of eight adjacent points. For example, if Mikhail is currently at the point $(0, 0)$, he can go to any of the following points in one move: $(1, 0)$; $(1, 1)$; $(0, 1)$; $(-1, 1)$; $(-1, 0)$; $(-1, -1)$; $(0, -1)$; $(1, -1)$.
If Mikhail goes from the point $(x1, y1)$ to the point $(x2, y2)$ in one move, and $x1 \ne x2$ and $y1 \ne y2$, then such a move is called a diagonal move.
Mikhail has $q$ queries. For the $i$-th query Mikhail's target is to go to the point $(n_i, m_i)$ from the point $(0, 0)$ in exactly $k_i$ moves. Among all possible movements he want to choose one with the maximum number of diagonal moves. Your task is to find the maximum number of diagonal moves or find that it is impossible to go from the point $(0, 0)$ to the point $(n_i, m_i)$ in $k_i$ moves.
Note that Mikhail can visit any point any number of times (even the destination point!).
-----Input-----
The first line of the input contains one integer $q$ ($1 \le q \le 10^4$) — the number of queries.
Then $q$ lines follow. The $i$-th of these $q$ lines contains three integers $n_i$, $m_i$ and $k_i$ ($1 \le n_i, m_i, k_i \le 10^{18}$) — $x$-coordinate of the destination point of the query, $y$-coordinate of the destination point of the query and the number of moves in the query, correspondingly.
-----Output-----
Print $q$ integers. The $i$-th integer should be equal to -1 if Mikhail cannot go from the point $(0, 0)$ to the point $(n_i, m_i)$ in exactly $k_i$ moves described above. Otherwise the $i$-th integer should be equal to the the maximum number of diagonal moves among all possible movements.
-----Example-----
Input
3
2 2 3
4 3 7
10 1 9
Output
1
6
-1
-----Note-----
One of the possible answers to the first test case: $(0, 0) \to (1, 0) \to (1, 1) \to (2, 2)$.
One of the possible answers to the second test case: $(0, 0) \to (0, 1) \to (1, 2) \to (0, 3) \to (1, 4) \to (2, 3) \to (3, 2) \to (4, 3)$.
In the third test case Mikhail cannot reach the point $(10, 1)$ in 9 moves.
|
def go():
n = int(input())
for i in range(n):
a, b, d = [int(i) for i in input().split(' ')]
if a > d or b > d:
print(-1)
elif a % 2 == b % 2:
if a % 2 == d % 2:
print(d)
else:
print(d - 2)
else:
if a % 2 == b % 2:
if d % 2 == a % 2:
print(d)
else:
print(d - 2)
else:
print(d - 1)
go()
|
Mikhail walks on a Cartesian plane. He starts at the point $(0, 0)$, and in one move he can go to any of eight adjacent points. For example, if Mikhail is currently at the point $(0, 0)$, he can go to any of the following points in one move: $(1, 0)$; $(1, 1)$; $(0, 1)$; $(-1, 1)$; $(-1, 0)$; $(-1, -1)$; $(0, -1)$; $(1, -1)$.
If Mikhail goes from the point $(x1, y1)$ to the point $(x2, y2)$ in one move, and $x1 \ne x2$ and $y1 \ne y2$, then such a move is called a diagonal move.
Mikhail has $q$ queries. For the $i$-th query Mikhail's target is to go to the point $(n_i, m_i)$ from the point $(0, 0)$ in exactly $k_i$ moves. Among all possible movements he want to choose one with the maximum number of diagonal moves. Your task is to find the maximum number of diagonal moves or find that it is impossible to go from the point $(0, 0)$ to the point $(n_i, m_i)$ in $k_i$ moves.
Note that Mikhail can visit any point any number of times (even the destination point!).
-----Input-----
The first line of the input contains one integer $q$ ($1 \le q \le 10^4$) — the number of queries.
Then $q$ lines follow. The $i$-th of these $q$ lines contains three integers $n_i$, $m_i$ and $k_i$ ($1 \le n_i, m_i, k_i \le 10^{18}$) — $x$-coordinate of the destination point of the query, $y$-coordinate of the destination point of the query and the number of moves in the query, correspondingly.
-----Output-----
Print $q$ integers. The $i$-th integer should be equal to -1 if Mikhail cannot go from the point $(0, 0)$ to the point $(n_i, m_i)$ in exactly $k_i$ moves described above. Otherwise the $i$-th integer should be equal to the the maximum number of diagonal moves among all possible movements.
-----Example-----
Input
3
2 2 3
4 3 7
10 1 9
Output
1
6
-1
-----Note-----
One of the possible answers to the first test case: $(0, 0) \to (1, 0) \to (1, 1) \to (2, 2)$.
One of the possible answers to the second test case: $(0, 0) \to (0, 1) \to (1, 2) \to (0, 3) \to (1, 4) \to (2, 3) \to (3, 2) \to (4, 3)$.
In the third test case Mikhail cannot reach the point $(10, 1)$ in 9 moves.
|
q = int(input())
for i in range(q):
n, m, k = map(int, input().split())
p = min(m, n)
r = max(n, m) - p
if (p+r) > k:
print(-1)
elif r % 2 == 1:
print(k - 1)
elif (k - p) % 2 == 0:
print(k)
else:
print(k - 2)
|
Mikhail walks on a Cartesian plane. He starts at the point $(0, 0)$, and in one move he can go to any of eight adjacent points. For example, if Mikhail is currently at the point $(0, 0)$, he can go to any of the following points in one move: $(1, 0)$; $(1, 1)$; $(0, 1)$; $(-1, 1)$; $(-1, 0)$; $(-1, -1)$; $(0, -1)$; $(1, -1)$.
If Mikhail goes from the point $(x1, y1)$ to the point $(x2, y2)$ in one move, and $x1 \ne x2$ and $y1 \ne y2$, then such a move is called a diagonal move.
Mikhail has $q$ queries. For the $i$-th query Mikhail's target is to go to the point $(n_i, m_i)$ from the point $(0, 0)$ in exactly $k_i$ moves. Among all possible movements he want to choose one with the maximum number of diagonal moves. Your task is to find the maximum number of diagonal moves or find that it is impossible to go from the point $(0, 0)$ to the point $(n_i, m_i)$ in $k_i$ moves.
Note that Mikhail can visit any point any number of times (even the destination point!).
-----Input-----
The first line of the input contains one integer $q$ ($1 \le q \le 10^4$) — the number of queries.
Then $q$ lines follow. The $i$-th of these $q$ lines contains three integers $n_i$, $m_i$ and $k_i$ ($1 \le n_i, m_i, k_i \le 10^{18}$) — $x$-coordinate of the destination point of the query, $y$-coordinate of the destination point of the query and the number of moves in the query, correspondingly.
-----Output-----
Print $q$ integers. The $i$-th integer should be equal to -1 if Mikhail cannot go from the point $(0, 0)$ to the point $(n_i, m_i)$ in exactly $k_i$ moves described above. Otherwise the $i$-th integer should be equal to the the maximum number of diagonal moves among all possible movements.
-----Example-----
Input
3
2 2 3
4 3 7
10 1 9
Output
1
6
-1
-----Note-----
One of the possible answers to the first test case: $(0, 0) \to (1, 0) \to (1, 1) \to (2, 2)$.
One of the possible answers to the second test case: $(0, 0) \to (0, 1) \to (1, 2) \to (0, 3) \to (1, 4) \to (2, 3) \to (3, 2) \to (4, 3)$.
In the third test case Mikhail cannot reach the point $(10, 1)$ in 9 moves.
|
q = int(input())
for i in range(q):
n, m, k = map(int, input().split())
ost = max(n, m) - min(n, m)
plus = 0
if ost % 2 != 0:
plus = 1
ost -= 1
mini = min(n, m) + ost + plus
#print('mini: ' + str(mini))
if k < mini:
print(-1)
elif (k - mini) % 2 == 0 or plus == 1:
print(k - plus)
else:
print(k - plus - 2)
|
Mikhail walks on a Cartesian plane. He starts at the point $(0, 0)$, and in one move he can go to any of eight adjacent points. For example, if Mikhail is currently at the point $(0, 0)$, he can go to any of the following points in one move: $(1, 0)$; $(1, 1)$; $(0, 1)$; $(-1, 1)$; $(-1, 0)$; $(-1, -1)$; $(0, -1)$; $(1, -1)$.
If Mikhail goes from the point $(x1, y1)$ to the point $(x2, y2)$ in one move, and $x1 \ne x2$ and $y1 \ne y2$, then such a move is called a diagonal move.
Mikhail has $q$ queries. For the $i$-th query Mikhail's target is to go to the point $(n_i, m_i)$ from the point $(0, 0)$ in exactly $k_i$ moves. Among all possible movements he want to choose one with the maximum number of diagonal moves. Your task is to find the maximum number of diagonal moves or find that it is impossible to go from the point $(0, 0)$ to the point $(n_i, m_i)$ in $k_i$ moves.
Note that Mikhail can visit any point any number of times (even the destination point!).
-----Input-----
The first line of the input contains one integer $q$ ($1 \le q \le 10^4$) — the number of queries.
Then $q$ lines follow. The $i$-th of these $q$ lines contains three integers $n_i$, $m_i$ and $k_i$ ($1 \le n_i, m_i, k_i \le 10^{18}$) — $x$-coordinate of the destination point of the query, $y$-coordinate of the destination point of the query and the number of moves in the query, correspondingly.
-----Output-----
Print $q$ integers. The $i$-th integer should be equal to -1 if Mikhail cannot go from the point $(0, 0)$ to the point $(n_i, m_i)$ in exactly $k_i$ moves described above. Otherwise the $i$-th integer should be equal to the the maximum number of diagonal moves among all possible movements.
-----Example-----
Input
3
2 2 3
4 3 7
10 1 9
Output
1
6
-1
-----Note-----
One of the possible answers to the first test case: $(0, 0) \to (1, 0) \to (1, 1) \to (2, 2)$.
One of the possible answers to the second test case: $(0, 0) \to (0, 1) \to (1, 2) \to (0, 3) \to (1, 4) \to (2, 3) \to (3, 2) \to (4, 3)$.
In the third test case Mikhail cannot reach the point $(10, 1)$ in 9 moves.
|
q=int(input())
for i in range(q):
n,m,k=list(map(int,input().split()))
if n>k or m>k:
print(-1)
else:
if n%2==0 and m%2==0:
if k%2==0:
print(k)
else:
print(k-2)
elif (n%2==0 and m%2==1) or (n%2==1 and m%2==0):
print(k-1)
elif n%2==1 and m%2==1:
if k%2==0:
print(k-2)
else:
print(k)
|
Mikhail walks on a Cartesian plane. He starts at the point $(0, 0)$, and in one move he can go to any of eight adjacent points. For example, if Mikhail is currently at the point $(0, 0)$, he can go to any of the following points in one move: $(1, 0)$; $(1, 1)$; $(0, 1)$; $(-1, 1)$; $(-1, 0)$; $(-1, -1)$; $(0, -1)$; $(1, -1)$.
If Mikhail goes from the point $(x1, y1)$ to the point $(x2, y2)$ in one move, and $x1 \ne x2$ and $y1 \ne y2$, then such a move is called a diagonal move.
Mikhail has $q$ queries. For the $i$-th query Mikhail's target is to go to the point $(n_i, m_i)$ from the point $(0, 0)$ in exactly $k_i$ moves. Among all possible movements he want to choose one with the maximum number of diagonal moves. Your task is to find the maximum number of diagonal moves or find that it is impossible to go from the point $(0, 0)$ to the point $(n_i, m_i)$ in $k_i$ moves.
Note that Mikhail can visit any point any number of times (even the destination point!).
-----Input-----
The first line of the input contains one integer $q$ ($1 \le q \le 10^4$) — the number of queries.
Then $q$ lines follow. The $i$-th of these $q$ lines contains three integers $n_i$, $m_i$ and $k_i$ ($1 \le n_i, m_i, k_i \le 10^{18}$) — $x$-coordinate of the destination point of the query, $y$-coordinate of the destination point of the query and the number of moves in the query, correspondingly.
-----Output-----
Print $q$ integers. The $i$-th integer should be equal to -1 if Mikhail cannot go from the point $(0, 0)$ to the point $(n_i, m_i)$ in exactly $k_i$ moves described above. Otherwise the $i$-th integer should be equal to the the maximum number of diagonal moves among all possible movements.
-----Example-----
Input
3
2 2 3
4 3 7
10 1 9
Output
1
6
-1
-----Note-----
One of the possible answers to the first test case: $(0, 0) \to (1, 0) \to (1, 1) \to (2, 2)$.
One of the possible answers to the second test case: $(0, 0) \to (0, 1) \to (1, 2) \to (0, 3) \to (1, 4) \to (2, 3) \to (3, 2) \to (4, 3)$.
In the third test case Mikhail cannot reach the point $(10, 1)$ in 9 moves.
|
q=int(input())
for i in range(q):
n, m, k = map(int, input().split())
ans=max(n,m)
diff=k-ans
if diff<0:
print(-1)
else:
if (n%2==0 and m%2==0) or (n%2!=0 and m%2!=0):
if diff%2==0:
ans+=diff
else:
ans+=diff-2
else:
ans+=diff-1
print(ans)
|
Mikhail walks on a Cartesian plane. He starts at the point $(0, 0)$, and in one move he can go to any of eight adjacent points. For example, if Mikhail is currently at the point $(0, 0)$, he can go to any of the following points in one move: $(1, 0)$; $(1, 1)$; $(0, 1)$; $(-1, 1)$; $(-1, 0)$; $(-1, -1)$; $(0, -1)$; $(1, -1)$.
If Mikhail goes from the point $(x1, y1)$ to the point $(x2, y2)$ in one move, and $x1 \ne x2$ and $y1 \ne y2$, then such a move is called a diagonal move.
Mikhail has $q$ queries. For the $i$-th query Mikhail's target is to go to the point $(n_i, m_i)$ from the point $(0, 0)$ in exactly $k_i$ moves. Among all possible movements he want to choose one with the maximum number of diagonal moves. Your task is to find the maximum number of diagonal moves or find that it is impossible to go from the point $(0, 0)$ to the point $(n_i, m_i)$ in $k_i$ moves.
Note that Mikhail can visit any point any number of times (even the destination point!).
-----Input-----
The first line of the input contains one integer $q$ ($1 \le q \le 10^4$) — the number of queries.
Then $q$ lines follow. The $i$-th of these $q$ lines contains three integers $n_i$, $m_i$ and $k_i$ ($1 \le n_i, m_i, k_i \le 10^{18}$) — $x$-coordinate of the destination point of the query, $y$-coordinate of the destination point of the query and the number of moves in the query, correspondingly.
-----Output-----
Print $q$ integers. The $i$-th integer should be equal to -1 if Mikhail cannot go from the point $(0, 0)$ to the point $(n_i, m_i)$ in exactly $k_i$ moves described above. Otherwise the $i$-th integer should be equal to the the maximum number of diagonal moves among all possible movements.
-----Example-----
Input
3
2 2 3
4 3 7
10 1 9
Output
1
6
-1
-----Note-----
One of the possible answers to the first test case: $(0, 0) \to (1, 0) \to (1, 1) \to (2, 2)$.
One of the possible answers to the second test case: $(0, 0) \to (0, 1) \to (1, 2) \to (0, 3) \to (1, 4) \to (2, 3) \to (3, 2) \to (4, 3)$.
In the third test case Mikhail cannot reach the point $(10, 1)$ in 9 moves.
|
"""
KA YM KA AS KA ASKA YASK KA SKAYMA
KA KA SKAY SK SK AS AY AY SK SKAY AS AS
KA AS AS YM KA AS AS YM KA SK AS YM AS
KAYM MA MA AYMA AS MASK SK MA MA SKAYMA
KA AS YMASKAYMAS YM AS AS SK YMASKAYMAS AS
KA KA AY SK YM AS SK AY SK AS AS
KA YM KA KA YM AS SK KA KA SKAYMA
"""
n=int(input())
for i in range(n):
x,y,k=map(int,input().split())
x,y=abs(x),abs(y)
min_moves=max(x,y)
if min_moves>k:
print(-1)
else:
ans=min(x,y)
x-=ans
y-=ans
p=max(x,y)
k-=ans
if k==p and p%2==0:
print(ans+k)
elif k==p and p%2==1:
print(ans+k-1)
elif p%2==0 and k%2==0:
print(ans+k)
elif p%2==0 and k%2==1:
print(ans+k-2)
elif p%2==1:
print(ans+k-1)
|
You are given three sequences: $a_1, a_2, \ldots, a_n$; $b_1, b_2, \ldots, b_n$; $c_1, c_2, \ldots, c_n$.
For each $i$, $a_i \neq b_i$, $a_i \neq c_i$, $b_i \neq c_i$.
Find a sequence $p_1, p_2, \ldots, p_n$, that satisfy the following conditions:
$p_i \in \{a_i, b_i, c_i\}$
$p_i \neq p_{(i \mod n) + 1}$.
In other words, for each element, you need to choose one of the three possible values, such that no two adjacent elements (where we consider elements $i,i+1$ adjacent for $i<n$ and also elements $1$ and $n$) will have equal value.
It can be proved that in the given constraints solution always exists. You don't need to minimize/maximize anything, you need to find any proper sequence.
-----Input-----
The first line of input contains one integer $t$ ($1 \leq t \leq 100$): the number of test cases.
The first line of each test case contains one integer $n$ ($3 \leq n \leq 100$): the number of elements in the given sequences.
The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \leq a_i \leq 100$).
The third line contains $n$ integers $b_1, b_2, \ldots, b_n$ ($1 \leq b_i \leq 100$).
The fourth line contains $n$ integers $c_1, c_2, \ldots, c_n$ ($1 \leq c_i \leq 100$).
It is guaranteed that $a_i \neq b_i$, $a_i \neq c_i$, $b_i \neq c_i$ for all $i$.
-----Output-----
For each test case, print $n$ integers: $p_1, p_2, \ldots, p_n$ ($p_i \in \{a_i, b_i, c_i\}$, $p_i \neq p_{i \mod n + 1}$).
If there are several solutions, you can print any.
-----Example-----
Input
5
3
1 1 1
2 2 2
3 3 3
4
1 2 1 2
2 1 2 1
3 4 3 4
7
1 3 3 1 1 1 1
2 4 4 3 2 2 4
4 2 2 2 4 4 2
3
1 2 1
2 3 3
3 1 2
10
1 1 1 2 2 2 3 3 3 1
2 2 2 3 3 3 1 1 1 2
3 3 3 1 1 1 2 2 2 3
Output
1 2 3
1 2 1 2
1 3 4 3 2 4 2
1 3 2
1 2 3 1 2 3 1 2 3 2
-----Note-----
In the first test case $p = [1, 2, 3]$.
It is a correct answer, because:
$p_1 = 1 = a_1$, $p_2 = 2 = b_2$, $p_3 = 3 = c_3$ $p_1 \neq p_2 $, $p_2 \neq p_3 $, $p_3 \neq p_1$
All possible correct answers to this test case are: $[1, 2, 3]$, $[1, 3, 2]$, $[2, 1, 3]$, $[2, 3, 1]$, $[3, 1, 2]$, $[3, 2, 1]$.
In the second test case $p = [1, 2, 1, 2]$.
In this sequence $p_1 = a_1$, $p_2 = a_2$, $p_3 = a_3$, $p_4 = a_4$. Also we can see, that no two adjacent elements of the sequence are equal.
In the third test case $p = [1, 3, 4, 3, 2, 4, 2]$.
In this sequence $p_1 = a_1$, $p_2 = a_2$, $p_3 = b_3$, $p_4 = b_4$, $p_5 = b_5$, $p_6 = c_6$, $p_7 = c_7$. Also we can see, that no two adjacent elements of the sequence are equal.
|
import sys
import random
from fractions import Fraction
from math import *
def input():
return sys.stdin.readline().strip()
def iinput():
return int(input())
def finput():
return float(input())
def tinput():
return input().split()
def linput():
return list(input())
def rinput():
return list(map(int, tinput()))
def fiinput():
return list(map(float, tinput()))
def rlinput():
return list(map(int, input().split()))
def trinput():
return tuple(rinput())
def srlinput():
return sorted(list(map(int, input().split())))
def NOYES(fl):
if fl:
print("NO")
else:
print("YES")
def YESNO(fl):
if fl:
print("YES")
else:
print("NO")
def main():
n = iinput()
#k = iinput()
#m = iinput()
#n = int(sys.stdin.readline().strip())
#n, k = rinput()
#n, m = rinput()
#m, k = rinput()
#n, k, m = rinput()
#n, m, k = rinput()
#k, n, m = rinput()
#k, m, n = rinput()
#m, k, n = rinput()
#m, n, k = rinput()
q = [rlinput(), rlinput(), rlinput()]
#q = linput()
ans = q[0].copy()
for i in range(1, n):
if ans[i] == ans[i - 1]:
ans[i] = q[1][i]
if i == n - 1:
o = 0
while q[o][i] == ans[n - 2] or q[o][i] == ans[0]:
o += 1
ans[i] = q[o][i]
print(*ans)
for i in range(iinput()):
main()
|
You are given three sequences: $a_1, a_2, \ldots, a_n$; $b_1, b_2, \ldots, b_n$; $c_1, c_2, \ldots, c_n$.
For each $i$, $a_i \neq b_i$, $a_i \neq c_i$, $b_i \neq c_i$.
Find a sequence $p_1, p_2, \ldots, p_n$, that satisfy the following conditions:
$p_i \in \{a_i, b_i, c_i\}$
$p_i \neq p_{(i \mod n) + 1}$.
In other words, for each element, you need to choose one of the three possible values, such that no two adjacent elements (where we consider elements $i,i+1$ adjacent for $i<n$ and also elements $1$ and $n$) will have equal value.
It can be proved that in the given constraints solution always exists. You don't need to minimize/maximize anything, you need to find any proper sequence.
-----Input-----
The first line of input contains one integer $t$ ($1 \leq t \leq 100$): the number of test cases.
The first line of each test case contains one integer $n$ ($3 \leq n \leq 100$): the number of elements in the given sequences.
The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \leq a_i \leq 100$).
The third line contains $n$ integers $b_1, b_2, \ldots, b_n$ ($1 \leq b_i \leq 100$).
The fourth line contains $n$ integers $c_1, c_2, \ldots, c_n$ ($1 \leq c_i \leq 100$).
It is guaranteed that $a_i \neq b_i$, $a_i \neq c_i$, $b_i \neq c_i$ for all $i$.
-----Output-----
For each test case, print $n$ integers: $p_1, p_2, \ldots, p_n$ ($p_i \in \{a_i, b_i, c_i\}$, $p_i \neq p_{i \mod n + 1}$).
If there are several solutions, you can print any.
-----Example-----
Input
5
3
1 1 1
2 2 2
3 3 3
4
1 2 1 2
2 1 2 1
3 4 3 4
7
1 3 3 1 1 1 1
2 4 4 3 2 2 4
4 2 2 2 4 4 2
3
1 2 1
2 3 3
3 1 2
10
1 1 1 2 2 2 3 3 3 1
2 2 2 3 3 3 1 1 1 2
3 3 3 1 1 1 2 2 2 3
Output
1 2 3
1 2 1 2
1 3 4 3 2 4 2
1 3 2
1 2 3 1 2 3 1 2 3 2
-----Note-----
In the first test case $p = [1, 2, 3]$.
It is a correct answer, because:
$p_1 = 1 = a_1$, $p_2 = 2 = b_2$, $p_3 = 3 = c_3$ $p_1 \neq p_2 $, $p_2 \neq p_3 $, $p_3 \neq p_1$
All possible correct answers to this test case are: $[1, 2, 3]$, $[1, 3, 2]$, $[2, 1, 3]$, $[2, 3, 1]$, $[3, 1, 2]$, $[3, 2, 1]$.
In the second test case $p = [1, 2, 1, 2]$.
In this sequence $p_1 = a_1$, $p_2 = a_2$, $p_3 = a_3$, $p_4 = a_4$. Also we can see, that no two adjacent elements of the sequence are equal.
In the third test case $p = [1, 3, 4, 3, 2, 4, 2]$.
In this sequence $p_1 = a_1$, $p_2 = a_2$, $p_3 = b_3$, $p_4 = b_4$, $p_5 = b_5$, $p_6 = c_6$, $p_7 = c_7$. Also we can see, that no two adjacent elements of the sequence are equal.
|
for _ in range(int(input())):
n=int(input())
a=list(map(int,input().split()))
b=list(map(int,input().split()))
c=list(map(int,input().split()))
p=a
for i in range(n):
if p[i]==p[(i+1)%n]:
if p[i]!=b[i] and p[(i-1)%n]!=b[i]:p[i]=b[i]
else:p[i]=c[i]
print(*p)
|
You are given three sequences: $a_1, a_2, \ldots, a_n$; $b_1, b_2, \ldots, b_n$; $c_1, c_2, \ldots, c_n$.
For each $i$, $a_i \neq b_i$, $a_i \neq c_i$, $b_i \neq c_i$.
Find a sequence $p_1, p_2, \ldots, p_n$, that satisfy the following conditions:
$p_i \in \{a_i, b_i, c_i\}$
$p_i \neq p_{(i \mod n) + 1}$.
In other words, for each element, you need to choose one of the three possible values, such that no two adjacent elements (where we consider elements $i,i+1$ adjacent for $i<n$ and also elements $1$ and $n$) will have equal value.
It can be proved that in the given constraints solution always exists. You don't need to minimize/maximize anything, you need to find any proper sequence.
-----Input-----
The first line of input contains one integer $t$ ($1 \leq t \leq 100$): the number of test cases.
The first line of each test case contains one integer $n$ ($3 \leq n \leq 100$): the number of elements in the given sequences.
The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \leq a_i \leq 100$).
The third line contains $n$ integers $b_1, b_2, \ldots, b_n$ ($1 \leq b_i \leq 100$).
The fourth line contains $n$ integers $c_1, c_2, \ldots, c_n$ ($1 \leq c_i \leq 100$).
It is guaranteed that $a_i \neq b_i$, $a_i \neq c_i$, $b_i \neq c_i$ for all $i$.
-----Output-----
For each test case, print $n$ integers: $p_1, p_2, \ldots, p_n$ ($p_i \in \{a_i, b_i, c_i\}$, $p_i \neq p_{i \mod n + 1}$).
If there are several solutions, you can print any.
-----Example-----
Input
5
3
1 1 1
2 2 2
3 3 3
4
1 2 1 2
2 1 2 1
3 4 3 4
7
1 3 3 1 1 1 1
2 4 4 3 2 2 4
4 2 2 2 4 4 2
3
1 2 1
2 3 3
3 1 2
10
1 1 1 2 2 2 3 3 3 1
2 2 2 3 3 3 1 1 1 2
3 3 3 1 1 1 2 2 2 3
Output
1 2 3
1 2 1 2
1 3 4 3 2 4 2
1 3 2
1 2 3 1 2 3 1 2 3 2
-----Note-----
In the first test case $p = [1, 2, 3]$.
It is a correct answer, because:
$p_1 = 1 = a_1$, $p_2 = 2 = b_2$, $p_3 = 3 = c_3$ $p_1 \neq p_2 $, $p_2 \neq p_3 $, $p_3 \neq p_1$
All possible correct answers to this test case are: $[1, 2, 3]$, $[1, 3, 2]$, $[2, 1, 3]$, $[2, 3, 1]$, $[3, 1, 2]$, $[3, 2, 1]$.
In the second test case $p = [1, 2, 1, 2]$.
In this sequence $p_1 = a_1$, $p_2 = a_2$, $p_3 = a_3$, $p_4 = a_4$. Also we can see, that no two adjacent elements of the sequence are equal.
In the third test case $p = [1, 3, 4, 3, 2, 4, 2]$.
In this sequence $p_1 = a_1$, $p_2 = a_2$, $p_3 = b_3$, $p_4 = b_4$, $p_5 = b_5$, $p_6 = c_6$, $p_7 = c_7$. Also we can see, that no two adjacent elements of the sequence are equal.
|
for __ in range(int(input())):
n = int(input())
ar1 = list(map(int, input().split()))
ar2 = list(map(int, input().split()))
ar3 = list(map(int, input().split()))
ans = [ar1[0]]
for i in range(1, n - 1):
if ar1[i] != ans[-1]:
ans.append(ar1[i])
elif ar2[i] != ans[-1]:
ans.append(ar2[i])
elif ar3[i] != ans[-1]:
ans.append(ar3[i])
if ar1[-1] != ans[-1] and ar1[-1] != ans[0]:
ans.append(ar1[-1])
elif ar2[-1] != ans[-1] and ar2[-1] != ans[0]:
ans.append(ar2[-1])
elif ar3[-1] != ans[-1] and ar3[-1] != ans[0]:
ans.append(ar3[-1])
print(*ans)
|
You are given three sequences: $a_1, a_2, \ldots, a_n$; $b_1, b_2, \ldots, b_n$; $c_1, c_2, \ldots, c_n$.
For each $i$, $a_i \neq b_i$, $a_i \neq c_i$, $b_i \neq c_i$.
Find a sequence $p_1, p_2, \ldots, p_n$, that satisfy the following conditions:
$p_i \in \{a_i, b_i, c_i\}$
$p_i \neq p_{(i \mod n) + 1}$.
In other words, for each element, you need to choose one of the three possible values, such that no two adjacent elements (where we consider elements $i,i+1$ adjacent for $i<n$ and also elements $1$ and $n$) will have equal value.
It can be proved that in the given constraints solution always exists. You don't need to minimize/maximize anything, you need to find any proper sequence.
-----Input-----
The first line of input contains one integer $t$ ($1 \leq t \leq 100$): the number of test cases.
The first line of each test case contains one integer $n$ ($3 \leq n \leq 100$): the number of elements in the given sequences.
The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \leq a_i \leq 100$).
The third line contains $n$ integers $b_1, b_2, \ldots, b_n$ ($1 \leq b_i \leq 100$).
The fourth line contains $n$ integers $c_1, c_2, \ldots, c_n$ ($1 \leq c_i \leq 100$).
It is guaranteed that $a_i \neq b_i$, $a_i \neq c_i$, $b_i \neq c_i$ for all $i$.
-----Output-----
For each test case, print $n$ integers: $p_1, p_2, \ldots, p_n$ ($p_i \in \{a_i, b_i, c_i\}$, $p_i \neq p_{i \mod n + 1}$).
If there are several solutions, you can print any.
-----Example-----
Input
5
3
1 1 1
2 2 2
3 3 3
4
1 2 1 2
2 1 2 1
3 4 3 4
7
1 3 3 1 1 1 1
2 4 4 3 2 2 4
4 2 2 2 4 4 2
3
1 2 1
2 3 3
3 1 2
10
1 1 1 2 2 2 3 3 3 1
2 2 2 3 3 3 1 1 1 2
3 3 3 1 1 1 2 2 2 3
Output
1 2 3
1 2 1 2
1 3 4 3 2 4 2
1 3 2
1 2 3 1 2 3 1 2 3 2
-----Note-----
In the first test case $p = [1, 2, 3]$.
It is a correct answer, because:
$p_1 = 1 = a_1$, $p_2 = 2 = b_2$, $p_3 = 3 = c_3$ $p_1 \neq p_2 $, $p_2 \neq p_3 $, $p_3 \neq p_1$
All possible correct answers to this test case are: $[1, 2, 3]$, $[1, 3, 2]$, $[2, 1, 3]$, $[2, 3, 1]$, $[3, 1, 2]$, $[3, 2, 1]$.
In the second test case $p = [1, 2, 1, 2]$.
In this sequence $p_1 = a_1$, $p_2 = a_2$, $p_3 = a_3$, $p_4 = a_4$. Also we can see, that no two adjacent elements of the sequence are equal.
In the third test case $p = [1, 3, 4, 3, 2, 4, 2]$.
In this sequence $p_1 = a_1$, $p_2 = a_2$, $p_3 = b_3$, $p_4 = b_4$, $p_5 = b_5$, $p_6 = c_6$, $p_7 = c_7$. Also we can see, that no two adjacent elements of the sequence are equal.
|
T = int(input())
for t in range(T):
N = int(input())
A = [int(_) for _ in input().split()]
B = [int(_) for _ in input().split()]
C = [int(_) for _ in input().split()]
R = []
for i in range(N):
if i == 0:
R.append(A[i])
continue
if i == N-1:
if A[i] != R[0] and A[i] != R[-1]:
R.append(A[i])
elif B[i] != R[0] and B[i] != R[-1]:
R.append(B[i])
else:
R.append(C[i])
continue
if A[i] != R[-1]:
R.append(A[i])
else:
R.append(B[i])
print(' '.join(map(str, R)))
|
You are given three sequences: $a_1, a_2, \ldots, a_n$; $b_1, b_2, \ldots, b_n$; $c_1, c_2, \ldots, c_n$.
For each $i$, $a_i \neq b_i$, $a_i \neq c_i$, $b_i \neq c_i$.
Find a sequence $p_1, p_2, \ldots, p_n$, that satisfy the following conditions:
$p_i \in \{a_i, b_i, c_i\}$
$p_i \neq p_{(i \mod n) + 1}$.
In other words, for each element, you need to choose one of the three possible values, such that no two adjacent elements (where we consider elements $i,i+1$ adjacent for $i<n$ and also elements $1$ and $n$) will have equal value.
It can be proved that in the given constraints solution always exists. You don't need to minimize/maximize anything, you need to find any proper sequence.
-----Input-----
The first line of input contains one integer $t$ ($1 \leq t \leq 100$): the number of test cases.
The first line of each test case contains one integer $n$ ($3 \leq n \leq 100$): the number of elements in the given sequences.
The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \leq a_i \leq 100$).
The third line contains $n$ integers $b_1, b_2, \ldots, b_n$ ($1 \leq b_i \leq 100$).
The fourth line contains $n$ integers $c_1, c_2, \ldots, c_n$ ($1 \leq c_i \leq 100$).
It is guaranteed that $a_i \neq b_i$, $a_i \neq c_i$, $b_i \neq c_i$ for all $i$.
-----Output-----
For each test case, print $n$ integers: $p_1, p_2, \ldots, p_n$ ($p_i \in \{a_i, b_i, c_i\}$, $p_i \neq p_{i \mod n + 1}$).
If there are several solutions, you can print any.
-----Example-----
Input
5
3
1 1 1
2 2 2
3 3 3
4
1 2 1 2
2 1 2 1
3 4 3 4
7
1 3 3 1 1 1 1
2 4 4 3 2 2 4
4 2 2 2 4 4 2
3
1 2 1
2 3 3
3 1 2
10
1 1 1 2 2 2 3 3 3 1
2 2 2 3 3 3 1 1 1 2
3 3 3 1 1 1 2 2 2 3
Output
1 2 3
1 2 1 2
1 3 4 3 2 4 2
1 3 2
1 2 3 1 2 3 1 2 3 2
-----Note-----
In the first test case $p = [1, 2, 3]$.
It is a correct answer, because:
$p_1 = 1 = a_1$, $p_2 = 2 = b_2$, $p_3 = 3 = c_3$ $p_1 \neq p_2 $, $p_2 \neq p_3 $, $p_3 \neq p_1$
All possible correct answers to this test case are: $[1, 2, 3]$, $[1, 3, 2]$, $[2, 1, 3]$, $[2, 3, 1]$, $[3, 1, 2]$, $[3, 2, 1]$.
In the second test case $p = [1, 2, 1, 2]$.
In this sequence $p_1 = a_1$, $p_2 = a_2$, $p_3 = a_3$, $p_4 = a_4$. Also we can see, that no two adjacent elements of the sequence are equal.
In the third test case $p = [1, 3, 4, 3, 2, 4, 2]$.
In this sequence $p_1 = a_1$, $p_2 = a_2$, $p_3 = b_3$, $p_4 = b_4$, $p_5 = b_5$, $p_6 = c_6$, $p_7 = c_7$. Also we can see, that no two adjacent elements of the sequence are equal.
|
gans = []
for _ in range(int(input())):
n = int(input())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
c = list(map(int, input().split()))
ans = [a[0]]
for i in range(1, n - 1):
if a[i] != ans[i - 1]:
ans.append(a[i])
else:
ans.append(b[i])
if a[-1] != ans[-1] and a[-1] != ans[0]:
ans.append(a[-1])
elif b[-1] != ans[-1] and b[-1] != ans[0]:
ans.append(b[-1])
else:
ans.append(c[-1])
gans.append(' '.join(map(str, ans)))
print('\n'.join(gans))
|
You are given three sequences: $a_1, a_2, \ldots, a_n$; $b_1, b_2, \ldots, b_n$; $c_1, c_2, \ldots, c_n$.
For each $i$, $a_i \neq b_i$, $a_i \neq c_i$, $b_i \neq c_i$.
Find a sequence $p_1, p_2, \ldots, p_n$, that satisfy the following conditions:
$p_i \in \{a_i, b_i, c_i\}$
$p_i \neq p_{(i \mod n) + 1}$.
In other words, for each element, you need to choose one of the three possible values, such that no two adjacent elements (where we consider elements $i,i+1$ adjacent for $i<n$ and also elements $1$ and $n$) will have equal value.
It can be proved that in the given constraints solution always exists. You don't need to minimize/maximize anything, you need to find any proper sequence.
-----Input-----
The first line of input contains one integer $t$ ($1 \leq t \leq 100$): the number of test cases.
The first line of each test case contains one integer $n$ ($3 \leq n \leq 100$): the number of elements in the given sequences.
The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \leq a_i \leq 100$).
The third line contains $n$ integers $b_1, b_2, \ldots, b_n$ ($1 \leq b_i \leq 100$).
The fourth line contains $n$ integers $c_1, c_2, \ldots, c_n$ ($1 \leq c_i \leq 100$).
It is guaranteed that $a_i \neq b_i$, $a_i \neq c_i$, $b_i \neq c_i$ for all $i$.
-----Output-----
For each test case, print $n$ integers: $p_1, p_2, \ldots, p_n$ ($p_i \in \{a_i, b_i, c_i\}$, $p_i \neq p_{i \mod n + 1}$).
If there are several solutions, you can print any.
-----Example-----
Input
5
3
1 1 1
2 2 2
3 3 3
4
1 2 1 2
2 1 2 1
3 4 3 4
7
1 3 3 1 1 1 1
2 4 4 3 2 2 4
4 2 2 2 4 4 2
3
1 2 1
2 3 3
3 1 2
10
1 1 1 2 2 2 3 3 3 1
2 2 2 3 3 3 1 1 1 2
3 3 3 1 1 1 2 2 2 3
Output
1 2 3
1 2 1 2
1 3 4 3 2 4 2
1 3 2
1 2 3 1 2 3 1 2 3 2
-----Note-----
In the first test case $p = [1, 2, 3]$.
It is a correct answer, because:
$p_1 = 1 = a_1$, $p_2 = 2 = b_2$, $p_3 = 3 = c_3$ $p_1 \neq p_2 $, $p_2 \neq p_3 $, $p_3 \neq p_1$
All possible correct answers to this test case are: $[1, 2, 3]$, $[1, 3, 2]$, $[2, 1, 3]$, $[2, 3, 1]$, $[3, 1, 2]$, $[3, 2, 1]$.
In the second test case $p = [1, 2, 1, 2]$.
In this sequence $p_1 = a_1$, $p_2 = a_2$, $p_3 = a_3$, $p_4 = a_4$. Also we can see, that no two adjacent elements of the sequence are equal.
In the third test case $p = [1, 3, 4, 3, 2, 4, 2]$.
In this sequence $p_1 = a_1$, $p_2 = a_2$, $p_3 = b_3$, $p_4 = b_4$, $p_5 = b_5$, $p_6 = c_6$, $p_7 = c_7$. Also we can see, that no two adjacent elements of the sequence are equal.
|
from math import *
from bisect import *
from collections import *
from random import *
from decimal import *
import sys
input=sys.stdin.readline
def inp():
return int(input())
def st():
return input().rstrip('\n')
def lis():
return list(map(int,input().split()))
def ma():
return list(map(int,input().split()))
t=inp()
while(t):
t-=1
n=inp()
a=lis()
b=lis()
c=lis()
r=[a[0]]
for i in range(1,n):
if(i==n-1):
if(a[i]!=r[0] and a[i]!=r[-1]):
r.append(a[i])
continue
if(b[i]!=r[0] and b[i]!=r[-1]):
r.append(b[i])
continue
if(c[i]!=r[0] and c[i]!=r[-1]):
r.append(c[i])
continue
if(a[i]!=r[-1]):
r.append(a[i])
continue
if(b[i]!=r[-1]):
r.append(b[i])
continue
if(c[i]!=r[-1]):
r.append(c[i])
continue
print(*r)
|
You have $n$ barrels lined up in a row, numbered from left to right from one. Initially, the $i$-th barrel contains $a_i$ liters of water.
You can pour water from one barrel to another. In one act of pouring, you can choose two different barrels $x$ and $y$ (the $x$-th barrel shouldn't be empty) and pour any possible amount of water from barrel $x$ to barrel $y$ (possibly, all water). You may assume that barrels have infinite capacity, so you can pour any amount of water in each of them.
Calculate the maximum possible difference between the maximum and the minimum amount of water in the barrels, if you can pour water at most $k$ times.
Some examples: if you have four barrels, each containing $5$ liters of water, and $k = 1$, you may pour $5$ liters from the second barrel into the fourth, so the amounts of water in the barrels are $[5, 0, 5, 10]$, and the difference between the maximum and the minimum is $10$; if all barrels are empty, you can't make any operation, so the difference between the maximum and the minimum amount is still $0$.
-----Input-----
The first line contains one integer $t$ ($1 \le t \le 1000$) — the number of test cases.
The first line of each test case contains two integers $n$ and $k$ ($1 \le k < n \le 2 \cdot 10^5$) — the number of barrels and the number of pourings you can make.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($0 \le a_i \le 10^{9}$), where $a_i$ is the initial amount of water the $i$-th barrel has.
It's guaranteed that the total sum of $n$ over test cases doesn't exceed $2 \cdot 10^5$.
-----Output-----
For each test case, print the maximum possible difference between the maximum and the minimum amount of water in the barrels, if you can pour water at most $k$ times.
-----Example-----
Input
2
4 1
5 5 5 5
3 2
0 0 0
Output
10
0
|
def solve():
n, k = map(int,input().split())
lst = list(map(int,input().split()))
lst.sort()
ans = 0
for i in range(n - k - 1, n):
ans += lst[i]
print(ans)
for i in range(int(input())):
solve()
|
You have $n$ barrels lined up in a row, numbered from left to right from one. Initially, the $i$-th barrel contains $a_i$ liters of water.
You can pour water from one barrel to another. In one act of pouring, you can choose two different barrels $x$ and $y$ (the $x$-th barrel shouldn't be empty) and pour any possible amount of water from barrel $x$ to barrel $y$ (possibly, all water). You may assume that barrels have infinite capacity, so you can pour any amount of water in each of them.
Calculate the maximum possible difference between the maximum and the minimum amount of water in the barrels, if you can pour water at most $k$ times.
Some examples: if you have four barrels, each containing $5$ liters of water, and $k = 1$, you may pour $5$ liters from the second barrel into the fourth, so the amounts of water in the barrels are $[5, 0, 5, 10]$, and the difference between the maximum and the minimum is $10$; if all barrels are empty, you can't make any operation, so the difference between the maximum and the minimum amount is still $0$.
-----Input-----
The first line contains one integer $t$ ($1 \le t \le 1000$) — the number of test cases.
The first line of each test case contains two integers $n$ and $k$ ($1 \le k < n \le 2 \cdot 10^5$) — the number of barrels and the number of pourings you can make.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($0 \le a_i \le 10^{9}$), where $a_i$ is the initial amount of water the $i$-th barrel has.
It's guaranteed that the total sum of $n$ over test cases doesn't exceed $2 \cdot 10^5$.
-----Output-----
For each test case, print the maximum possible difference between the maximum and the minimum amount of water in the barrels, if you can pour water at most $k$ times.
-----Example-----
Input
2
4 1
5 5 5 5
3 2
0 0 0
Output
10
0
|
t=int(input())
for i in range(t):
n,k=[int(i) for i in input().split()]
a=[int(i) for i in input().split()]
a.sort(reverse=True)
print(sum(a[:k+1]))
|
You have $n$ barrels lined up in a row, numbered from left to right from one. Initially, the $i$-th barrel contains $a_i$ liters of water.
You can pour water from one barrel to another. In one act of pouring, you can choose two different barrels $x$ and $y$ (the $x$-th barrel shouldn't be empty) and pour any possible amount of water from barrel $x$ to barrel $y$ (possibly, all water). You may assume that barrels have infinite capacity, so you can pour any amount of water in each of them.
Calculate the maximum possible difference between the maximum and the minimum amount of water in the barrels, if you can pour water at most $k$ times.
Some examples: if you have four barrels, each containing $5$ liters of water, and $k = 1$, you may pour $5$ liters from the second barrel into the fourth, so the amounts of water in the barrels are $[5, 0, 5, 10]$, and the difference between the maximum and the minimum is $10$; if all barrels are empty, you can't make any operation, so the difference between the maximum and the minimum amount is still $0$.
-----Input-----
The first line contains one integer $t$ ($1 \le t \le 1000$) — the number of test cases.
The first line of each test case contains two integers $n$ and $k$ ($1 \le k < n \le 2 \cdot 10^5$) — the number of barrels and the number of pourings you can make.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($0 \le a_i \le 10^{9}$), where $a_i$ is the initial amount of water the $i$-th barrel has.
It's guaranteed that the total sum of $n$ over test cases doesn't exceed $2 \cdot 10^5$.
-----Output-----
For each test case, print the maximum possible difference between the maximum and the minimum amount of water in the barrels, if you can pour water at most $k$ times.
-----Example-----
Input
2
4 1
5 5 5 5
3 2
0 0 0
Output
10
0
|
# map(int, input().split())
rw = int(input())
for wewq in range(rw):
n, k = list(map(int, input().split()))
a = list(map(int, input().split()))
a.sort()
a.reverse()
f = 0
for i in range(k + 1):
f += a[i]
print(f)
|
You have $n$ barrels lined up in a row, numbered from left to right from one. Initially, the $i$-th barrel contains $a_i$ liters of water.
You can pour water from one barrel to another. In one act of pouring, you can choose two different barrels $x$ and $y$ (the $x$-th barrel shouldn't be empty) and pour any possible amount of water from barrel $x$ to barrel $y$ (possibly, all water). You may assume that barrels have infinite capacity, so you can pour any amount of water in each of them.
Calculate the maximum possible difference between the maximum and the minimum amount of water in the barrels, if you can pour water at most $k$ times.
Some examples: if you have four barrels, each containing $5$ liters of water, and $k = 1$, you may pour $5$ liters from the second barrel into the fourth, so the amounts of water in the barrels are $[5, 0, 5, 10]$, and the difference between the maximum and the minimum is $10$; if all barrels are empty, you can't make any operation, so the difference between the maximum and the minimum amount is still $0$.
-----Input-----
The first line contains one integer $t$ ($1 \le t \le 1000$) — the number of test cases.
The first line of each test case contains two integers $n$ and $k$ ($1 \le k < n \le 2 \cdot 10^5$) — the number of barrels and the number of pourings you can make.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($0 \le a_i \le 10^{9}$), where $a_i$ is the initial amount of water the $i$-th barrel has.
It's guaranteed that the total sum of $n$ over test cases doesn't exceed $2 \cdot 10^5$.
-----Output-----
For each test case, print the maximum possible difference between the maximum and the minimum amount of water in the barrels, if you can pour water at most $k$ times.
-----Example-----
Input
2
4 1
5 5 5 5
3 2
0 0 0
Output
10
0
|
t=int(input())
for you in range(t):
l=input().split()
n=int(l[0])
k=int(l[1])
l=input().split()
li=[int(i) for i in l]
if(k==0):
print(max(li)-min(li))
continue
z=0
li.sort()
li.reverse()
for i in range(k+1):
z+=li[i]
print(z)
|
You have $n$ barrels lined up in a row, numbered from left to right from one. Initially, the $i$-th barrel contains $a_i$ liters of water.
You can pour water from one barrel to another. In one act of pouring, you can choose two different barrels $x$ and $y$ (the $x$-th barrel shouldn't be empty) and pour any possible amount of water from barrel $x$ to barrel $y$ (possibly, all water). You may assume that barrels have infinite capacity, so you can pour any amount of water in each of them.
Calculate the maximum possible difference between the maximum and the minimum amount of water in the barrels, if you can pour water at most $k$ times.
Some examples: if you have four barrels, each containing $5$ liters of water, and $k = 1$, you may pour $5$ liters from the second barrel into the fourth, so the amounts of water in the barrels are $[5, 0, 5, 10]$, and the difference between the maximum and the minimum is $10$; if all barrels are empty, you can't make any operation, so the difference between the maximum and the minimum amount is still $0$.
-----Input-----
The first line contains one integer $t$ ($1 \le t \le 1000$) — the number of test cases.
The first line of each test case contains two integers $n$ and $k$ ($1 \le k < n \le 2 \cdot 10^5$) — the number of barrels and the number of pourings you can make.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($0 \le a_i \le 10^{9}$), where $a_i$ is the initial amount of water the $i$-th barrel has.
It's guaranteed that the total sum of $n$ over test cases doesn't exceed $2 \cdot 10^5$.
-----Output-----
For each test case, print the maximum possible difference between the maximum and the minimum amount of water in the barrels, if you can pour water at most $k$ times.
-----Example-----
Input
2
4 1
5 5 5 5
3 2
0 0 0
Output
10
0
|
for _ in range (int(input())):
n,k=map(int,input().split())
a=list(map(int,input().split()))
a.sort(reverse=True)
for i in range (1,k+1):
a[0]+=a[i]
a[i]=0
print(a[0]-a[1])
|
You have $n$ barrels lined up in a row, numbered from left to right from one. Initially, the $i$-th barrel contains $a_i$ liters of water.
You can pour water from one barrel to another. In one act of pouring, you can choose two different barrels $x$ and $y$ (the $x$-th barrel shouldn't be empty) and pour any possible amount of water from barrel $x$ to barrel $y$ (possibly, all water). You may assume that barrels have infinite capacity, so you can pour any amount of water in each of them.
Calculate the maximum possible difference between the maximum and the minimum amount of water in the barrels, if you can pour water at most $k$ times.
Some examples: if you have four barrels, each containing $5$ liters of water, and $k = 1$, you may pour $5$ liters from the second barrel into the fourth, so the amounts of water in the barrels are $[5, 0, 5, 10]$, and the difference between the maximum and the minimum is $10$; if all barrels are empty, you can't make any operation, so the difference between the maximum and the minimum amount is still $0$.
-----Input-----
The first line contains one integer $t$ ($1 \le t \le 1000$) — the number of test cases.
The first line of each test case contains two integers $n$ and $k$ ($1 \le k < n \le 2 \cdot 10^5$) — the number of barrels and the number of pourings you can make.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($0 \le a_i \le 10^{9}$), where $a_i$ is the initial amount of water the $i$-th barrel has.
It's guaranteed that the total sum of $n$ over test cases doesn't exceed $2 \cdot 10^5$.
-----Output-----
For each test case, print the maximum possible difference between the maximum and the minimum amount of water in the barrels, if you can pour water at most $k$ times.
-----Example-----
Input
2
4 1
5 5 5 5
3 2
0 0 0
Output
10
0
|
for __ in range(int(input())):
n, k = list(map(int, input().split()))
ar = list(map(int, input().split()))
ar.sort(reverse=True)
ans = 0
for i in range(min(n, k + 1)):
ans += ar[i]
print(ans)
|
You have $n$ barrels lined up in a row, numbered from left to right from one. Initially, the $i$-th barrel contains $a_i$ liters of water.
You can pour water from one barrel to another. In one act of pouring, you can choose two different barrels $x$ and $y$ (the $x$-th barrel shouldn't be empty) and pour any possible amount of water from barrel $x$ to barrel $y$ (possibly, all water). You may assume that barrels have infinite capacity, so you can pour any amount of water in each of them.
Calculate the maximum possible difference between the maximum and the minimum amount of water in the barrels, if you can pour water at most $k$ times.
Some examples: if you have four barrels, each containing $5$ liters of water, and $k = 1$, you may pour $5$ liters from the second barrel into the fourth, so the amounts of water in the barrels are $[5, 0, 5, 10]$, and the difference between the maximum and the minimum is $10$; if all barrels are empty, you can't make any operation, so the difference between the maximum and the minimum amount is still $0$.
-----Input-----
The first line contains one integer $t$ ($1 \le t \le 1000$) — the number of test cases.
The first line of each test case contains two integers $n$ and $k$ ($1 \le k < n \le 2 \cdot 10^5$) — the number of barrels and the number of pourings you can make.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($0 \le a_i \le 10^{9}$), where $a_i$ is the initial amount of water the $i$-th barrel has.
It's guaranteed that the total sum of $n$ over test cases doesn't exceed $2 \cdot 10^5$.
-----Output-----
For each test case, print the maximum possible difference between the maximum and the minimum amount of water in the barrels, if you can pour water at most $k$ times.
-----Example-----
Input
2
4 1
5 5 5 5
3 2
0 0 0
Output
10
0
|
import sys, math
import io, os
#data = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
from bisect import bisect_left as bl, bisect_right as br, insort
from heapq import heapify, heappush, heappop
from collections import defaultdict as dd, deque, Counter
#from itertools import permutations,combinations
def data(): return sys.stdin.readline().strip()
def mdata(): return list(map(int, data().split()))
def outl(var) : sys.stdout.write('\n'.join(map(str, var))+'\n')
def out(var) : sys.stdout.write(str(var)+'\n')
#from decimal import Decimal
#from fractions import Fraction
#sys.setrecursionlimit(100000)
INF = float('inf')
mod=10**9+7
for t in range(int(data())):
n,k=mdata()
a=sorted(mdata(),reverse=True)
s=sum(a[:k+1])
out(s)
|
You have $n$ barrels lined up in a row, numbered from left to right from one. Initially, the $i$-th barrel contains $a_i$ liters of water.
You can pour water from one barrel to another. In one act of pouring, you can choose two different barrels $x$ and $y$ (the $x$-th barrel shouldn't be empty) and pour any possible amount of water from barrel $x$ to barrel $y$ (possibly, all water). You may assume that barrels have infinite capacity, so you can pour any amount of water in each of them.
Calculate the maximum possible difference between the maximum and the minimum amount of water in the barrels, if you can pour water at most $k$ times.
Some examples: if you have four barrels, each containing $5$ liters of water, and $k = 1$, you may pour $5$ liters from the second barrel into the fourth, so the amounts of water in the barrels are $[5, 0, 5, 10]$, and the difference between the maximum and the minimum is $10$; if all barrels are empty, you can't make any operation, so the difference between the maximum and the minimum amount is still $0$.
-----Input-----
The first line contains one integer $t$ ($1 \le t \le 1000$) — the number of test cases.
The first line of each test case contains two integers $n$ and $k$ ($1 \le k < n \le 2 \cdot 10^5$) — the number of barrels and the number of pourings you can make.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($0 \le a_i \le 10^{9}$), where $a_i$ is the initial amount of water the $i$-th barrel has.
It's guaranteed that the total sum of $n$ over test cases doesn't exceed $2 \cdot 10^5$.
-----Output-----
For each test case, print the maximum possible difference between the maximum and the minimum amount of water in the barrels, if you can pour water at most $k$ times.
-----Example-----
Input
2
4 1
5 5 5 5
3 2
0 0 0
Output
10
0
|
import sys
input = sys.stdin.readline
t = int(input())
for i in range(t):
n,k = map(int,input().split())
a = list(map(int,input().split()))
a.sort()
a.reverse()
cum = [a[0]]
for i in range(n-1):
cum.append(cum[i]+a[i+1])
cum.append(cum[-1])
print(cum[k])
|
You have $n$ barrels lined up in a row, numbered from left to right from one. Initially, the $i$-th barrel contains $a_i$ liters of water.
You can pour water from one barrel to another. In one act of pouring, you can choose two different barrels $x$ and $y$ (the $x$-th barrel shouldn't be empty) and pour any possible amount of water from barrel $x$ to barrel $y$ (possibly, all water). You may assume that barrels have infinite capacity, so you can pour any amount of water in each of them.
Calculate the maximum possible difference between the maximum and the minimum amount of water in the barrels, if you can pour water at most $k$ times.
Some examples: if you have four barrels, each containing $5$ liters of water, and $k = 1$, you may pour $5$ liters from the second barrel into the fourth, so the amounts of water in the barrels are $[5, 0, 5, 10]$, and the difference between the maximum and the minimum is $10$; if all barrels are empty, you can't make any operation, so the difference between the maximum and the minimum amount is still $0$.
-----Input-----
The first line contains one integer $t$ ($1 \le t \le 1000$) — the number of test cases.
The first line of each test case contains two integers $n$ and $k$ ($1 \le k < n \le 2 \cdot 10^5$) — the number of barrels and the number of pourings you can make.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($0 \le a_i \le 10^{9}$), where $a_i$ is the initial amount of water the $i$-th barrel has.
It's guaranteed that the total sum of $n$ over test cases doesn't exceed $2 \cdot 10^5$.
-----Output-----
For each test case, print the maximum possible difference between the maximum and the minimum amount of water in the barrels, if you can pour water at most $k$ times.
-----Example-----
Input
2
4 1
5 5 5 5
3 2
0 0 0
Output
10
0
|
t = int(input())
for _ in range(t):
#n = int(input())
n, k=map(int, input().split())
a = list(map(int, input().split()))
a.sort()
s=0
for i in range(k+1):
s+=a[n-1-i]
print(s)
|
You have $n$ barrels lined up in a row, numbered from left to right from one. Initially, the $i$-th barrel contains $a_i$ liters of water.
You can pour water from one barrel to another. In one act of pouring, you can choose two different barrels $x$ and $y$ (the $x$-th barrel shouldn't be empty) and pour any possible amount of water from barrel $x$ to barrel $y$ (possibly, all water). You may assume that barrels have infinite capacity, so you can pour any amount of water in each of them.
Calculate the maximum possible difference between the maximum and the minimum amount of water in the barrels, if you can pour water at most $k$ times.
Some examples: if you have four barrels, each containing $5$ liters of water, and $k = 1$, you may pour $5$ liters from the second barrel into the fourth, so the amounts of water in the barrels are $[5, 0, 5, 10]$, and the difference between the maximum and the minimum is $10$; if all barrels are empty, you can't make any operation, so the difference between the maximum and the minimum amount is still $0$.
-----Input-----
The first line contains one integer $t$ ($1 \le t \le 1000$) — the number of test cases.
The first line of each test case contains two integers $n$ and $k$ ($1 \le k < n \le 2 \cdot 10^5$) — the number of barrels and the number of pourings you can make.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($0 \le a_i \le 10^{9}$), where $a_i$ is the initial amount of water the $i$-th barrel has.
It's guaranteed that the total sum of $n$ over test cases doesn't exceed $2 \cdot 10^5$.
-----Output-----
For each test case, print the maximum possible difference between the maximum and the minimum amount of water in the barrels, if you can pour water at most $k$ times.
-----Example-----
Input
2
4 1
5 5 5 5
3 2
0 0 0
Output
10
0
|
def main():
N, K = list(map(int, input().split()))
*A, = list(map(int, input().split()))
A.sort()
print(A[-1] + sum(A[-K-1:-1]))
def __starting_point():
for __ in [0]*int(input()):
main()
__starting_point()
|
You have $n$ barrels lined up in a row, numbered from left to right from one. Initially, the $i$-th barrel contains $a_i$ liters of water.
You can pour water from one barrel to another. In one act of pouring, you can choose two different barrels $x$ and $y$ (the $x$-th barrel shouldn't be empty) and pour any possible amount of water from barrel $x$ to barrel $y$ (possibly, all water). You may assume that barrels have infinite capacity, so you can pour any amount of water in each of them.
Calculate the maximum possible difference between the maximum and the minimum amount of water in the barrels, if you can pour water at most $k$ times.
Some examples: if you have four barrels, each containing $5$ liters of water, and $k = 1$, you may pour $5$ liters from the second barrel into the fourth, so the amounts of water in the barrels are $[5, 0, 5, 10]$, and the difference between the maximum and the minimum is $10$; if all barrels are empty, you can't make any operation, so the difference between the maximum and the minimum amount is still $0$.
-----Input-----
The first line contains one integer $t$ ($1 \le t \le 1000$) — the number of test cases.
The first line of each test case contains two integers $n$ and $k$ ($1 \le k < n \le 2 \cdot 10^5$) — the number of barrels and the number of pourings you can make.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($0 \le a_i \le 10^{9}$), where $a_i$ is the initial amount of water the $i$-th barrel has.
It's guaranteed that the total sum of $n$ over test cases doesn't exceed $2 \cdot 10^5$.
-----Output-----
For each test case, print the maximum possible difference between the maximum and the minimum amount of water in the barrels, if you can pour water at most $k$ times.
-----Example-----
Input
2
4 1
5 5 5 5
3 2
0 0 0
Output
10
0
|
import sys
import random
# import numpy as np
import math
import copy
from heapq import heappush, heappop, heapify
from functools import cmp_to_key
from bisect import bisect_left, bisect_right
from collections import defaultdict, deque, Counter
# sys.setrecursionlimit(1000000)
# input aliases
input = sys.stdin.readline
getS = lambda: input().strip()
getN = lambda: int(input())
getList = lambda: list(map(int, input().split()))
getZList = lambda: [int(x) - 1 for x in input().split()]
INF = float("inf")
MOD = 10 ** 9 + 7
divide = lambda x: pow(x, MOD-2, MOD)
def judge(at, ax, ay, bt, bx, by):
if abs(at - bt) >= abs(ax - bx) + abs(ay - by):
return True
else:
return False
def solve():
n, k = getList()
li = getList()
if k >= n:
print(sum(li))
return
li.sort(reverse=True)
print(sum(li[:k+1]))
return
def main():
n = getN()
for _ in range(n):
solve()
return
def __starting_point():
main()
# solve()
__starting_point()
|
You have $n$ barrels lined up in a row, numbered from left to right from one. Initially, the $i$-th barrel contains $a_i$ liters of water.
You can pour water from one barrel to another. In one act of pouring, you can choose two different barrels $x$ and $y$ (the $x$-th barrel shouldn't be empty) and pour any possible amount of water from barrel $x$ to barrel $y$ (possibly, all water). You may assume that barrels have infinite capacity, so you can pour any amount of water in each of them.
Calculate the maximum possible difference between the maximum and the minimum amount of water in the barrels, if you can pour water at most $k$ times.
Some examples: if you have four barrels, each containing $5$ liters of water, and $k = 1$, you may pour $5$ liters from the second barrel into the fourth, so the amounts of water in the barrels are $[5, 0, 5, 10]$, and the difference between the maximum and the minimum is $10$; if all barrels are empty, you can't make any operation, so the difference between the maximum and the minimum amount is still $0$.
-----Input-----
The first line contains one integer $t$ ($1 \le t \le 1000$) — the number of test cases.
The first line of each test case contains two integers $n$ and $k$ ($1 \le k < n \le 2 \cdot 10^5$) — the number of barrels and the number of pourings you can make.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($0 \le a_i \le 10^{9}$), where $a_i$ is the initial amount of water the $i$-th barrel has.
It's guaranteed that the total sum of $n$ over test cases doesn't exceed $2 \cdot 10^5$.
-----Output-----
For each test case, print the maximum possible difference between the maximum and the minimum amount of water in the barrels, if you can pour water at most $k$ times.
-----Example-----
Input
2
4 1
5 5 5 5
3 2
0 0 0
Output
10
0
|
from sys import stdin
t = int(stdin.readline())
for _ in range(t):
n, k = tuple(int(x) for x in stdin.readline().split())
lst = sorted(int(x) for x in stdin.readline().split())
print(sum(lst[-k-1:]))
|
You have $n$ barrels lined up in a row, numbered from left to right from one. Initially, the $i$-th barrel contains $a_i$ liters of water.
You can pour water from one barrel to another. In one act of pouring, you can choose two different barrels $x$ and $y$ (the $x$-th barrel shouldn't be empty) and pour any possible amount of water from barrel $x$ to barrel $y$ (possibly, all water). You may assume that barrels have infinite capacity, so you can pour any amount of water in each of them.
Calculate the maximum possible difference between the maximum and the minimum amount of water in the barrels, if you can pour water at most $k$ times.
Some examples: if you have four barrels, each containing $5$ liters of water, and $k = 1$, you may pour $5$ liters from the second barrel into the fourth, so the amounts of water in the barrels are $[5, 0, 5, 10]$, and the difference between the maximum and the minimum is $10$; if all barrels are empty, you can't make any operation, so the difference between the maximum and the minimum amount is still $0$.
-----Input-----
The first line contains one integer $t$ ($1 \le t \le 1000$) — the number of test cases.
The first line of each test case contains two integers $n$ and $k$ ($1 \le k < n \le 2 \cdot 10^5$) — the number of barrels and the number of pourings you can make.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($0 \le a_i \le 10^{9}$), where $a_i$ is the initial amount of water the $i$-th barrel has.
It's guaranteed that the total sum of $n$ over test cases doesn't exceed $2 \cdot 10^5$.
-----Output-----
For each test case, print the maximum possible difference between the maximum and the minimum amount of water in the barrels, if you can pour water at most $k$ times.
-----Example-----
Input
2
4 1
5 5 5 5
3 2
0 0 0
Output
10
0
|
t = int(input())
for _ in range(t):
n,k = [int(x) for x in input().split()]
l = [int(x) for x in input().split()]
l.sort()
l.reverse()
print(sum(l[:min(k+1,n)]))
|
You have $n$ barrels lined up in a row, numbered from left to right from one. Initially, the $i$-th barrel contains $a_i$ liters of water.
You can pour water from one barrel to another. In one act of pouring, you can choose two different barrels $x$ and $y$ (the $x$-th barrel shouldn't be empty) and pour any possible amount of water from barrel $x$ to barrel $y$ (possibly, all water). You may assume that barrels have infinite capacity, so you can pour any amount of water in each of them.
Calculate the maximum possible difference between the maximum and the minimum amount of water in the barrels, if you can pour water at most $k$ times.
Some examples: if you have four barrels, each containing $5$ liters of water, and $k = 1$, you may pour $5$ liters from the second barrel into the fourth, so the amounts of water in the barrels are $[5, 0, 5, 10]$, and the difference between the maximum and the minimum is $10$; if all barrels are empty, you can't make any operation, so the difference between the maximum and the minimum amount is still $0$.
-----Input-----
The first line contains one integer $t$ ($1 \le t \le 1000$) — the number of test cases.
The first line of each test case contains two integers $n$ and $k$ ($1 \le k < n \le 2 \cdot 10^5$) — the number of barrels and the number of pourings you can make.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($0 \le a_i \le 10^{9}$), where $a_i$ is the initial amount of water the $i$-th barrel has.
It's guaranteed that the total sum of $n$ over test cases doesn't exceed $2 \cdot 10^5$.
-----Output-----
For each test case, print the maximum possible difference between the maximum and the minimum amount of water in the barrels, if you can pour water at most $k$ times.
-----Example-----
Input
2
4 1
5 5 5 5
3 2
0 0 0
Output
10
0
|
for _ in range(int(input())):
n, k = list(map(int, input().split()))
A = list(map(int, input().split()))
A.sort(reverse=True)
if k == 0:
print(max(A) - min(A))
else:
print(A[0] + sum(A[1:k+1]))
|
You have $n$ barrels lined up in a row, numbered from left to right from one. Initially, the $i$-th barrel contains $a_i$ liters of water.
You can pour water from one barrel to another. In one act of pouring, you can choose two different barrels $x$ and $y$ (the $x$-th barrel shouldn't be empty) and pour any possible amount of water from barrel $x$ to barrel $y$ (possibly, all water). You may assume that barrels have infinite capacity, so you can pour any amount of water in each of them.
Calculate the maximum possible difference between the maximum and the minimum amount of water in the barrels, if you can pour water at most $k$ times.
Some examples: if you have four barrels, each containing $5$ liters of water, and $k = 1$, you may pour $5$ liters from the second barrel into the fourth, so the amounts of water in the barrels are $[5, 0, 5, 10]$, and the difference between the maximum and the minimum is $10$; if all barrels are empty, you can't make any operation, so the difference between the maximum and the minimum amount is still $0$.
-----Input-----
The first line contains one integer $t$ ($1 \le t \le 1000$) — the number of test cases.
The first line of each test case contains two integers $n$ and $k$ ($1 \le k < n \le 2 \cdot 10^5$) — the number of barrels and the number of pourings you can make.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($0 \le a_i \le 10^{9}$), where $a_i$ is the initial amount of water the $i$-th barrel has.
It's guaranteed that the total sum of $n$ over test cases doesn't exceed $2 \cdot 10^5$.
-----Output-----
For each test case, print the maximum possible difference between the maximum and the minimum amount of water in the barrels, if you can pour water at most $k$ times.
-----Example-----
Input
2
4 1
5 5 5 5
3 2
0 0 0
Output
10
0
|
n = int(input())
for _ in range(n):
n, k = list(map(int, input().split()))
arr = list(map(int, input().split()))
arr.sort(reverse=True)
print(sum(arr[:k+1]))
|
You have $n$ barrels lined up in a row, numbered from left to right from one. Initially, the $i$-th barrel contains $a_i$ liters of water.
You can pour water from one barrel to another. In one act of pouring, you can choose two different barrels $x$ and $y$ (the $x$-th barrel shouldn't be empty) and pour any possible amount of water from barrel $x$ to barrel $y$ (possibly, all water). You may assume that barrels have infinite capacity, so you can pour any amount of water in each of them.
Calculate the maximum possible difference between the maximum and the minimum amount of water in the barrels, if you can pour water at most $k$ times.
Some examples: if you have four barrels, each containing $5$ liters of water, and $k = 1$, you may pour $5$ liters from the second barrel into the fourth, so the amounts of water in the barrels are $[5, 0, 5, 10]$, and the difference between the maximum and the minimum is $10$; if all barrels are empty, you can't make any operation, so the difference between the maximum and the minimum amount is still $0$.
-----Input-----
The first line contains one integer $t$ ($1 \le t \le 1000$) — the number of test cases.
The first line of each test case contains two integers $n$ and $k$ ($1 \le k < n \le 2 \cdot 10^5$) — the number of barrels and the number of pourings you can make.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($0 \le a_i \le 10^{9}$), where $a_i$ is the initial amount of water the $i$-th barrel has.
It's guaranteed that the total sum of $n$ over test cases doesn't exceed $2 \cdot 10^5$.
-----Output-----
For each test case, print the maximum possible difference between the maximum and the minimum amount of water in the barrels, if you can pour water at most $k$ times.
-----Example-----
Input
2
4 1
5 5 5 5
3 2
0 0 0
Output
10
0
|
"""T=int(input())
for _ in range(0,T):
n=int(input())
a,b=map(int,input().split())
s=input()
s=[int(x) for x in input().split()]
for i in range(0,len(s)):
a,b=map(int,input().split())"""
T=int(input())
for _ in range(0,T):
n,k=list(map(int,input().split()))
s=[int(x) for x in input().split()]
s.sort()
s=s[::-1]
for i in range(1,min(k+1,len(s))):
s[0]+=s[i]
print(s[0])
|
You have $n$ barrels lined up in a row, numbered from left to right from one. Initially, the $i$-th barrel contains $a_i$ liters of water.
You can pour water from one barrel to another. In one act of pouring, you can choose two different barrels $x$ and $y$ (the $x$-th barrel shouldn't be empty) and pour any possible amount of water from barrel $x$ to barrel $y$ (possibly, all water). You may assume that barrels have infinite capacity, so you can pour any amount of water in each of them.
Calculate the maximum possible difference between the maximum and the minimum amount of water in the barrels, if you can pour water at most $k$ times.
Some examples: if you have four barrels, each containing $5$ liters of water, and $k = 1$, you may pour $5$ liters from the second barrel into the fourth, so the amounts of water in the barrels are $[5, 0, 5, 10]$, and the difference between the maximum and the minimum is $10$; if all barrels are empty, you can't make any operation, so the difference between the maximum and the minimum amount is still $0$.
-----Input-----
The first line contains one integer $t$ ($1 \le t \le 1000$) — the number of test cases.
The first line of each test case contains two integers $n$ and $k$ ($1 \le k < n \le 2 \cdot 10^5$) — the number of barrels and the number of pourings you can make.
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($0 \le a_i \le 10^{9}$), where $a_i$ is the initial amount of water the $i$-th barrel has.
It's guaranteed that the total sum of $n$ over test cases doesn't exceed $2 \cdot 10^5$.
-----Output-----
For each test case, print the maximum possible difference between the maximum and the minimum amount of water in the barrels, if you can pour water at most $k$ times.
-----Example-----
Input
2
4 1
5 5 5 5
3 2
0 0 0
Output
10
0
|
t=int(input())
while t:
t-=1
n,k=list(map(int,input().split()))
a=[int(i) for i in input().split()]
a.sort()
ans=0
a.reverse()
for i in range(k+1):
ans+=a[i]
print(ans)
|
You are given a permutation $p=[p_1, p_2, \ldots, p_n]$ of integers from $1$ to $n$. Let's call the number $m$ ($1 \le m \le n$) beautiful, if there exists two indices $l, r$ ($1 \le l \le r \le n$), such that the numbers $[p_l, p_{l+1}, \ldots, p_r]$ is a permutation of numbers $1, 2, \ldots, m$.
For example, let $p = [4, 5, 1, 3, 2, 6]$. In this case, the numbers $1, 3, 5, 6$ are beautiful and $2, 4$ are not. It is because: if $l = 3$ and $r = 3$ we will have a permutation $[1]$ for $m = 1$; if $l = 3$ and $r = 5$ we will have a permutation $[1, 3, 2]$ for $m = 3$; if $l = 1$ and $r = 5$ we will have a permutation $[4, 5, 1, 3, 2]$ for $m = 5$; if $l = 1$ and $r = 6$ we will have a permutation $[4, 5, 1, 3, 2, 6]$ for $m = 6$; it is impossible to take some $l$ and $r$, such that $[p_l, p_{l+1}, \ldots, p_r]$ is a permutation of numbers $1, 2, \ldots, m$ for $m = 2$ and for $m = 4$.
You are given a permutation $p=[p_1, p_2, \ldots, p_n]$. For all $m$ ($1 \le m \le n$) determine if it is a beautiful number or not.
-----Input-----
The first line contains the only integer $t$ ($1 \le t \le 1000$) — the number of test cases in the input. The next lines contain the description of test cases.
The first line of a test case contains a number $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of the given permutation $p$. The next line contains $n$ integers $p_1, p_2, \ldots, p_n$ ($1 \le p_i \le n$, all $p_i$ are different) — the given permutation $p$.
It is guaranteed, that the sum of $n$ from all test cases in the input doesn't exceed $2 \cdot 10^5$.
-----Output-----
Print $t$ lines — the answers to test cases in the order they are given in the input.
The answer to a test case is the string of length $n$, there the $i$-th character is equal to $1$ if $i$ is a beautiful number and is equal to $0$ if $i$ is not a beautiful number.
-----Example-----
Input
3
6
4 5 1 3 2 6
5
5 3 1 2 4
4
1 4 3 2
Output
101011
11111
1001
-----Note-----
The first test case is described in the problem statement.
In the second test case all numbers from $1$ to $5$ are beautiful: if $l = 3$ and $r = 3$ we will have a permutation $[1]$ for $m = 1$; if $l = 3$ and $r = 4$ we will have a permutation $[1, 2]$ for $m = 2$; if $l = 2$ and $r = 4$ we will have a permutation $[3, 1, 2]$ for $m = 3$; if $l = 2$ and $r = 5$ we will have a permutation $[3, 1, 2, 4]$ for $m = 4$; if $l = 1$ and $r = 5$ we will have a permutation $[5, 3, 1, 2, 4]$ for $m = 5$.
|
for _ in range(int(input())):
input()
nums = [int(x) for x in input().split()]
new_ar = list(zip(nums,[i for i in range(len(nums))]))
new_ar.sort()
maxx = new_ar[0][1]
minn = new_ar[0][1]
s="1"
for j in range(1,len(new_ar)):
if(new_ar[j][1]>maxx):
maxx = new_ar[j][1]
if(new_ar[j][1]<minn):
minn = new_ar[j][1]
if(maxx-minn<j+1):
s+="1"
else:
s+="0"
print(s)
|
You are given a permutation $p=[p_1, p_2, \ldots, p_n]$ of integers from $1$ to $n$. Let's call the number $m$ ($1 \le m \le n$) beautiful, if there exists two indices $l, r$ ($1 \le l \le r \le n$), such that the numbers $[p_l, p_{l+1}, \ldots, p_r]$ is a permutation of numbers $1, 2, \ldots, m$.
For example, let $p = [4, 5, 1, 3, 2, 6]$. In this case, the numbers $1, 3, 5, 6$ are beautiful and $2, 4$ are not. It is because: if $l = 3$ and $r = 3$ we will have a permutation $[1]$ for $m = 1$; if $l = 3$ and $r = 5$ we will have a permutation $[1, 3, 2]$ for $m = 3$; if $l = 1$ and $r = 5$ we will have a permutation $[4, 5, 1, 3, 2]$ for $m = 5$; if $l = 1$ and $r = 6$ we will have a permutation $[4, 5, 1, 3, 2, 6]$ for $m = 6$; it is impossible to take some $l$ and $r$, such that $[p_l, p_{l+1}, \ldots, p_r]$ is a permutation of numbers $1, 2, \ldots, m$ for $m = 2$ and for $m = 4$.
You are given a permutation $p=[p_1, p_2, \ldots, p_n]$. For all $m$ ($1 \le m \le n$) determine if it is a beautiful number or not.
-----Input-----
The first line contains the only integer $t$ ($1 \le t \le 1000$) — the number of test cases in the input. The next lines contain the description of test cases.
The first line of a test case contains a number $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of the given permutation $p$. The next line contains $n$ integers $p_1, p_2, \ldots, p_n$ ($1 \le p_i \le n$, all $p_i$ are different) — the given permutation $p$.
It is guaranteed, that the sum of $n$ from all test cases in the input doesn't exceed $2 \cdot 10^5$.
-----Output-----
Print $t$ lines — the answers to test cases in the order they are given in the input.
The answer to a test case is the string of length $n$, there the $i$-th character is equal to $1$ if $i$ is a beautiful number and is equal to $0$ if $i$ is not a beautiful number.
-----Example-----
Input
3
6
4 5 1 3 2 6
5
5 3 1 2 4
4
1 4 3 2
Output
101011
11111
1001
-----Note-----
The first test case is described in the problem statement.
In the second test case all numbers from $1$ to $5$ are beautiful: if $l = 3$ and $r = 3$ we will have a permutation $[1]$ for $m = 1$; if $l = 3$ and $r = 4$ we will have a permutation $[1, 2]$ for $m = 2$; if $l = 2$ and $r = 4$ we will have a permutation $[3, 1, 2]$ for $m = 3$; if $l = 2$ and $r = 5$ we will have a permutation $[3, 1, 2, 4]$ for $m = 4$; if $l = 1$ and $r = 5$ we will have a permutation $[5, 3, 1, 2, 4]$ for $m = 5$.
|
import sys
def I():
return sys.stdin.readline().rstrip()
for _ in range(int(I())):
n = int(I())
l = list(map(int,I().split()))
r = list(range(n))
r.sort(key=lambda x: l[x])
mn, mx = None, None
for i in range(n):
if mn is None:
mn = mx = r[ i ]
else:
mn = min( mn, r[ i ] )
mx = max( mx, r[ i ] )
l[ i ] = '1' if mx - mn == i else '0'
print("".join(l))
|
You are given a permutation $p=[p_1, p_2, \ldots, p_n]$ of integers from $1$ to $n$. Let's call the number $m$ ($1 \le m \le n$) beautiful, if there exists two indices $l, r$ ($1 \le l \le r \le n$), such that the numbers $[p_l, p_{l+1}, \ldots, p_r]$ is a permutation of numbers $1, 2, \ldots, m$.
For example, let $p = [4, 5, 1, 3, 2, 6]$. In this case, the numbers $1, 3, 5, 6$ are beautiful and $2, 4$ are not. It is because: if $l = 3$ and $r = 3$ we will have a permutation $[1]$ for $m = 1$; if $l = 3$ and $r = 5$ we will have a permutation $[1, 3, 2]$ for $m = 3$; if $l = 1$ and $r = 5$ we will have a permutation $[4, 5, 1, 3, 2]$ for $m = 5$; if $l = 1$ and $r = 6$ we will have a permutation $[4, 5, 1, 3, 2, 6]$ for $m = 6$; it is impossible to take some $l$ and $r$, such that $[p_l, p_{l+1}, \ldots, p_r]$ is a permutation of numbers $1, 2, \ldots, m$ for $m = 2$ and for $m = 4$.
You are given a permutation $p=[p_1, p_2, \ldots, p_n]$. For all $m$ ($1 \le m \le n$) determine if it is a beautiful number or not.
-----Input-----
The first line contains the only integer $t$ ($1 \le t \le 1000$) — the number of test cases in the input. The next lines contain the description of test cases.
The first line of a test case contains a number $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of the given permutation $p$. The next line contains $n$ integers $p_1, p_2, \ldots, p_n$ ($1 \le p_i \le n$, all $p_i$ are different) — the given permutation $p$.
It is guaranteed, that the sum of $n$ from all test cases in the input doesn't exceed $2 \cdot 10^5$.
-----Output-----
Print $t$ lines — the answers to test cases in the order they are given in the input.
The answer to a test case is the string of length $n$, there the $i$-th character is equal to $1$ if $i$ is a beautiful number and is equal to $0$ if $i$ is not a beautiful number.
-----Example-----
Input
3
6
4 5 1 3 2 6
5
5 3 1 2 4
4
1 4 3 2
Output
101011
11111
1001
-----Note-----
The first test case is described in the problem statement.
In the second test case all numbers from $1$ to $5$ are beautiful: if $l = 3$ and $r = 3$ we will have a permutation $[1]$ for $m = 1$; if $l = 3$ and $r = 4$ we will have a permutation $[1, 2]$ for $m = 2$; if $l = 2$ and $r = 4$ we will have a permutation $[3, 1, 2]$ for $m = 3$; if $l = 2$ and $r = 5$ we will have a permutation $[3, 1, 2, 4]$ for $m = 4$; if $l = 1$ and $r = 5$ we will have a permutation $[5, 3, 1, 2, 4]$ for $m = 5$.
|
from sys import stdin
def rl():
return [int(w) for w in stdin.readline().split()]
k, = rl()
for _ in range(k):
n, = rl()
p = rl()
q = [0] * n
for i, x in enumerate(p):
q[x-1] = i
l = r = q[0]
m = []
for k, i in enumerate(q):
if i < l:
l = i
elif i > r:
r = i
m.append('1' if r - l == k else '0')
print(''.join(m))
|
You are given a permutation $p=[p_1, p_2, \ldots, p_n]$ of integers from $1$ to $n$. Let's call the number $m$ ($1 \le m \le n$) beautiful, if there exists two indices $l, r$ ($1 \le l \le r \le n$), such that the numbers $[p_l, p_{l+1}, \ldots, p_r]$ is a permutation of numbers $1, 2, \ldots, m$.
For example, let $p = [4, 5, 1, 3, 2, 6]$. In this case, the numbers $1, 3, 5, 6$ are beautiful and $2, 4$ are not. It is because: if $l = 3$ and $r = 3$ we will have a permutation $[1]$ for $m = 1$; if $l = 3$ and $r = 5$ we will have a permutation $[1, 3, 2]$ for $m = 3$; if $l = 1$ and $r = 5$ we will have a permutation $[4, 5, 1, 3, 2]$ for $m = 5$; if $l = 1$ and $r = 6$ we will have a permutation $[4, 5, 1, 3, 2, 6]$ for $m = 6$; it is impossible to take some $l$ and $r$, such that $[p_l, p_{l+1}, \ldots, p_r]$ is a permutation of numbers $1, 2, \ldots, m$ for $m = 2$ and for $m = 4$.
You are given a permutation $p=[p_1, p_2, \ldots, p_n]$. For all $m$ ($1 \le m \le n$) determine if it is a beautiful number or not.
-----Input-----
The first line contains the only integer $t$ ($1 \le t \le 1000$) — the number of test cases in the input. The next lines contain the description of test cases.
The first line of a test case contains a number $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of the given permutation $p$. The next line contains $n$ integers $p_1, p_2, \ldots, p_n$ ($1 \le p_i \le n$, all $p_i$ are different) — the given permutation $p$.
It is guaranteed, that the sum of $n$ from all test cases in the input doesn't exceed $2 \cdot 10^5$.
-----Output-----
Print $t$ lines — the answers to test cases in the order they are given in the input.
The answer to a test case is the string of length $n$, there the $i$-th character is equal to $1$ if $i$ is a beautiful number and is equal to $0$ if $i$ is not a beautiful number.
-----Example-----
Input
3
6
4 5 1 3 2 6
5
5 3 1 2 4
4
1 4 3 2
Output
101011
11111
1001
-----Note-----
The first test case is described in the problem statement.
In the second test case all numbers from $1$ to $5$ are beautiful: if $l = 3$ and $r = 3$ we will have a permutation $[1]$ for $m = 1$; if $l = 3$ and $r = 4$ we will have a permutation $[1, 2]$ for $m = 2$; if $l = 2$ and $r = 4$ we will have a permutation $[3, 1, 2]$ for $m = 3$; if $l = 2$ and $r = 5$ we will have a permutation $[3, 1, 2, 4]$ for $m = 4$; if $l = 1$ and $r = 5$ we will have a permutation $[5, 3, 1, 2, 4]$ for $m = 5$.
|
# @author
import sys
class BBeautifulNumbers:
def solve(self):
for _ in range(int(input())):
n = int(input())
p = [int(_) - 1 for _ in input().split()]
mn_index = [float('inf')] * n
mx_index = [-float('inf')] * n
prev = [0] * n
for i in range(n):
prev[p[i]] = i
# print(prev)
for i in range(n):
mn_index[i] = min(mn_index[i - 1], prev[i])
mx_index[i] = max(mx_index[i - 1], prev[i])
ans = ['0'] * n
# print(mn_index, mx_index)
for i in range(n):
l, r = mn_index[i], mx_index[i]
ans[i] = '1' if r - l + 1 == i + 1 else '0'
print(''.join(ans))
solver = BBeautifulNumbers()
input = sys.stdin.readline
solver.solve()
|
You are given a permutation $p=[p_1, p_2, \ldots, p_n]$ of integers from $1$ to $n$. Let's call the number $m$ ($1 \le m \le n$) beautiful, if there exists two indices $l, r$ ($1 \le l \le r \le n$), such that the numbers $[p_l, p_{l+1}, \ldots, p_r]$ is a permutation of numbers $1, 2, \ldots, m$.
For example, let $p = [4, 5, 1, 3, 2, 6]$. In this case, the numbers $1, 3, 5, 6$ are beautiful and $2, 4$ are not. It is because: if $l = 3$ and $r = 3$ we will have a permutation $[1]$ for $m = 1$; if $l = 3$ and $r = 5$ we will have a permutation $[1, 3, 2]$ for $m = 3$; if $l = 1$ and $r = 5$ we will have a permutation $[4, 5, 1, 3, 2]$ for $m = 5$; if $l = 1$ and $r = 6$ we will have a permutation $[4, 5, 1, 3, 2, 6]$ for $m = 6$; it is impossible to take some $l$ and $r$, such that $[p_l, p_{l+1}, \ldots, p_r]$ is a permutation of numbers $1, 2, \ldots, m$ for $m = 2$ and for $m = 4$.
You are given a permutation $p=[p_1, p_2, \ldots, p_n]$. For all $m$ ($1 \le m \le n$) determine if it is a beautiful number or not.
-----Input-----
The first line contains the only integer $t$ ($1 \le t \le 1000$) — the number of test cases in the input. The next lines contain the description of test cases.
The first line of a test case contains a number $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of the given permutation $p$. The next line contains $n$ integers $p_1, p_2, \ldots, p_n$ ($1 \le p_i \le n$, all $p_i$ are different) — the given permutation $p$.
It is guaranteed, that the sum of $n$ from all test cases in the input doesn't exceed $2 \cdot 10^5$.
-----Output-----
Print $t$ lines — the answers to test cases in the order they are given in the input.
The answer to a test case is the string of length $n$, there the $i$-th character is equal to $1$ if $i$ is a beautiful number and is equal to $0$ if $i$ is not a beautiful number.
-----Example-----
Input
3
6
4 5 1 3 2 6
5
5 3 1 2 4
4
1 4 3 2
Output
101011
11111
1001
-----Note-----
The first test case is described in the problem statement.
In the second test case all numbers from $1$ to $5$ are beautiful: if $l = 3$ and $r = 3$ we will have a permutation $[1]$ for $m = 1$; if $l = 3$ and $r = 4$ we will have a permutation $[1, 2]$ for $m = 2$; if $l = 2$ and $r = 4$ we will have a permutation $[3, 1, 2]$ for $m = 3$; if $l = 2$ and $r = 5$ we will have a permutation $[3, 1, 2, 4]$ for $m = 4$; if $l = 1$ and $r = 5$ we will have a permutation $[5, 3, 1, 2, 4]$ for $m = 5$.
|
def f(L):
n=len(L)
M=[0]*(len(L)+1)
for i in range(len(L)):
M[L[i]]=i
s=[0]*len(L)
s[0]=1
sumof=M[1]
mx=M[1]
mi=M[1]
for i in range(2,n):
k=M[i]
if k>mx:mx=k
if k<mi:mi=k
sumof+=k
if sumof==(mx*(mx+1))//2-((mi-1)*mi)//2:
s[i-1]=1
s[n-1]=1
return s
for i in ' '*int(input()):
n=int(input())
s=f(list(map(int,input().split())))
for i in s:print(i,end='')
print()
|
You are given a permutation $p=[p_1, p_2, \ldots, p_n]$ of integers from $1$ to $n$. Let's call the number $m$ ($1 \le m \le n$) beautiful, if there exists two indices $l, r$ ($1 \le l \le r \le n$), such that the numbers $[p_l, p_{l+1}, \ldots, p_r]$ is a permutation of numbers $1, 2, \ldots, m$.
For example, let $p = [4, 5, 1, 3, 2, 6]$. In this case, the numbers $1, 3, 5, 6$ are beautiful and $2, 4$ are not. It is because: if $l = 3$ and $r = 3$ we will have a permutation $[1]$ for $m = 1$; if $l = 3$ and $r = 5$ we will have a permutation $[1, 3, 2]$ for $m = 3$; if $l = 1$ and $r = 5$ we will have a permutation $[4, 5, 1, 3, 2]$ for $m = 5$; if $l = 1$ and $r = 6$ we will have a permutation $[4, 5, 1, 3, 2, 6]$ for $m = 6$; it is impossible to take some $l$ and $r$, such that $[p_l, p_{l+1}, \ldots, p_r]$ is a permutation of numbers $1, 2, \ldots, m$ for $m = 2$ and for $m = 4$.
You are given a permutation $p=[p_1, p_2, \ldots, p_n]$. For all $m$ ($1 \le m \le n$) determine if it is a beautiful number or not.
-----Input-----
The first line contains the only integer $t$ ($1 \le t \le 1000$) — the number of test cases in the input. The next lines contain the description of test cases.
The first line of a test case contains a number $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of the given permutation $p$. The next line contains $n$ integers $p_1, p_2, \ldots, p_n$ ($1 \le p_i \le n$, all $p_i$ are different) — the given permutation $p$.
It is guaranteed, that the sum of $n$ from all test cases in the input doesn't exceed $2 \cdot 10^5$.
-----Output-----
Print $t$ lines — the answers to test cases in the order they are given in the input.
The answer to a test case is the string of length $n$, there the $i$-th character is equal to $1$ if $i$ is a beautiful number and is equal to $0$ if $i$ is not a beautiful number.
-----Example-----
Input
3
6
4 5 1 3 2 6
5
5 3 1 2 4
4
1 4 3 2
Output
101011
11111
1001
-----Note-----
The first test case is described in the problem statement.
In the second test case all numbers from $1$ to $5$ are beautiful: if $l = 3$ and $r = 3$ we will have a permutation $[1]$ for $m = 1$; if $l = 3$ and $r = 4$ we will have a permutation $[1, 2]$ for $m = 2$; if $l = 2$ and $r = 4$ we will have a permutation $[3, 1, 2]$ for $m = 3$; if $l = 2$ and $r = 5$ we will have a permutation $[3, 1, 2, 4]$ for $m = 4$; if $l = 1$ and $r = 5$ we will have a permutation $[5, 3, 1, 2, 4]$ for $m = 5$.
|
t=int(input())
for _ in range(t):
n=int(input())
a=list(map(int,input().split()))
pos=[0 for i in range(n+1)]
for i in range(n):
pos[a[i]]=i
ans=[-1 for i in range(n)]
ans[0]=1
l,r=pos[1],pos[1]
for i in range(2,n+1):
l=min(l,pos[i])
r=max(r,pos[i])
if r-l==i-1:
ans[i-1]=1
else:
ans[i-1]=0
print("".join(map(str,ans)))
|
You are given a permutation $p=[p_1, p_2, \ldots, p_n]$ of integers from $1$ to $n$. Let's call the number $m$ ($1 \le m \le n$) beautiful, if there exists two indices $l, r$ ($1 \le l \le r \le n$), such that the numbers $[p_l, p_{l+1}, \ldots, p_r]$ is a permutation of numbers $1, 2, \ldots, m$.
For example, let $p = [4, 5, 1, 3, 2, 6]$. In this case, the numbers $1, 3, 5, 6$ are beautiful and $2, 4$ are not. It is because: if $l = 3$ and $r = 3$ we will have a permutation $[1]$ for $m = 1$; if $l = 3$ and $r = 5$ we will have a permutation $[1, 3, 2]$ for $m = 3$; if $l = 1$ and $r = 5$ we will have a permutation $[4, 5, 1, 3, 2]$ for $m = 5$; if $l = 1$ and $r = 6$ we will have a permutation $[4, 5, 1, 3, 2, 6]$ for $m = 6$; it is impossible to take some $l$ and $r$, such that $[p_l, p_{l+1}, \ldots, p_r]$ is a permutation of numbers $1, 2, \ldots, m$ for $m = 2$ and for $m = 4$.
You are given a permutation $p=[p_1, p_2, \ldots, p_n]$. For all $m$ ($1 \le m \le n$) determine if it is a beautiful number or not.
-----Input-----
The first line contains the only integer $t$ ($1 \le t \le 1000$) — the number of test cases in the input. The next lines contain the description of test cases.
The first line of a test case contains a number $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of the given permutation $p$. The next line contains $n$ integers $p_1, p_2, \ldots, p_n$ ($1 \le p_i \le n$, all $p_i$ are different) — the given permutation $p$.
It is guaranteed, that the sum of $n$ from all test cases in the input doesn't exceed $2 \cdot 10^5$.
-----Output-----
Print $t$ lines — the answers to test cases in the order they are given in the input.
The answer to a test case is the string of length $n$, there the $i$-th character is equal to $1$ if $i$ is a beautiful number and is equal to $0$ if $i$ is not a beautiful number.
-----Example-----
Input
3
6
4 5 1 3 2 6
5
5 3 1 2 4
4
1 4 3 2
Output
101011
11111
1001
-----Note-----
The first test case is described in the problem statement.
In the second test case all numbers from $1$ to $5$ are beautiful: if $l = 3$ and $r = 3$ we will have a permutation $[1]$ for $m = 1$; if $l = 3$ and $r = 4$ we will have a permutation $[1, 2]$ for $m = 2$; if $l = 2$ and $r = 4$ we will have a permutation $[3, 1, 2]$ for $m = 3$; if $l = 2$ and $r = 5$ we will have a permutation $[3, 1, 2, 4]$ for $m = 4$; if $l = 1$ and $r = 5$ we will have a permutation $[5, 3, 1, 2, 4]$ for $m = 5$.
|
t = int(input())
for t_i in range(t):
n = int(input())
P = input().split()
l, r = -1, -1
for i in range(n):
P[i] = int(P[i])
if P[i] == 1:
l = i
r = i
max_seen = 1
beaut = ['1']
for _ in range(n - 1):
if l == 0:
l_cand = 10**8
else:
l_cand = P[l - 1]
if r == n - 1:
r_cand = 10**8
else:
r_cand = P[r + 1]
if r_cand > l_cand:
l -= 1
max_seen = max(l_cand, max_seen)
else:
r += 1
max_seen = max(r_cand, max_seen)
beaut.append('1' if max_seen == r - l + 1 else '0')
print(''.join(beaut))
|
You are given a permutation $p=[p_1, p_2, \ldots, p_n]$ of integers from $1$ to $n$. Let's call the number $m$ ($1 \le m \le n$) beautiful, if there exists two indices $l, r$ ($1 \le l \le r \le n$), such that the numbers $[p_l, p_{l+1}, \ldots, p_r]$ is a permutation of numbers $1, 2, \ldots, m$.
For example, let $p = [4, 5, 1, 3, 2, 6]$. In this case, the numbers $1, 3, 5, 6$ are beautiful and $2, 4$ are not. It is because: if $l = 3$ and $r = 3$ we will have a permutation $[1]$ for $m = 1$; if $l = 3$ and $r = 5$ we will have a permutation $[1, 3, 2]$ for $m = 3$; if $l = 1$ and $r = 5$ we will have a permutation $[4, 5, 1, 3, 2]$ for $m = 5$; if $l = 1$ and $r = 6$ we will have a permutation $[4, 5, 1, 3, 2, 6]$ for $m = 6$; it is impossible to take some $l$ and $r$, such that $[p_l, p_{l+1}, \ldots, p_r]$ is a permutation of numbers $1, 2, \ldots, m$ for $m = 2$ and for $m = 4$.
You are given a permutation $p=[p_1, p_2, \ldots, p_n]$. For all $m$ ($1 \le m \le n$) determine if it is a beautiful number or not.
-----Input-----
The first line contains the only integer $t$ ($1 \le t \le 1000$) — the number of test cases in the input. The next lines contain the description of test cases.
The first line of a test case contains a number $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of the given permutation $p$. The next line contains $n$ integers $p_1, p_2, \ldots, p_n$ ($1 \le p_i \le n$, all $p_i$ are different) — the given permutation $p$.
It is guaranteed, that the sum of $n$ from all test cases in the input doesn't exceed $2 \cdot 10^5$.
-----Output-----
Print $t$ lines — the answers to test cases in the order they are given in the input.
The answer to a test case is the string of length $n$, there the $i$-th character is equal to $1$ if $i$ is a beautiful number and is equal to $0$ if $i$ is not a beautiful number.
-----Example-----
Input
3
6
4 5 1 3 2 6
5
5 3 1 2 4
4
1 4 3 2
Output
101011
11111
1001
-----Note-----
The first test case is described in the problem statement.
In the second test case all numbers from $1$ to $5$ are beautiful: if $l = 3$ and $r = 3$ we will have a permutation $[1]$ for $m = 1$; if $l = 3$ and $r = 4$ we will have a permutation $[1, 2]$ for $m = 2$; if $l = 2$ and $r = 4$ we will have a permutation $[3, 1, 2]$ for $m = 3$; if $l = 2$ and $r = 5$ we will have a permutation $[3, 1, 2, 4]$ for $m = 4$; if $l = 1$ and $r = 5$ we will have a permutation $[5, 3, 1, 2, 4]$ for $m = 5$.
|
for _ in range(int(input())):
n = int(input())
a = list(map(int,input().split()))
d = {}
for i in range(n):
d[a[i]] = i
ans = ''
mn = 200001
mx = -1
for i in range(1,n+1):
if(mn > d[i]):
mn = d[i]
if(mx < d[i]):
mx = d[i]
if(mx - mn + 1 > i):
ans += '0'
else:
ans += '1'
print(ans)
|
You are given a permutation $p=[p_1, p_2, \ldots, p_n]$ of integers from $1$ to $n$. Let's call the number $m$ ($1 \le m \le n$) beautiful, if there exists two indices $l, r$ ($1 \le l \le r \le n$), such that the numbers $[p_l, p_{l+1}, \ldots, p_r]$ is a permutation of numbers $1, 2, \ldots, m$.
For example, let $p = [4, 5, 1, 3, 2, 6]$. In this case, the numbers $1, 3, 5, 6$ are beautiful and $2, 4$ are not. It is because: if $l = 3$ and $r = 3$ we will have a permutation $[1]$ for $m = 1$; if $l = 3$ and $r = 5$ we will have a permutation $[1, 3, 2]$ for $m = 3$; if $l = 1$ and $r = 5$ we will have a permutation $[4, 5, 1, 3, 2]$ for $m = 5$; if $l = 1$ and $r = 6$ we will have a permutation $[4, 5, 1, 3, 2, 6]$ for $m = 6$; it is impossible to take some $l$ and $r$, such that $[p_l, p_{l+1}, \ldots, p_r]$ is a permutation of numbers $1, 2, \ldots, m$ for $m = 2$ and for $m = 4$.
You are given a permutation $p=[p_1, p_2, \ldots, p_n]$. For all $m$ ($1 \le m \le n$) determine if it is a beautiful number or not.
-----Input-----
The first line contains the only integer $t$ ($1 \le t \le 1000$) — the number of test cases in the input. The next lines contain the description of test cases.
The first line of a test case contains a number $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of the given permutation $p$. The next line contains $n$ integers $p_1, p_2, \ldots, p_n$ ($1 \le p_i \le n$, all $p_i$ are different) — the given permutation $p$.
It is guaranteed, that the sum of $n$ from all test cases in the input doesn't exceed $2 \cdot 10^5$.
-----Output-----
Print $t$ lines — the answers to test cases in the order they are given in the input.
The answer to a test case is the string of length $n$, there the $i$-th character is equal to $1$ if $i$ is a beautiful number and is equal to $0$ if $i$ is not a beautiful number.
-----Example-----
Input
3
6
4 5 1 3 2 6
5
5 3 1 2 4
4
1 4 3 2
Output
101011
11111
1001
-----Note-----
The first test case is described in the problem statement.
In the second test case all numbers from $1$ to $5$ are beautiful: if $l = 3$ and $r = 3$ we will have a permutation $[1]$ for $m = 1$; if $l = 3$ and $r = 4$ we will have a permutation $[1, 2]$ for $m = 2$; if $l = 2$ and $r = 4$ we will have a permutation $[3, 1, 2]$ for $m = 3$; if $l = 2$ and $r = 5$ we will have a permutation $[3, 1, 2, 4]$ for $m = 4$; if $l = 1$ and $r = 5$ we will have a permutation $[5, 3, 1, 2, 4]$ for $m = 5$.
|
from math import *
from collections import *
import sys
sys.setrecursionlimit(10**9)
t = int(input())
for y in range(t):
n = int(input())
a = list(map(int,input().split()))
ans = ['1']
le = 1
l = a.index(1)
l -= 1
r = l + 2
m = 1
while(le < n):
if(l != -1 and r != n):
if(a[l] > a[r]):
m = max(m,a[r])
r += 1
if(m == le+1):
ans.append('1')
else:
ans.append('0')
else:
m = max(m,a[l])
l -= 1
if(m == le+1):
ans.append('1')
else:
ans.append('0')
elif(l != -1):
m = max(m,a[l])
l -= 1
if(m == le+1):
ans.append('1')
else:
ans.append('0')
else:
m = max(m,a[r])
r += 1
if(m == le+1):
ans.append('1')
else:
ans.append('0')
le += 1
print("".join(ans))
|
You are given a permutation $p=[p_1, p_2, \ldots, p_n]$ of integers from $1$ to $n$. Let's call the number $m$ ($1 \le m \le n$) beautiful, if there exists two indices $l, r$ ($1 \le l \le r \le n$), such that the numbers $[p_l, p_{l+1}, \ldots, p_r]$ is a permutation of numbers $1, 2, \ldots, m$.
For example, let $p = [4, 5, 1, 3, 2, 6]$. In this case, the numbers $1, 3, 5, 6$ are beautiful and $2, 4$ are not. It is because: if $l = 3$ and $r = 3$ we will have a permutation $[1]$ for $m = 1$; if $l = 3$ and $r = 5$ we will have a permutation $[1, 3, 2]$ for $m = 3$; if $l = 1$ and $r = 5$ we will have a permutation $[4, 5, 1, 3, 2]$ for $m = 5$; if $l = 1$ and $r = 6$ we will have a permutation $[4, 5, 1, 3, 2, 6]$ for $m = 6$; it is impossible to take some $l$ and $r$, such that $[p_l, p_{l+1}, \ldots, p_r]$ is a permutation of numbers $1, 2, \ldots, m$ for $m = 2$ and for $m = 4$.
You are given a permutation $p=[p_1, p_2, \ldots, p_n]$. For all $m$ ($1 \le m \le n$) determine if it is a beautiful number or not.
-----Input-----
The first line contains the only integer $t$ ($1 \le t \le 1000$) — the number of test cases in the input. The next lines contain the description of test cases.
The first line of a test case contains a number $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of the given permutation $p$. The next line contains $n$ integers $p_1, p_2, \ldots, p_n$ ($1 \le p_i \le n$, all $p_i$ are different) — the given permutation $p$.
It is guaranteed, that the sum of $n$ from all test cases in the input doesn't exceed $2 \cdot 10^5$.
-----Output-----
Print $t$ lines — the answers to test cases in the order they are given in the input.
The answer to a test case is the string of length $n$, there the $i$-th character is equal to $1$ if $i$ is a beautiful number and is equal to $0$ if $i$ is not a beautiful number.
-----Example-----
Input
3
6
4 5 1 3 2 6
5
5 3 1 2 4
4
1 4 3 2
Output
101011
11111
1001
-----Note-----
The first test case is described in the problem statement.
In the second test case all numbers from $1$ to $5$ are beautiful: if $l = 3$ and $r = 3$ we will have a permutation $[1]$ for $m = 1$; if $l = 3$ and $r = 4$ we will have a permutation $[1, 2]$ for $m = 2$; if $l = 2$ and $r = 4$ we will have a permutation $[3, 1, 2]$ for $m = 3$; if $l = 2$ and $r = 5$ we will have a permutation $[3, 1, 2, 4]$ for $m = 4$; if $l = 1$ and $r = 5$ we will have a permutation $[5, 3, 1, 2, 4]$ for $m = 5$.
|
t = int(input())
for _ in range(t):
n = int(input())
a = list(map(int, input().split()))
pos = [0]*(n+1)
for i, x in enumerate(a):
pos[x] = i
used = [0, 1] + [0]*n
ans = [0]*n
l, r = pos[1], pos[1]
count = 1
for x in range(1, n+1):
if not used[x]:
if pos[x] < l:
while not used[x]:
l -= 1
used[a[l]] = 1
count += 1
else:
while not used[x]:
r += 1
used[a[r]] = 1
count += 1
if count == x:
ans[x-1] = 1
print(*ans, sep='')
|
You are given a permutation $p=[p_1, p_2, \ldots, p_n]$ of integers from $1$ to $n$. Let's call the number $m$ ($1 \le m \le n$) beautiful, if there exists two indices $l, r$ ($1 \le l \le r \le n$), such that the numbers $[p_l, p_{l+1}, \ldots, p_r]$ is a permutation of numbers $1, 2, \ldots, m$.
For example, let $p = [4, 5, 1, 3, 2, 6]$. In this case, the numbers $1, 3, 5, 6$ are beautiful and $2, 4$ are not. It is because: if $l = 3$ and $r = 3$ we will have a permutation $[1]$ for $m = 1$; if $l = 3$ and $r = 5$ we will have a permutation $[1, 3, 2]$ for $m = 3$; if $l = 1$ and $r = 5$ we will have a permutation $[4, 5, 1, 3, 2]$ for $m = 5$; if $l = 1$ and $r = 6$ we will have a permutation $[4, 5, 1, 3, 2, 6]$ for $m = 6$; it is impossible to take some $l$ and $r$, such that $[p_l, p_{l+1}, \ldots, p_r]$ is a permutation of numbers $1, 2, \ldots, m$ for $m = 2$ and for $m = 4$.
You are given a permutation $p=[p_1, p_2, \ldots, p_n]$. For all $m$ ($1 \le m \le n$) determine if it is a beautiful number or not.
-----Input-----
The first line contains the only integer $t$ ($1 \le t \le 1000$) — the number of test cases in the input. The next lines contain the description of test cases.
The first line of a test case contains a number $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of the given permutation $p$. The next line contains $n$ integers $p_1, p_2, \ldots, p_n$ ($1 \le p_i \le n$, all $p_i$ are different) — the given permutation $p$.
It is guaranteed, that the sum of $n$ from all test cases in the input doesn't exceed $2 \cdot 10^5$.
-----Output-----
Print $t$ lines — the answers to test cases in the order they are given in the input.
The answer to a test case is the string of length $n$, there the $i$-th character is equal to $1$ if $i$ is a beautiful number and is equal to $0$ if $i$ is not a beautiful number.
-----Example-----
Input
3
6
4 5 1 3 2 6
5
5 3 1 2 4
4
1 4 3 2
Output
101011
11111
1001
-----Note-----
The first test case is described in the problem statement.
In the second test case all numbers from $1$ to $5$ are beautiful: if $l = 3$ and $r = 3$ we will have a permutation $[1]$ for $m = 1$; if $l = 3$ and $r = 4$ we will have a permutation $[1, 2]$ for $m = 2$; if $l = 2$ and $r = 4$ we will have a permutation $[3, 1, 2]$ for $m = 3$; if $l = 2$ and $r = 5$ we will have a permutation $[3, 1, 2, 4]$ for $m = 4$; if $l = 1$ and $r = 5$ we will have a permutation $[5, 3, 1, 2, 4]$ for $m = 5$.
|
def mi():
return map(int, input().split())
'''
3
6
4 5 1 3 2 6
5
5 3 1 2 4
4
1 4 3 2
3
6
4 5 1 3 2 6
5
5 3 1 2 4
4
1 4 3 2
'''
for _ in range(int(input())):
n = int(input())
a = list(mi())
t = a.index(1)
dist = [0]*(n+1)
dic = [0]*n
for i in range(n):
dist[a[i]] = abs(t-i)
dic[i] = [a[i], i]
dic.sort()
lm = dic[0][1]
rm = dic[0][1]
print (1, end = '')
for i in range(1, n):
if (dic[i][1]<lm):
lm = dic[i][1]
if (dic[i][1]>rm):
rm = dic[i][1]
if rm-lm<i+1:
print (1, end = '')
else:
print (0, end = '')
print()
|
You are given a permutation $p=[p_1, p_2, \ldots, p_n]$ of integers from $1$ to $n$. Let's call the number $m$ ($1 \le m \le n$) beautiful, if there exists two indices $l, r$ ($1 \le l \le r \le n$), such that the numbers $[p_l, p_{l+1}, \ldots, p_r]$ is a permutation of numbers $1, 2, \ldots, m$.
For example, let $p = [4, 5, 1, 3, 2, 6]$. In this case, the numbers $1, 3, 5, 6$ are beautiful and $2, 4$ are not. It is because: if $l = 3$ and $r = 3$ we will have a permutation $[1]$ for $m = 1$; if $l = 3$ and $r = 5$ we will have a permutation $[1, 3, 2]$ for $m = 3$; if $l = 1$ and $r = 5$ we will have a permutation $[4, 5, 1, 3, 2]$ for $m = 5$; if $l = 1$ and $r = 6$ we will have a permutation $[4, 5, 1, 3, 2, 6]$ for $m = 6$; it is impossible to take some $l$ and $r$, such that $[p_l, p_{l+1}, \ldots, p_r]$ is a permutation of numbers $1, 2, \ldots, m$ for $m = 2$ and for $m = 4$.
You are given a permutation $p=[p_1, p_2, \ldots, p_n]$. For all $m$ ($1 \le m \le n$) determine if it is a beautiful number or not.
-----Input-----
The first line contains the only integer $t$ ($1 \le t \le 1000$) — the number of test cases in the input. The next lines contain the description of test cases.
The first line of a test case contains a number $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of the given permutation $p$. The next line contains $n$ integers $p_1, p_2, \ldots, p_n$ ($1 \le p_i \le n$, all $p_i$ are different) — the given permutation $p$.
It is guaranteed, that the sum of $n$ from all test cases in the input doesn't exceed $2 \cdot 10^5$.
-----Output-----
Print $t$ lines — the answers to test cases in the order they are given in the input.
The answer to a test case is the string of length $n$, there the $i$-th character is equal to $1$ if $i$ is a beautiful number and is equal to $0$ if $i$ is not a beautiful number.
-----Example-----
Input
3
6
4 5 1 3 2 6
5
5 3 1 2 4
4
1 4 3 2
Output
101011
11111
1001
-----Note-----
The first test case is described in the problem statement.
In the second test case all numbers from $1$ to $5$ are beautiful: if $l = 3$ and $r = 3$ we will have a permutation $[1]$ for $m = 1$; if $l = 3$ and $r = 4$ we will have a permutation $[1, 2]$ for $m = 2$; if $l = 2$ and $r = 4$ we will have a permutation $[3, 1, 2]$ for $m = 3$; if $l = 2$ and $r = 5$ we will have a permutation $[3, 1, 2, 4]$ for $m = 4$; if $l = 1$ and $r = 5$ we will have a permutation $[5, 3, 1, 2, 4]$ for $m = 5$.
|
from sys import stdin
input = stdin.readline
t = int(input())
for _ in range(t):
n = int(input())
a = list(map(int,input().split()))
start = 0
for i,v in enumerate(a):
if v == 1:
start = i
break
ans = [0]*-~n
ans[n-1] = 1
mx = 1
l = start
r = start
def move(x):
nonlocal l,r,mx
if x:
mx = max(a[r+1],mx)
r += 1
else:
mx = max(a[l-1],mx)
l -= 1
while mx < n:
if mx == r-l+1:
ans[mx-1] = 1
if l == 0:
move(1)
elif r == n-1:
move(0)
else:
if a[l-1] > a[r+1]:
move(1)
else:
move(0)
print("".join(map(str,ans[:n])))
|
You are given a permutation $p=[p_1, p_2, \ldots, p_n]$ of integers from $1$ to $n$. Let's call the number $m$ ($1 \le m \le n$) beautiful, if there exists two indices $l, r$ ($1 \le l \le r \le n$), such that the numbers $[p_l, p_{l+1}, \ldots, p_r]$ is a permutation of numbers $1, 2, \ldots, m$.
For example, let $p = [4, 5, 1, 3, 2, 6]$. In this case, the numbers $1, 3, 5, 6$ are beautiful and $2, 4$ are not. It is because: if $l = 3$ and $r = 3$ we will have a permutation $[1]$ for $m = 1$; if $l = 3$ and $r = 5$ we will have a permutation $[1, 3, 2]$ for $m = 3$; if $l = 1$ and $r = 5$ we will have a permutation $[4, 5, 1, 3, 2]$ for $m = 5$; if $l = 1$ and $r = 6$ we will have a permutation $[4, 5, 1, 3, 2, 6]$ for $m = 6$; it is impossible to take some $l$ and $r$, such that $[p_l, p_{l+1}, \ldots, p_r]$ is a permutation of numbers $1, 2, \ldots, m$ for $m = 2$ and for $m = 4$.
You are given a permutation $p=[p_1, p_2, \ldots, p_n]$. For all $m$ ($1 \le m \le n$) determine if it is a beautiful number or not.
-----Input-----
The first line contains the only integer $t$ ($1 \le t \le 1000$) — the number of test cases in the input. The next lines contain the description of test cases.
The first line of a test case contains a number $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of the given permutation $p$. The next line contains $n$ integers $p_1, p_2, \ldots, p_n$ ($1 \le p_i \le n$, all $p_i$ are different) — the given permutation $p$.
It is guaranteed, that the sum of $n$ from all test cases in the input doesn't exceed $2 \cdot 10^5$.
-----Output-----
Print $t$ lines — the answers to test cases in the order they are given in the input.
The answer to a test case is the string of length $n$, there the $i$-th character is equal to $1$ if $i$ is a beautiful number and is equal to $0$ if $i$ is not a beautiful number.
-----Example-----
Input
3
6
4 5 1 3 2 6
5
5 3 1 2 4
4
1 4 3 2
Output
101011
11111
1001
-----Note-----
The first test case is described in the problem statement.
In the second test case all numbers from $1$ to $5$ are beautiful: if $l = 3$ and $r = 3$ we will have a permutation $[1]$ for $m = 1$; if $l = 3$ and $r = 4$ we will have a permutation $[1, 2]$ for $m = 2$; if $l = 2$ and $r = 4$ we will have a permutation $[3, 1, 2]$ for $m = 3$; if $l = 2$ and $r = 5$ we will have a permutation $[3, 1, 2, 4]$ for $m = 4$; if $l = 1$ and $r = 5$ we will have a permutation $[5, 3, 1, 2, 4]$ for $m = 5$.
|
#!/usr/bin/env python3
from itertools import combinations
import sys
input = sys.stdin.readline
INF = 10**9
t = int(input())
for i in range(t):
n = int(input())
a = [INF] + [int(item) for item in input().split()] + [INF]
ans = [1]
l = r = a.index(1)
max_val = 1
for i in range(2, n+1):
if i == max(max_val, a[l-1]):
ans.append(1)
l -= 1
max_val = i
elif i == max(max_val, a[r+1]):
ans.append(1)
r += 1
max_val = i
elif a[l-1] < a[r+1]:
ans.append(0)
max_val = max(max_val, a[l-1])
l -= 1
else:
ans.append(0)
max_val = max(max_val, a[r+1])
r += 1
print("".join([str(item) for item in ans]))
|
You are given a permutation $p=[p_1, p_2, \ldots, p_n]$ of integers from $1$ to $n$. Let's call the number $m$ ($1 \le m \le n$) beautiful, if there exists two indices $l, r$ ($1 \le l \le r \le n$), such that the numbers $[p_l, p_{l+1}, \ldots, p_r]$ is a permutation of numbers $1, 2, \ldots, m$.
For example, let $p = [4, 5, 1, 3, 2, 6]$. In this case, the numbers $1, 3, 5, 6$ are beautiful and $2, 4$ are not. It is because: if $l = 3$ and $r = 3$ we will have a permutation $[1]$ for $m = 1$; if $l = 3$ and $r = 5$ we will have a permutation $[1, 3, 2]$ for $m = 3$; if $l = 1$ and $r = 5$ we will have a permutation $[4, 5, 1, 3, 2]$ for $m = 5$; if $l = 1$ and $r = 6$ we will have a permutation $[4, 5, 1, 3, 2, 6]$ for $m = 6$; it is impossible to take some $l$ and $r$, such that $[p_l, p_{l+1}, \ldots, p_r]$ is a permutation of numbers $1, 2, \ldots, m$ for $m = 2$ and for $m = 4$.
You are given a permutation $p=[p_1, p_2, \ldots, p_n]$. For all $m$ ($1 \le m \le n$) determine if it is a beautiful number or not.
-----Input-----
The first line contains the only integer $t$ ($1 \le t \le 1000$) — the number of test cases in the input. The next lines contain the description of test cases.
The first line of a test case contains a number $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of the given permutation $p$. The next line contains $n$ integers $p_1, p_2, \ldots, p_n$ ($1 \le p_i \le n$, all $p_i$ are different) — the given permutation $p$.
It is guaranteed, that the sum of $n$ from all test cases in the input doesn't exceed $2 \cdot 10^5$.
-----Output-----
Print $t$ lines — the answers to test cases in the order they are given in the input.
The answer to a test case is the string of length $n$, there the $i$-th character is equal to $1$ if $i$ is a beautiful number and is equal to $0$ if $i$ is not a beautiful number.
-----Example-----
Input
3
6
4 5 1 3 2 6
5
5 3 1 2 4
4
1 4 3 2
Output
101011
11111
1001
-----Note-----
The first test case is described in the problem statement.
In the second test case all numbers from $1$ to $5$ are beautiful: if $l = 3$ and $r = 3$ we will have a permutation $[1]$ for $m = 1$; if $l = 3$ and $r = 4$ we will have a permutation $[1, 2]$ for $m = 2$; if $l = 2$ and $r = 4$ we will have a permutation $[3, 1, 2]$ for $m = 3$; if $l = 2$ and $r = 5$ we will have a permutation $[3, 1, 2, 4]$ for $m = 4$; if $l = 1$ and $r = 5$ we will have a permutation $[5, 3, 1, 2, 4]$ for $m = 5$.
|
for j in range(int(input())):
n = int(input())
c = list(map(int,input().split()))
index = [0]*n
for i in range(n):
index[c[i]-1]=i
ma = 0
mi = n
ans = ['0']*n
# print(index)
for k in range(n):
ma = max(index[k],ma)
mi = min(index[k],mi)
#print(k,mr,index[k]-index[0])
if ma-mi<=k:
ans[k]='1'
print(''.join(ans))
|
You are given a permutation $p=[p_1, p_2, \ldots, p_n]$ of integers from $1$ to $n$. Let's call the number $m$ ($1 \le m \le n$) beautiful, if there exists two indices $l, r$ ($1 \le l \le r \le n$), such that the numbers $[p_l, p_{l+1}, \ldots, p_r]$ is a permutation of numbers $1, 2, \ldots, m$.
For example, let $p = [4, 5, 1, 3, 2, 6]$. In this case, the numbers $1, 3, 5, 6$ are beautiful and $2, 4$ are not. It is because: if $l = 3$ and $r = 3$ we will have a permutation $[1]$ for $m = 1$; if $l = 3$ and $r = 5$ we will have a permutation $[1, 3, 2]$ for $m = 3$; if $l = 1$ and $r = 5$ we will have a permutation $[4, 5, 1, 3, 2]$ for $m = 5$; if $l = 1$ and $r = 6$ we will have a permutation $[4, 5, 1, 3, 2, 6]$ for $m = 6$; it is impossible to take some $l$ and $r$, such that $[p_l, p_{l+1}, \ldots, p_r]$ is a permutation of numbers $1, 2, \ldots, m$ for $m = 2$ and for $m = 4$.
You are given a permutation $p=[p_1, p_2, \ldots, p_n]$. For all $m$ ($1 \le m \le n$) determine if it is a beautiful number or not.
-----Input-----
The first line contains the only integer $t$ ($1 \le t \le 1000$) — the number of test cases in the input. The next lines contain the description of test cases.
The first line of a test case contains a number $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of the given permutation $p$. The next line contains $n$ integers $p_1, p_2, \ldots, p_n$ ($1 \le p_i \le n$, all $p_i$ are different) — the given permutation $p$.
It is guaranteed, that the sum of $n$ from all test cases in the input doesn't exceed $2 \cdot 10^5$.
-----Output-----
Print $t$ lines — the answers to test cases in the order they are given in the input.
The answer to a test case is the string of length $n$, there the $i$-th character is equal to $1$ if $i$ is a beautiful number and is equal to $0$ if $i$ is not a beautiful number.
-----Example-----
Input
3
6
4 5 1 3 2 6
5
5 3 1 2 4
4
1 4 3 2
Output
101011
11111
1001
-----Note-----
The first test case is described in the problem statement.
In the second test case all numbers from $1$ to $5$ are beautiful: if $l = 3$ and $r = 3$ we will have a permutation $[1]$ for $m = 1$; if $l = 3$ and $r = 4$ we will have a permutation $[1, 2]$ for $m = 2$; if $l = 2$ and $r = 4$ we will have a permutation $[3, 1, 2]$ for $m = 3$; if $l = 2$ and $r = 5$ we will have a permutation $[3, 1, 2, 4]$ for $m = 4$; if $l = 1$ and $r = 5$ we will have a permutation $[5, 3, 1, 2, 4]$ for $m = 5$.
|
q=int(input())
for t in range(q):
n=int(input())
a=list(map(int,input().split()))
ma=1
ans='1'
uk1=a.index(1)
uk2=uk1
while uk2-uk1+1!=n:
if uk2==n-1:
uk1-=1
ma=max(ma,a[uk1])
if ma==uk2-uk1+1:
ans=ans+'1'
else:
ans=ans+'0'
else:
if uk1==0:
uk2+=1
ma=max(ma,a[uk2])
if ma == uk2 - uk1 + 1:
ans = ans + '1'
else:
ans=ans+'0'
else:
if a[uk1-1]<a[uk2+1]:
uk1 -= 1
ma = max(ma, a[uk1])
if ma == uk2 - uk1 + 1:
ans = ans + '1'
else:
ans = ans + '0'
else:
uk2 += 1
ma = max(ma, a[uk2])
if ma == uk2 - uk1 + 1:
ans = ans + '1'
else:
ans = ans + '0'
print(ans)
|
You are given a permutation $p=[p_1, p_2, \ldots, p_n]$ of integers from $1$ to $n$. Let's call the number $m$ ($1 \le m \le n$) beautiful, if there exists two indices $l, r$ ($1 \le l \le r \le n$), such that the numbers $[p_l, p_{l+1}, \ldots, p_r]$ is a permutation of numbers $1, 2, \ldots, m$.
For example, let $p = [4, 5, 1, 3, 2, 6]$. In this case, the numbers $1, 3, 5, 6$ are beautiful and $2, 4$ are not. It is because: if $l = 3$ and $r = 3$ we will have a permutation $[1]$ for $m = 1$; if $l = 3$ and $r = 5$ we will have a permutation $[1, 3, 2]$ for $m = 3$; if $l = 1$ and $r = 5$ we will have a permutation $[4, 5, 1, 3, 2]$ for $m = 5$; if $l = 1$ and $r = 6$ we will have a permutation $[4, 5, 1, 3, 2, 6]$ for $m = 6$; it is impossible to take some $l$ and $r$, such that $[p_l, p_{l+1}, \ldots, p_r]$ is a permutation of numbers $1, 2, \ldots, m$ for $m = 2$ and for $m = 4$.
You are given a permutation $p=[p_1, p_2, \ldots, p_n]$. For all $m$ ($1 \le m \le n$) determine if it is a beautiful number or not.
-----Input-----
The first line contains the only integer $t$ ($1 \le t \le 1000$) — the number of test cases in the input. The next lines contain the description of test cases.
The first line of a test case contains a number $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of the given permutation $p$. The next line contains $n$ integers $p_1, p_2, \ldots, p_n$ ($1 \le p_i \le n$, all $p_i$ are different) — the given permutation $p$.
It is guaranteed, that the sum of $n$ from all test cases in the input doesn't exceed $2 \cdot 10^5$.
-----Output-----
Print $t$ lines — the answers to test cases in the order they are given in the input.
The answer to a test case is the string of length $n$, there the $i$-th character is equal to $1$ if $i$ is a beautiful number and is equal to $0$ if $i$ is not a beautiful number.
-----Example-----
Input
3
6
4 5 1 3 2 6
5
5 3 1 2 4
4
1 4 3 2
Output
101011
11111
1001
-----Note-----
The first test case is described in the problem statement.
In the second test case all numbers from $1$ to $5$ are beautiful: if $l = 3$ and $r = 3$ we will have a permutation $[1]$ for $m = 1$; if $l = 3$ and $r = 4$ we will have a permutation $[1, 2]$ for $m = 2$; if $l = 2$ and $r = 4$ we will have a permutation $[3, 1, 2]$ for $m = 3$; if $l = 2$ and $r = 5$ we will have a permutation $[3, 1, 2, 4]$ for $m = 4$; if $l = 1$ and $r = 5$ we will have a permutation $[5, 3, 1, 2, 4]$ for $m = 5$.
|
lpn = int(input())
for loop in range(lpn):
n = int(input())
p = list(map(int,input().split()))
for i in range(n):
if p[i] == 1:
oneind = i
break
l = oneind
r = oneind
nmax = 1
ans = [0] * n
ans[0] = 1
for i in range(n-1):
if l == 0 or( r != n-1 and p[l-1] > p[r+1]):
r += 1
nmax = max(nmax,p[r])
if i+2 == nmax:
ans[i+1] = 1
else:
l -= 1
nmax = max(nmax,p[l])
if i+2 == nmax:
ans[i+1] = 1
print("".join(map(str,ans)))
|
You are given a permutation $p=[p_1, p_2, \ldots, p_n]$ of integers from $1$ to $n$. Let's call the number $m$ ($1 \le m \le n$) beautiful, if there exists two indices $l, r$ ($1 \le l \le r \le n$), such that the numbers $[p_l, p_{l+1}, \ldots, p_r]$ is a permutation of numbers $1, 2, \ldots, m$.
For example, let $p = [4, 5, 1, 3, 2, 6]$. In this case, the numbers $1, 3, 5, 6$ are beautiful and $2, 4$ are not. It is because: if $l = 3$ and $r = 3$ we will have a permutation $[1]$ for $m = 1$; if $l = 3$ and $r = 5$ we will have a permutation $[1, 3, 2]$ for $m = 3$; if $l = 1$ and $r = 5$ we will have a permutation $[4, 5, 1, 3, 2]$ for $m = 5$; if $l = 1$ and $r = 6$ we will have a permutation $[4, 5, 1, 3, 2, 6]$ for $m = 6$; it is impossible to take some $l$ and $r$, such that $[p_l, p_{l+1}, \ldots, p_r]$ is a permutation of numbers $1, 2, \ldots, m$ for $m = 2$ and for $m = 4$.
You are given a permutation $p=[p_1, p_2, \ldots, p_n]$. For all $m$ ($1 \le m \le n$) determine if it is a beautiful number or not.
-----Input-----
The first line contains the only integer $t$ ($1 \le t \le 1000$) — the number of test cases in the input. The next lines contain the description of test cases.
The first line of a test case contains a number $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of the given permutation $p$. The next line contains $n$ integers $p_1, p_2, \ldots, p_n$ ($1 \le p_i \le n$, all $p_i$ are different) — the given permutation $p$.
It is guaranteed, that the sum of $n$ from all test cases in the input doesn't exceed $2 \cdot 10^5$.
-----Output-----
Print $t$ lines — the answers to test cases in the order they are given in the input.
The answer to a test case is the string of length $n$, there the $i$-th character is equal to $1$ if $i$ is a beautiful number and is equal to $0$ if $i$ is not a beautiful number.
-----Example-----
Input
3
6
4 5 1 3 2 6
5
5 3 1 2 4
4
1 4 3 2
Output
101011
11111
1001
-----Note-----
The first test case is described in the problem statement.
In the second test case all numbers from $1$ to $5$ are beautiful: if $l = 3$ and $r = 3$ we will have a permutation $[1]$ for $m = 1$; if $l = 3$ and $r = 4$ we will have a permutation $[1, 2]$ for $m = 2$; if $l = 2$ and $r = 4$ we will have a permutation $[3, 1, 2]$ for $m = 3$; if $l = 2$ and $r = 5$ we will have a permutation $[3, 1, 2, 4]$ for $m = 4$; if $l = 1$ and $r = 5$ we will have a permutation $[5, 3, 1, 2, 4]$ for $m = 5$.
|
t = int(input())
for i in range(t):
n = int(input())
a = [int(i) for i in input().split()]
ans = ['0'] * n
ans[0] = '1'
ans[-1] = '1'
l = 0
r = n - 1
now = n
while (r - l) > 1:
if a[r] > now:
r -= 1
continue
if a[l] > now:
l += 1
continue
if (r - l + 1) == now:
ans[r - l] = '1'
now -= 1
if (r - l + 1) == now:
ans[r - l] = '1'
print(''.join(ans))
|
You are given a permutation $p=[p_1, p_2, \ldots, p_n]$ of integers from $1$ to $n$. Let's call the number $m$ ($1 \le m \le n$) beautiful, if there exists two indices $l, r$ ($1 \le l \le r \le n$), such that the numbers $[p_l, p_{l+1}, \ldots, p_r]$ is a permutation of numbers $1, 2, \ldots, m$.
For example, let $p = [4, 5, 1, 3, 2, 6]$. In this case, the numbers $1, 3, 5, 6$ are beautiful and $2, 4$ are not. It is because: if $l = 3$ and $r = 3$ we will have a permutation $[1]$ for $m = 1$; if $l = 3$ and $r = 5$ we will have a permutation $[1, 3, 2]$ for $m = 3$; if $l = 1$ and $r = 5$ we will have a permutation $[4, 5, 1, 3, 2]$ for $m = 5$; if $l = 1$ and $r = 6$ we will have a permutation $[4, 5, 1, 3, 2, 6]$ for $m = 6$; it is impossible to take some $l$ and $r$, such that $[p_l, p_{l+1}, \ldots, p_r]$ is a permutation of numbers $1, 2, \ldots, m$ for $m = 2$ and for $m = 4$.
You are given a permutation $p=[p_1, p_2, \ldots, p_n]$. For all $m$ ($1 \le m \le n$) determine if it is a beautiful number or not.
-----Input-----
The first line contains the only integer $t$ ($1 \le t \le 1000$) — the number of test cases in the input. The next lines contain the description of test cases.
The first line of a test case contains a number $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of the given permutation $p$. The next line contains $n$ integers $p_1, p_2, \ldots, p_n$ ($1 \le p_i \le n$, all $p_i$ are different) — the given permutation $p$.
It is guaranteed, that the sum of $n$ from all test cases in the input doesn't exceed $2 \cdot 10^5$.
-----Output-----
Print $t$ lines — the answers to test cases in the order they are given in the input.
The answer to a test case is the string of length $n$, there the $i$-th character is equal to $1$ if $i$ is a beautiful number and is equal to $0$ if $i$ is not a beautiful number.
-----Example-----
Input
3
6
4 5 1 3 2 6
5
5 3 1 2 4
4
1 4 3 2
Output
101011
11111
1001
-----Note-----
The first test case is described in the problem statement.
In the second test case all numbers from $1$ to $5$ are beautiful: if $l = 3$ and $r = 3$ we will have a permutation $[1]$ for $m = 1$; if $l = 3$ and $r = 4$ we will have a permutation $[1, 2]$ for $m = 2$; if $l = 2$ and $r = 4$ we will have a permutation $[3, 1, 2]$ for $m = 3$; if $l = 2$ and $r = 5$ we will have a permutation $[3, 1, 2, 4]$ for $m = 4$; if $l = 1$ and $r = 5$ we will have a permutation $[5, 3, 1, 2, 4]$ for $m = 5$.
|
# https://codeforces.com/contest/1265/problem/B
def main():
n = int(input())
p = list(map(int, input().split()))
idx = [0] * n
for i in range(n):
idx[p[i]-1] = i
ans = ''
left = n
right = 0
for i in range(n):
left = min(left, idx[i])
right = max(right, idx[i])
if right - left == i:
ans += '1'
else:
ans += '0'
return ans
t = int(input())
for i in range(t):
print(main())
|
You are given a permutation $p=[p_1, p_2, \ldots, p_n]$ of integers from $1$ to $n$. Let's call the number $m$ ($1 \le m \le n$) beautiful, if there exists two indices $l, r$ ($1 \le l \le r \le n$), such that the numbers $[p_l, p_{l+1}, \ldots, p_r]$ is a permutation of numbers $1, 2, \ldots, m$.
For example, let $p = [4, 5, 1, 3, 2, 6]$. In this case, the numbers $1, 3, 5, 6$ are beautiful and $2, 4$ are not. It is because: if $l = 3$ and $r = 3$ we will have a permutation $[1]$ for $m = 1$; if $l = 3$ and $r = 5$ we will have a permutation $[1, 3, 2]$ for $m = 3$; if $l = 1$ and $r = 5$ we will have a permutation $[4, 5, 1, 3, 2]$ for $m = 5$; if $l = 1$ and $r = 6$ we will have a permutation $[4, 5, 1, 3, 2, 6]$ for $m = 6$; it is impossible to take some $l$ and $r$, such that $[p_l, p_{l+1}, \ldots, p_r]$ is a permutation of numbers $1, 2, \ldots, m$ for $m = 2$ and for $m = 4$.
You are given a permutation $p=[p_1, p_2, \ldots, p_n]$. For all $m$ ($1 \le m \le n$) determine if it is a beautiful number or not.
-----Input-----
The first line contains the only integer $t$ ($1 \le t \le 1000$) — the number of test cases in the input. The next lines contain the description of test cases.
The first line of a test case contains a number $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of the given permutation $p$. The next line contains $n$ integers $p_1, p_2, \ldots, p_n$ ($1 \le p_i \le n$, all $p_i$ are different) — the given permutation $p$.
It is guaranteed, that the sum of $n$ from all test cases in the input doesn't exceed $2 \cdot 10^5$.
-----Output-----
Print $t$ lines — the answers to test cases in the order they are given in the input.
The answer to a test case is the string of length $n$, there the $i$-th character is equal to $1$ if $i$ is a beautiful number and is equal to $0$ if $i$ is not a beautiful number.
-----Example-----
Input
3
6
4 5 1 3 2 6
5
5 3 1 2 4
4
1 4 3 2
Output
101011
11111
1001
-----Note-----
The first test case is described in the problem statement.
In the second test case all numbers from $1$ to $5$ are beautiful: if $l = 3$ and $r = 3$ we will have a permutation $[1]$ for $m = 1$; if $l = 3$ and $r = 4$ we will have a permutation $[1, 2]$ for $m = 2$; if $l = 2$ and $r = 4$ we will have a permutation $[3, 1, 2]$ for $m = 3$; if $l = 2$ and $r = 5$ we will have a permutation $[3, 1, 2, 4]$ for $m = 4$; if $l = 1$ and $r = 5$ we will have a permutation $[5, 3, 1, 2, 4]$ for $m = 5$.
|
def f():
n = int(input())
A = [int(s) for s in input().split()]
ans = [0]*n
ans[0] = 1
ans[n-1] = 1
i = 0
j = n-1
outMin = n+1
while j>i:
if A[i] > A[j]:
if A[i] < outMin:
outMin = A[i]
i += 1
else:
if A[j] < outMin:
outMin = A[j]
j -= 1
if j-i == outMin-2:
ans[j-i] = 1
print(''.join(str(i) for i in ans))
t = int(input())
for i in range(t):
f()
|
You are given a permutation $p=[p_1, p_2, \ldots, p_n]$ of integers from $1$ to $n$. Let's call the number $m$ ($1 \le m \le n$) beautiful, if there exists two indices $l, r$ ($1 \le l \le r \le n$), such that the numbers $[p_l, p_{l+1}, \ldots, p_r]$ is a permutation of numbers $1, 2, \ldots, m$.
For example, let $p = [4, 5, 1, 3, 2, 6]$. In this case, the numbers $1, 3, 5, 6$ are beautiful and $2, 4$ are not. It is because: if $l = 3$ and $r = 3$ we will have a permutation $[1]$ for $m = 1$; if $l = 3$ and $r = 5$ we will have a permutation $[1, 3, 2]$ for $m = 3$; if $l = 1$ and $r = 5$ we will have a permutation $[4, 5, 1, 3, 2]$ for $m = 5$; if $l = 1$ and $r = 6$ we will have a permutation $[4, 5, 1, 3, 2, 6]$ for $m = 6$; it is impossible to take some $l$ and $r$, such that $[p_l, p_{l+1}, \ldots, p_r]$ is a permutation of numbers $1, 2, \ldots, m$ for $m = 2$ and for $m = 4$.
You are given a permutation $p=[p_1, p_2, \ldots, p_n]$. For all $m$ ($1 \le m \le n$) determine if it is a beautiful number or not.
-----Input-----
The first line contains the only integer $t$ ($1 \le t \le 1000$) — the number of test cases in the input. The next lines contain the description of test cases.
The first line of a test case contains a number $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of the given permutation $p$. The next line contains $n$ integers $p_1, p_2, \ldots, p_n$ ($1 \le p_i \le n$, all $p_i$ are different) — the given permutation $p$.
It is guaranteed, that the sum of $n$ from all test cases in the input doesn't exceed $2 \cdot 10^5$.
-----Output-----
Print $t$ lines — the answers to test cases in the order they are given in the input.
The answer to a test case is the string of length $n$, there the $i$-th character is equal to $1$ if $i$ is a beautiful number and is equal to $0$ if $i$ is not a beautiful number.
-----Example-----
Input
3
6
4 5 1 3 2 6
5
5 3 1 2 4
4
1 4 3 2
Output
101011
11111
1001
-----Note-----
The first test case is described in the problem statement.
In the second test case all numbers from $1$ to $5$ are beautiful: if $l = 3$ and $r = 3$ we will have a permutation $[1]$ for $m = 1$; if $l = 3$ and $r = 4$ we will have a permutation $[1, 2]$ for $m = 2$; if $l = 2$ and $r = 4$ we will have a permutation $[3, 1, 2]$ for $m = 3$; if $l = 2$ and $r = 5$ we will have a permutation $[3, 1, 2, 4]$ for $m = 4$; if $l = 1$ and $r = 5$ we will have a permutation $[5, 3, 1, 2, 4]$ for $m = 5$.
|
n = int(input())
for _ in range(n):
k = int(input())
pos = [0] * k
arr = list(map(int, input().split(' ')))
for i in range(k):
pos[arr[i] - 1] = i
#print(pos)
left, right = [0] * k, [0] * k
left[0], right[0] = pos[0], pos[0]
for i in range(1, k):
left[i] = min(left[i - 1], pos[i])
right[i] = max(right[i - 1], pos[i])
#print(left)
#print(right)
for i in range(k):
if right[i] - left[i] == i:
print(1, end="")
else:
print(0, end="")
print()
|
You are given a permutation $p=[p_1, p_2, \ldots, p_n]$ of integers from $1$ to $n$. Let's call the number $m$ ($1 \le m \le n$) beautiful, if there exists two indices $l, r$ ($1 \le l \le r \le n$), such that the numbers $[p_l, p_{l+1}, \ldots, p_r]$ is a permutation of numbers $1, 2, \ldots, m$.
For example, let $p = [4, 5, 1, 3, 2, 6]$. In this case, the numbers $1, 3, 5, 6$ are beautiful and $2, 4$ are not. It is because: if $l = 3$ and $r = 3$ we will have a permutation $[1]$ for $m = 1$; if $l = 3$ and $r = 5$ we will have a permutation $[1, 3, 2]$ for $m = 3$; if $l = 1$ and $r = 5$ we will have a permutation $[4, 5, 1, 3, 2]$ for $m = 5$; if $l = 1$ and $r = 6$ we will have a permutation $[4, 5, 1, 3, 2, 6]$ for $m = 6$; it is impossible to take some $l$ and $r$, such that $[p_l, p_{l+1}, \ldots, p_r]$ is a permutation of numbers $1, 2, \ldots, m$ for $m = 2$ and for $m = 4$.
You are given a permutation $p=[p_1, p_2, \ldots, p_n]$. For all $m$ ($1 \le m \le n$) determine if it is a beautiful number or not.
-----Input-----
The first line contains the only integer $t$ ($1 \le t \le 1000$) — the number of test cases in the input. The next lines contain the description of test cases.
The first line of a test case contains a number $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of the given permutation $p$. The next line contains $n$ integers $p_1, p_2, \ldots, p_n$ ($1 \le p_i \le n$, all $p_i$ are different) — the given permutation $p$.
It is guaranteed, that the sum of $n$ from all test cases in the input doesn't exceed $2 \cdot 10^5$.
-----Output-----
Print $t$ lines — the answers to test cases in the order they are given in the input.
The answer to a test case is the string of length $n$, there the $i$-th character is equal to $1$ if $i$ is a beautiful number and is equal to $0$ if $i$ is not a beautiful number.
-----Example-----
Input
3
6
4 5 1 3 2 6
5
5 3 1 2 4
4
1 4 3 2
Output
101011
11111
1001
-----Note-----
The first test case is described in the problem statement.
In the second test case all numbers from $1$ to $5$ are beautiful: if $l = 3$ and $r = 3$ we will have a permutation $[1]$ for $m = 1$; if $l = 3$ and $r = 4$ we will have a permutation $[1, 2]$ for $m = 2$; if $l = 2$ and $r = 4$ we will have a permutation $[3, 1, 2]$ for $m = 3$; if $l = 2$ and $r = 5$ we will have a permutation $[3, 1, 2, 4]$ for $m = 4$; if $l = 1$ and $r = 5$ we will have a permutation $[5, 3, 1, 2, 4]$ for $m = 5$.
|
for kkk in range(int(input())):
n = int(input())
l = list(map(int, input().split()))
d = {}
for i in range(n):
d[l[i]] = i
ans = ["0" for i in range(n+1)]
ans[1] = "1"
posleft = d[1]
posright = d[1]
for j in range(2, n+1):
if(d[j]==posleft-1 or d[j]==posright+1):
if(ans[j-1]=="1"):
ans[j] = "1"
elif(d[j]<posright and d[j]>posleft):
if(posright - posleft + 1 == j):
ans[j] = "1"
if(d[j]<posleft):
posleft = d[j]
if(d[j]>posright):
posright = d[j]
print(''.join(ans[1:]))
|
You are given a permutation $p=[p_1, p_2, \ldots, p_n]$ of integers from $1$ to $n$. Let's call the number $m$ ($1 \le m \le n$) beautiful, if there exists two indices $l, r$ ($1 \le l \le r \le n$), such that the numbers $[p_l, p_{l+1}, \ldots, p_r]$ is a permutation of numbers $1, 2, \ldots, m$.
For example, let $p = [4, 5, 1, 3, 2, 6]$. In this case, the numbers $1, 3, 5, 6$ are beautiful and $2, 4$ are not. It is because: if $l = 3$ and $r = 3$ we will have a permutation $[1]$ for $m = 1$; if $l = 3$ and $r = 5$ we will have a permutation $[1, 3, 2]$ for $m = 3$; if $l = 1$ and $r = 5$ we will have a permutation $[4, 5, 1, 3, 2]$ for $m = 5$; if $l = 1$ and $r = 6$ we will have a permutation $[4, 5, 1, 3, 2, 6]$ for $m = 6$; it is impossible to take some $l$ and $r$, such that $[p_l, p_{l+1}, \ldots, p_r]$ is a permutation of numbers $1, 2, \ldots, m$ for $m = 2$ and for $m = 4$.
You are given a permutation $p=[p_1, p_2, \ldots, p_n]$. For all $m$ ($1 \le m \le n$) determine if it is a beautiful number or not.
-----Input-----
The first line contains the only integer $t$ ($1 \le t \le 1000$) — the number of test cases in the input. The next lines contain the description of test cases.
The first line of a test case contains a number $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of the given permutation $p$. The next line contains $n$ integers $p_1, p_2, \ldots, p_n$ ($1 \le p_i \le n$, all $p_i$ are different) — the given permutation $p$.
It is guaranteed, that the sum of $n$ from all test cases in the input doesn't exceed $2 \cdot 10^5$.
-----Output-----
Print $t$ lines — the answers to test cases in the order they are given in the input.
The answer to a test case is the string of length $n$, there the $i$-th character is equal to $1$ if $i$ is a beautiful number and is equal to $0$ if $i$ is not a beautiful number.
-----Example-----
Input
3
6
4 5 1 3 2 6
5
5 3 1 2 4
4
1 4 3 2
Output
101011
11111
1001
-----Note-----
The first test case is described in the problem statement.
In the second test case all numbers from $1$ to $5$ are beautiful: if $l = 3$ and $r = 3$ we will have a permutation $[1]$ for $m = 1$; if $l = 3$ and $r = 4$ we will have a permutation $[1, 2]$ for $m = 2$; if $l = 2$ and $r = 4$ we will have a permutation $[3, 1, 2]$ for $m = 3$; if $l = 2$ and $r = 5$ we will have a permutation $[3, 1, 2, 4]$ for $m = 4$; if $l = 1$ and $r = 5$ we will have a permutation $[5, 3, 1, 2, 4]$ for $m = 5$.
|
import sys
import math
import bisect
sys.setrecursionlimit(1000000000)
def input():
return sys.stdin.readline().strip()
def iinput():
return int(input())
def finput():
return float(input())
def tinput():
return input().split()
def rinput():
return map(int, tinput())
def rlinput():
return list(rinput())
def main():
n = iinput()
c = rlinput()
q, res, w, e = [0] * n, ['0'] * n, 0, n
for i in range(n):
q[c[i] - 1] = i
for i in range(n):
w = max(q[i], w)
e = min(q[i], e)
if w <= i + e:
res[i] = '1'
print(''.join(res))
for j in range(int(input())):
main()
|
You are given a permutation $p=[p_1, p_2, \ldots, p_n]$ of integers from $1$ to $n$. Let's call the number $m$ ($1 \le m \le n$) beautiful, if there exists two indices $l, r$ ($1 \le l \le r \le n$), such that the numbers $[p_l, p_{l+1}, \ldots, p_r]$ is a permutation of numbers $1, 2, \ldots, m$.
For example, let $p = [4, 5, 1, 3, 2, 6]$. In this case, the numbers $1, 3, 5, 6$ are beautiful and $2, 4$ are not. It is because: if $l = 3$ and $r = 3$ we will have a permutation $[1]$ for $m = 1$; if $l = 3$ and $r = 5$ we will have a permutation $[1, 3, 2]$ for $m = 3$; if $l = 1$ and $r = 5$ we will have a permutation $[4, 5, 1, 3, 2]$ for $m = 5$; if $l = 1$ and $r = 6$ we will have a permutation $[4, 5, 1, 3, 2, 6]$ for $m = 6$; it is impossible to take some $l$ and $r$, such that $[p_l, p_{l+1}, \ldots, p_r]$ is a permutation of numbers $1, 2, \ldots, m$ for $m = 2$ and for $m = 4$.
You are given a permutation $p=[p_1, p_2, \ldots, p_n]$. For all $m$ ($1 \le m \le n$) determine if it is a beautiful number or not.
-----Input-----
The first line contains the only integer $t$ ($1 \le t \le 1000$) — the number of test cases in the input. The next lines contain the description of test cases.
The first line of a test case contains a number $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of the given permutation $p$. The next line contains $n$ integers $p_1, p_2, \ldots, p_n$ ($1 \le p_i \le n$, all $p_i$ are different) — the given permutation $p$.
It is guaranteed, that the sum of $n$ from all test cases in the input doesn't exceed $2 \cdot 10^5$.
-----Output-----
Print $t$ lines — the answers to test cases in the order they are given in the input.
The answer to a test case is the string of length $n$, there the $i$-th character is equal to $1$ if $i$ is a beautiful number and is equal to $0$ if $i$ is not a beautiful number.
-----Example-----
Input
3
6
4 5 1 3 2 6
5
5 3 1 2 4
4
1 4 3 2
Output
101011
11111
1001
-----Note-----
The first test case is described in the problem statement.
In the second test case all numbers from $1$ to $5$ are beautiful: if $l = 3$ and $r = 3$ we will have a permutation $[1]$ for $m = 1$; if $l = 3$ and $r = 4$ we will have a permutation $[1, 2]$ for $m = 2$; if $l = 2$ and $r = 4$ we will have a permutation $[3, 1, 2]$ for $m = 3$; if $l = 2$ and $r = 5$ we will have a permutation $[3, 1, 2, 4]$ for $m = 4$; if $l = 1$ and $r = 5$ we will have a permutation $[5, 3, 1, 2, 4]$ for $m = 5$.
|
from math import floor, ceil
t = int(input())
for _ in range(t):
n = int(input())
a = list(map(int, input().split()))
pos = dict()
for p, i in enumerate(a):
pos[i] = p
minpos = [None] + [pos[1]] + [None]*(n-1)
maxpos = [None] + [pos[1]] + [None]*(n-1)
for i in range(2, n+1):
minpos[i] = min(minpos[i-1], pos[i])
maxpos[i] = max(maxpos[i-1], pos[i])
good = ['0']*n
for i in range(1, n+1):
if maxpos[i] - minpos[i] + 1 == i:
good[i-1] = '1'
print(''.join(good))
|
The sequence of $m$ integers is called the permutation if it contains all integers from $1$ to $m$ exactly once. The number $m$ is called the length of the permutation.
Dreamoon has two permutations $p_1$ and $p_2$ of non-zero lengths $l_1$ and $l_2$.
Now Dreamoon concatenates these two permutations into another sequence $a$ of length $l_1 + l_2$. First $l_1$ elements of $a$ is the permutation $p_1$ and next $l_2$ elements of $a$ is the permutation $p_2$.
You are given the sequence $a$, and you need to find two permutations $p_1$ and $p_2$. If there are several possible ways to restore them, you should find all of them. (Note that it is also possible that there will be no ways.)
-----Input-----
The first line contains an integer $t$ ($1 \le t \le 10\,000$) denoting the number of test cases in the input.
Each test case contains two lines. The first line contains one integer $n$ ($2 \leq n \leq 200\,000$): the length of $a$. The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \leq a_i \leq n-1$).
The total sum of $n$ is less than $200\,000$.
-----Output-----
For each test case, the first line of output should contain one integer $k$: the number of ways to divide $a$ into permutations $p_1$ and $p_2$.
Each of the next $k$ lines should contain two integers $l_1$ and $l_2$ ($1 \leq l_1, l_2 \leq n, l_1 + l_2 = n$), denoting, that it is possible to divide $a$ into two permutations of length $l_1$ and $l_2$ ($p_1$ is the first $l_1$ elements of $a$, and $p_2$ is the last $l_2$ elements of $a$). You can print solutions in any order.
-----Example-----
Input
6
5
1 4 3 2 1
6
2 4 1 3 2 1
4
2 1 1 3
4
1 3 3 1
12
2 1 3 4 5 6 7 8 9 1 10 2
3
1 1 1
Output
2
1 4
4 1
1
4 2
0
0
1
2 10
0
-----Note-----
In the first example, two possible ways to divide $a$ into permutations are $\{1\} + \{4, 3, 2, 1\}$ and $\{1,4,3,2\} + \{1\}$.
In the second example, the only way to divide $a$ into permutations is $\{2,4,1,3\} + \{2,1\}$.
In the third example, there are no possible ways.
|
def possible(a):
ans = set()
s = set()
lmax = 0
for i in range(len(a)):
lmax = max(lmax, a[i])
s.add(a[i])
if lmax == i + 1 and len(s) == i + 1:
ans.add(i + 1)
return ans
t = int(input())
for case_num in range(t):
n = int(input())
a = list(map(int, input().split(' ')))
left = possible(a)
a.reverse()
right = possible(a)
ans = []
for l in left:
if n - l in right:
ans.append(l)
print(len(ans))
for l in ans:
print(l, n - l)
|
The sequence of $m$ integers is called the permutation if it contains all integers from $1$ to $m$ exactly once. The number $m$ is called the length of the permutation.
Dreamoon has two permutations $p_1$ and $p_2$ of non-zero lengths $l_1$ and $l_2$.
Now Dreamoon concatenates these two permutations into another sequence $a$ of length $l_1 + l_2$. First $l_1$ elements of $a$ is the permutation $p_1$ and next $l_2$ elements of $a$ is the permutation $p_2$.
You are given the sequence $a$, and you need to find two permutations $p_1$ and $p_2$. If there are several possible ways to restore them, you should find all of them. (Note that it is also possible that there will be no ways.)
-----Input-----
The first line contains an integer $t$ ($1 \le t \le 10\,000$) denoting the number of test cases in the input.
Each test case contains two lines. The first line contains one integer $n$ ($2 \leq n \leq 200\,000$): the length of $a$. The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \leq a_i \leq n-1$).
The total sum of $n$ is less than $200\,000$.
-----Output-----
For each test case, the first line of output should contain one integer $k$: the number of ways to divide $a$ into permutations $p_1$ and $p_2$.
Each of the next $k$ lines should contain two integers $l_1$ and $l_2$ ($1 \leq l_1, l_2 \leq n, l_1 + l_2 = n$), denoting, that it is possible to divide $a$ into two permutations of length $l_1$ and $l_2$ ($p_1$ is the first $l_1$ elements of $a$, and $p_2$ is the last $l_2$ elements of $a$). You can print solutions in any order.
-----Example-----
Input
6
5
1 4 3 2 1
6
2 4 1 3 2 1
4
2 1 1 3
4
1 3 3 1
12
2 1 3 4 5 6 7 8 9 1 10 2
3
1 1 1
Output
2
1 4
4 1
1
4 2
0
0
1
2 10
0
-----Note-----
In the first example, two possible ways to divide $a$ into permutations are $\{1\} + \{4, 3, 2, 1\}$ and $\{1,4,3,2\} + \{1\}$.
In the second example, the only way to divide $a$ into permutations is $\{2,4,1,3\} + \{2,1\}$.
In the third example, there are no possible ways.
|
import sys
input=sys.stdin.readline
t=int(input())
for _ in range(t):
n=int(input())
aa=list(map(int,input().split()))
ss=set()
st=0
ind=1
pre=[0 for i in range(n)]
for i in range(n):
if aa[i] in ss:
break
ss.add(aa[i])
while ind<=len(ss):
if ind in ss:
ind+=1
else:
break
if len(ss)!=ind-1:
pre[i]=0
else:
pre[i]=ind
ind=1
# print(pre)
ss=set()
suff=[0 for i in range(n)]
for i in range(n-1,-1,-1):
if aa[i] in ss:
break
ss.add(aa[i])
while ind<=len(ss):
if ind in ss:
ind+=1
else:
break
if len(ss)!=ind-1:
suff[i]=0
else:
suff[i]=ind
tot=0
ans=[]
for i in range(n-1):
if pre[i]>0 and suff[i+1]>0:
tot+=1
ans.append([i+1,n-i-1])
print(tot)
for i in ans:
print(i[0],i[1])
|
The sequence of $m$ integers is called the permutation if it contains all integers from $1$ to $m$ exactly once. The number $m$ is called the length of the permutation.
Dreamoon has two permutations $p_1$ and $p_2$ of non-zero lengths $l_1$ and $l_2$.
Now Dreamoon concatenates these two permutations into another sequence $a$ of length $l_1 + l_2$. First $l_1$ elements of $a$ is the permutation $p_1$ and next $l_2$ elements of $a$ is the permutation $p_2$.
You are given the sequence $a$, and you need to find two permutations $p_1$ and $p_2$. If there are several possible ways to restore them, you should find all of them. (Note that it is also possible that there will be no ways.)
-----Input-----
The first line contains an integer $t$ ($1 \le t \le 10\,000$) denoting the number of test cases in the input.
Each test case contains two lines. The first line contains one integer $n$ ($2 \leq n \leq 200\,000$): the length of $a$. The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \leq a_i \leq n-1$).
The total sum of $n$ is less than $200\,000$.
-----Output-----
For each test case, the first line of output should contain one integer $k$: the number of ways to divide $a$ into permutations $p_1$ and $p_2$.
Each of the next $k$ lines should contain two integers $l_1$ and $l_2$ ($1 \leq l_1, l_2 \leq n, l_1 + l_2 = n$), denoting, that it is possible to divide $a$ into two permutations of length $l_1$ and $l_2$ ($p_1$ is the first $l_1$ elements of $a$, and $p_2$ is the last $l_2$ elements of $a$). You can print solutions in any order.
-----Example-----
Input
6
5
1 4 3 2 1
6
2 4 1 3 2 1
4
2 1 1 3
4
1 3 3 1
12
2 1 3 4 5 6 7 8 9 1 10 2
3
1 1 1
Output
2
1 4
4 1
1
4 2
0
0
1
2 10
0
-----Note-----
In the first example, two possible ways to divide $a$ into permutations are $\{1\} + \{4, 3, 2, 1\}$ and $\{1,4,3,2\} + \{1\}$.
In the second example, the only way to divide $a$ into permutations is $\{2,4,1,3\} + \{2,1\}$.
In the third example, there are no possible ways.
|
# @author
import sys
class BDreamoonLikesPermutations:
def solve(self):
for _ in range(int(input())):
def is_perm(a):
return len(set(a)) == len(a) and min(a) == 1 and max(a) == len(a)
n = int(input())
a = [int(_) for _ in input().split()]
done = set()
ans = set()
i = 0
for i in range(n):
if a[i] in done:
break
done.add(a[i])
if is_perm(a[:i]) and is_perm(a[i:]):
ans.add((i, n - i))
done = set()
for i in range(n - 1, -1, -1):
if a[i] in done:
break
done.add(a[i])
if is_perm(a[:i + 1]) and is_perm(a[i + 1:]):
ans.add((i + 1, n - i - 1))
print(len(ans))
for sol in ans:
print(*sol)
solver = BDreamoonLikesPermutations()
input = sys.stdin.readline
solver.solve()
|
The sequence of $m$ integers is called the permutation if it contains all integers from $1$ to $m$ exactly once. The number $m$ is called the length of the permutation.
Dreamoon has two permutations $p_1$ and $p_2$ of non-zero lengths $l_1$ and $l_2$.
Now Dreamoon concatenates these two permutations into another sequence $a$ of length $l_1 + l_2$. First $l_1$ elements of $a$ is the permutation $p_1$ and next $l_2$ elements of $a$ is the permutation $p_2$.
You are given the sequence $a$, and you need to find two permutations $p_1$ and $p_2$. If there are several possible ways to restore them, you should find all of them. (Note that it is also possible that there will be no ways.)
-----Input-----
The first line contains an integer $t$ ($1 \le t \le 10\,000$) denoting the number of test cases in the input.
Each test case contains two lines. The first line contains one integer $n$ ($2 \leq n \leq 200\,000$): the length of $a$. The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \leq a_i \leq n-1$).
The total sum of $n$ is less than $200\,000$.
-----Output-----
For each test case, the first line of output should contain one integer $k$: the number of ways to divide $a$ into permutations $p_1$ and $p_2$.
Each of the next $k$ lines should contain two integers $l_1$ and $l_2$ ($1 \leq l_1, l_2 \leq n, l_1 + l_2 = n$), denoting, that it is possible to divide $a$ into two permutations of length $l_1$ and $l_2$ ($p_1$ is the first $l_1$ elements of $a$, and $p_2$ is the last $l_2$ elements of $a$). You can print solutions in any order.
-----Example-----
Input
6
5
1 4 3 2 1
6
2 4 1 3 2 1
4
2 1 1 3
4
1 3 3 1
12
2 1 3 4 5 6 7 8 9 1 10 2
3
1 1 1
Output
2
1 4
4 1
1
4 2
0
0
1
2 10
0
-----Note-----
In the first example, two possible ways to divide $a$ into permutations are $\{1\} + \{4, 3, 2, 1\}$ and $\{1,4,3,2\} + \{1\}$.
In the second example, the only way to divide $a$ into permutations is $\{2,4,1,3\} + \{2,1\}$.
In the third example, there are no possible ways.
|
def readIntArray():
return list(map(int,input().split()))
t = int(input())
for _ in range(t):
n = int(input())
a = readIntArray()
mp = {}
for val in a:
if val not in mp:
mp[val] = 0
mp[val] += 1
l1 = max(a)
l2 = n - l1
if l2 <= 0:
print(0)
continue
good = True
for i in range(1, l2 + 1):
if i not in mp or mp[i] != 2:
good = False
break
for i in range(l2 + 1, l1 + 1):
if i not in mp or mp[i] != 1:
good = False
break
if not good:
print(0)
continue
mp = {}
ans = set()
cur = 0
st = set()
used = set()
for i in range(n):
if a[i] in used:
break
st.add(a[i])
used.add(a[i])
while cur + 1 in st:
st.remove(cur + 1)
cur += 1
if cur == l1 or cur == l2 and len(st) == 0:
ans.add((cur, n - cur))
print(len(ans))
for val in ans:
print(val[0], val[1])
|
The sequence of $m$ integers is called the permutation if it contains all integers from $1$ to $m$ exactly once. The number $m$ is called the length of the permutation.
Dreamoon has two permutations $p_1$ and $p_2$ of non-zero lengths $l_1$ and $l_2$.
Now Dreamoon concatenates these two permutations into another sequence $a$ of length $l_1 + l_2$. First $l_1$ elements of $a$ is the permutation $p_1$ and next $l_2$ elements of $a$ is the permutation $p_2$.
You are given the sequence $a$, and you need to find two permutations $p_1$ and $p_2$. If there are several possible ways to restore them, you should find all of them. (Note that it is also possible that there will be no ways.)
-----Input-----
The first line contains an integer $t$ ($1 \le t \le 10\,000$) denoting the number of test cases in the input.
Each test case contains two lines. The first line contains one integer $n$ ($2 \leq n \leq 200\,000$): the length of $a$. The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \leq a_i \leq n-1$).
The total sum of $n$ is less than $200\,000$.
-----Output-----
For each test case, the first line of output should contain one integer $k$: the number of ways to divide $a$ into permutations $p_1$ and $p_2$.
Each of the next $k$ lines should contain two integers $l_1$ and $l_2$ ($1 \leq l_1, l_2 \leq n, l_1 + l_2 = n$), denoting, that it is possible to divide $a$ into two permutations of length $l_1$ and $l_2$ ($p_1$ is the first $l_1$ elements of $a$, and $p_2$ is the last $l_2$ elements of $a$). You can print solutions in any order.
-----Example-----
Input
6
5
1 4 3 2 1
6
2 4 1 3 2 1
4
2 1 1 3
4
1 3 3 1
12
2 1 3 4 5 6 7 8 9 1 10 2
3
1 1 1
Output
2
1 4
4 1
1
4 2
0
0
1
2 10
0
-----Note-----
In the first example, two possible ways to divide $a$ into permutations are $\{1\} + \{4, 3, 2, 1\}$ and $\{1,4,3,2\} + \{1\}$.
In the second example, the only way to divide $a$ into permutations is $\{2,4,1,3\} + \{2,1\}$.
In the third example, there are no possible ways.
|
t = int(input())
for _ in range(t):
n = int(input())
a = [int(x) for x in input().split()]
mx = max(a)
sols = []
if mx < n:
l1 = list(sorted(a[:mx]))
l2 = list(sorted(a[mx:]))
rl1 = list(range(1, mx+1))
rl2 = list(range(1, n-mx+1))
if l1 == rl1 and l2 == rl2:
sols.append((mx, n - mx))
l1 = list(sorted(a[:n-mx]))
l2 = list(sorted(a[n-mx:]))
if mx*2 != n and l1 == rl2 and l2 == rl1:
sols.append((n-mx, mx))
print(len(sols))
for p in sols:
print(*p)
|
The sequence of $m$ integers is called the permutation if it contains all integers from $1$ to $m$ exactly once. The number $m$ is called the length of the permutation.
Dreamoon has two permutations $p_1$ and $p_2$ of non-zero lengths $l_1$ and $l_2$.
Now Dreamoon concatenates these two permutations into another sequence $a$ of length $l_1 + l_2$. First $l_1$ elements of $a$ is the permutation $p_1$ and next $l_2$ elements of $a$ is the permutation $p_2$.
You are given the sequence $a$, and you need to find two permutations $p_1$ and $p_2$. If there are several possible ways to restore them, you should find all of them. (Note that it is also possible that there will be no ways.)
-----Input-----
The first line contains an integer $t$ ($1 \le t \le 10\,000$) denoting the number of test cases in the input.
Each test case contains two lines. The first line contains one integer $n$ ($2 \leq n \leq 200\,000$): the length of $a$. The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \leq a_i \leq n-1$).
The total sum of $n$ is less than $200\,000$.
-----Output-----
For each test case, the first line of output should contain one integer $k$: the number of ways to divide $a$ into permutations $p_1$ and $p_2$.
Each of the next $k$ lines should contain two integers $l_1$ and $l_2$ ($1 \leq l_1, l_2 \leq n, l_1 + l_2 = n$), denoting, that it is possible to divide $a$ into two permutations of length $l_1$ and $l_2$ ($p_1$ is the first $l_1$ elements of $a$, and $p_2$ is the last $l_2$ elements of $a$). You can print solutions in any order.
-----Example-----
Input
6
5
1 4 3 2 1
6
2 4 1 3 2 1
4
2 1 1 3
4
1 3 3 1
12
2 1 3 4 5 6 7 8 9 1 10 2
3
1 1 1
Output
2
1 4
4 1
1
4 2
0
0
1
2 10
0
-----Note-----
In the first example, two possible ways to divide $a$ into permutations are $\{1\} + \{4, 3, 2, 1\}$ and $\{1,4,3,2\} + \{1\}$.
In the second example, the only way to divide $a$ into permutations is $\{2,4,1,3\} + \{2,1\}$.
In the third example, there are no possible ways.
|
from collections import deque
t = int(input())
for _ in range(t):
n = int(input())
liste = list(map(int, input().split()))
vis = [0 for i in range(n)]
can = [0 for i in range(n)]
can2 = [0 for i in range(n)]
maxi = 0
for i in range(1, n):
if (vis[liste[i-1]]):
break
vis[liste[i-1]] = 1
maxi = max(maxi, liste[i-1])
if (maxi == i):
can[maxi] = 1
liste = liste[::-1]
maxi = 0
vis = [0 for i in range(n)]
for i in range(1, n):
if (vis[liste[i-1]]):
break
vis[liste[i-1]] = 1
maxi = max(maxi, liste[i-1])
if (maxi == i):
can2[maxi] = 1
count = 0
for i in range(1, n):
if (can[i] and can2[n-i]):
count += 1
print(count)
for i in range(1, n):
if (can[i] and can2[n-i]):
print(i, n-i)
|
The sequence of $m$ integers is called the permutation if it contains all integers from $1$ to $m$ exactly once. The number $m$ is called the length of the permutation.
Dreamoon has two permutations $p_1$ and $p_2$ of non-zero lengths $l_1$ and $l_2$.
Now Dreamoon concatenates these two permutations into another sequence $a$ of length $l_1 + l_2$. First $l_1$ elements of $a$ is the permutation $p_1$ and next $l_2$ elements of $a$ is the permutation $p_2$.
You are given the sequence $a$, and you need to find two permutations $p_1$ and $p_2$. If there are several possible ways to restore them, you should find all of them. (Note that it is also possible that there will be no ways.)
-----Input-----
The first line contains an integer $t$ ($1 \le t \le 10\,000$) denoting the number of test cases in the input.
Each test case contains two lines. The first line contains one integer $n$ ($2 \leq n \leq 200\,000$): the length of $a$. The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \leq a_i \leq n-1$).
The total sum of $n$ is less than $200\,000$.
-----Output-----
For each test case, the first line of output should contain one integer $k$: the number of ways to divide $a$ into permutations $p_1$ and $p_2$.
Each of the next $k$ lines should contain two integers $l_1$ and $l_2$ ($1 \leq l_1, l_2 \leq n, l_1 + l_2 = n$), denoting, that it is possible to divide $a$ into two permutations of length $l_1$ and $l_2$ ($p_1$ is the first $l_1$ elements of $a$, and $p_2$ is the last $l_2$ elements of $a$). You can print solutions in any order.
-----Example-----
Input
6
5
1 4 3 2 1
6
2 4 1 3 2 1
4
2 1 1 3
4
1 3 3 1
12
2 1 3 4 5 6 7 8 9 1 10 2
3
1 1 1
Output
2
1 4
4 1
1
4 2
0
0
1
2 10
0
-----Note-----
In the first example, two possible ways to divide $a$ into permutations are $\{1\} + \{4, 3, 2, 1\}$ and $\{1,4,3,2\} + \{1\}$.
In the second example, the only way to divide $a$ into permutations is $\{2,4,1,3\} + \{2,1\}$.
In the third example, there are no possible ways.
|
for _ in range(int(input())):
n = int(input())
a = list(map(int, input().split()))
dpF = [0 for i in range(n)]
dpB = [0 for i in range(n)]
noRep = 1
r = {}
m = 0
for i in range(n):
if r.get(a[i]) == None:
r[a[i]] = 1
m = max(m, a[i])
if m == i + 1:
dpF[i] = 1
else:
break
r = {}
m = 0
for i in range(n - 1, -1, -1):
if r.get(a[i]) == None:
r[a[i]] = 1
m = max(m, a[i])
if m == n - i:
dpB[i] = 1
else:
break
# print(dpF)
# print(dpB)
ans = 0
ansList = []
for i in range(n - 1):
if dpF[i] == 1 and dpB[i + 1] == 1:
ans += 1
ansList.append([i + 1, n - i - 1])
print(ans)
for i in ansList:
print(i[0], i[1])
|
The sequence of $m$ integers is called the permutation if it contains all integers from $1$ to $m$ exactly once. The number $m$ is called the length of the permutation.
Dreamoon has two permutations $p_1$ and $p_2$ of non-zero lengths $l_1$ and $l_2$.
Now Dreamoon concatenates these two permutations into another sequence $a$ of length $l_1 + l_2$. First $l_1$ elements of $a$ is the permutation $p_1$ and next $l_2$ elements of $a$ is the permutation $p_2$.
You are given the sequence $a$, and you need to find two permutations $p_1$ and $p_2$. If there are several possible ways to restore them, you should find all of them. (Note that it is also possible that there will be no ways.)
-----Input-----
The first line contains an integer $t$ ($1 \le t \le 10\,000$) denoting the number of test cases in the input.
Each test case contains two lines. The first line contains one integer $n$ ($2 \leq n \leq 200\,000$): the length of $a$. The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \leq a_i \leq n-1$).
The total sum of $n$ is less than $200\,000$.
-----Output-----
For each test case, the first line of output should contain one integer $k$: the number of ways to divide $a$ into permutations $p_1$ and $p_2$.
Each of the next $k$ lines should contain two integers $l_1$ and $l_2$ ($1 \leq l_1, l_2 \leq n, l_1 + l_2 = n$), denoting, that it is possible to divide $a$ into two permutations of length $l_1$ and $l_2$ ($p_1$ is the first $l_1$ elements of $a$, and $p_2$ is the last $l_2$ elements of $a$). You can print solutions in any order.
-----Example-----
Input
6
5
1 4 3 2 1
6
2 4 1 3 2 1
4
2 1 1 3
4
1 3 3 1
12
2 1 3 4 5 6 7 8 9 1 10 2
3
1 1 1
Output
2
1 4
4 1
1
4 2
0
0
1
2 10
0
-----Note-----
In the first example, two possible ways to divide $a$ into permutations are $\{1\} + \{4, 3, 2, 1\}$ and $\{1,4,3,2\} + \{1\}$.
In the second example, the only way to divide $a$ into permutations is $\{2,4,1,3\} + \{2,1\}$.
In the third example, there are no possible ways.
|
from math import *
mod = 1000000007
for zz in range(int(input())):
n = int(input())
a = [int(i) for i in input().split()]
ans = []
cs = set()
d = {}
c = 0
for i in range(n):
if a[i] not in d:
c += 1
d[a[i]] = 0
d[a[i]] += 1
mv = 0
m = [0] * n
m[-1] = a[-1]
for i in range(n - 2, -1, -1):
m[i] = max(m[i + 1], a[i])
for i in range(n):
mv = max(a[i], mv)
if a[i] in cs:
break
cs.add(a[i])
d[a[i]] -= 1
if d[a[i]] <= 0:
c -= 1
if mv == i + 1 and c == n - i - 1 and m[i + 1] == n - i - 1:
ans.append(i)
print(len(ans))
for i in ans:
print(i + 1, n - i - 1)
|
The sequence of $m$ integers is called the permutation if it contains all integers from $1$ to $m$ exactly once. The number $m$ is called the length of the permutation.
Dreamoon has two permutations $p_1$ and $p_2$ of non-zero lengths $l_1$ and $l_2$.
Now Dreamoon concatenates these two permutations into another sequence $a$ of length $l_1 + l_2$. First $l_1$ elements of $a$ is the permutation $p_1$ and next $l_2$ elements of $a$ is the permutation $p_2$.
You are given the sequence $a$, and you need to find two permutations $p_1$ and $p_2$. If there are several possible ways to restore them, you should find all of them. (Note that it is also possible that there will be no ways.)
-----Input-----
The first line contains an integer $t$ ($1 \le t \le 10\,000$) denoting the number of test cases in the input.
Each test case contains two lines. The first line contains one integer $n$ ($2 \leq n \leq 200\,000$): the length of $a$. The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \leq a_i \leq n-1$).
The total sum of $n$ is less than $200\,000$.
-----Output-----
For each test case, the first line of output should contain one integer $k$: the number of ways to divide $a$ into permutations $p_1$ and $p_2$.
Each of the next $k$ lines should contain two integers $l_1$ and $l_2$ ($1 \leq l_1, l_2 \leq n, l_1 + l_2 = n$), denoting, that it is possible to divide $a$ into two permutations of length $l_1$ and $l_2$ ($p_1$ is the first $l_1$ elements of $a$, and $p_2$ is the last $l_2$ elements of $a$). You can print solutions in any order.
-----Example-----
Input
6
5
1 4 3 2 1
6
2 4 1 3 2 1
4
2 1 1 3
4
1 3 3 1
12
2 1 3 4 5 6 7 8 9 1 10 2
3
1 1 1
Output
2
1 4
4 1
1
4 2
0
0
1
2 10
0
-----Note-----
In the first example, two possible ways to divide $a$ into permutations are $\{1\} + \{4, 3, 2, 1\}$ and $\{1,4,3,2\} + \{1\}$.
In the second example, the only way to divide $a$ into permutations is $\{2,4,1,3\} + \{2,1\}$.
In the third example, there are no possible ways.
|
def per(X):
S=set(X)
if not len(X)==len(S):
return False
for i in range(1,len(X)+1):
if i not in S: return False
return True
for y in range(int(input())):
n=int(input())
L=list(map(int,input().split()))
m=max(L)
r=[]
if n!=m:
if per(L[:m]) and per(L[m:]):
r.append((m,n-m))
if per(L[-m:]) and per(L[:-m]):
r.append((n-m,m))
r=list(set(r))
print(len(r))
for a,b in r:
print(a,b)
|
The sequence of $m$ integers is called the permutation if it contains all integers from $1$ to $m$ exactly once. The number $m$ is called the length of the permutation.
Dreamoon has two permutations $p_1$ and $p_2$ of non-zero lengths $l_1$ and $l_2$.
Now Dreamoon concatenates these two permutations into another sequence $a$ of length $l_1 + l_2$. First $l_1$ elements of $a$ is the permutation $p_1$ and next $l_2$ elements of $a$ is the permutation $p_2$.
You are given the sequence $a$, and you need to find two permutations $p_1$ and $p_2$. If there are several possible ways to restore them, you should find all of them. (Note that it is also possible that there will be no ways.)
-----Input-----
The first line contains an integer $t$ ($1 \le t \le 10\,000$) denoting the number of test cases in the input.
Each test case contains two lines. The first line contains one integer $n$ ($2 \leq n \leq 200\,000$): the length of $a$. The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \leq a_i \leq n-1$).
The total sum of $n$ is less than $200\,000$.
-----Output-----
For each test case, the first line of output should contain one integer $k$: the number of ways to divide $a$ into permutations $p_1$ and $p_2$.
Each of the next $k$ lines should contain two integers $l_1$ and $l_2$ ($1 \leq l_1, l_2 \leq n, l_1 + l_2 = n$), denoting, that it is possible to divide $a$ into two permutations of length $l_1$ and $l_2$ ($p_1$ is the first $l_1$ elements of $a$, and $p_2$ is the last $l_2$ elements of $a$). You can print solutions in any order.
-----Example-----
Input
6
5
1 4 3 2 1
6
2 4 1 3 2 1
4
2 1 1 3
4
1 3 3 1
12
2 1 3 4 5 6 7 8 9 1 10 2
3
1 1 1
Output
2
1 4
4 1
1
4 2
0
0
1
2 10
0
-----Note-----
In the first example, two possible ways to divide $a$ into permutations are $\{1\} + \{4, 3, 2, 1\}$ and $\{1,4,3,2\} + \{1\}$.
In the second example, the only way to divide $a$ into permutations is $\{2,4,1,3\} + \{2,1\}$.
In the third example, there are no possible ways.
|
import sys
input = sys.stdin.readline
t = int(input())
for _ in range(t):
n = int(input())
a = list(map(int, input().split()))
seen = [False] * (n+1)
ans = set()
for i, x in enumerate(a):
if seen[x]:
if sorted(a[:i]) == list(range(1, i+1)) and sorted(a[i:]) == list(range(1, n-i+1)):
ans.add((i, n-i))
break
seen[x] = True
seen = [False] * (n+1)
for i, x in list(enumerate(a))[::-1]:
if seen[x]:
if sorted(a[:i+1]) == list(range(1, i+2)) and sorted(a[i+1:]) == list(range(1, n-i)):
ans.add((i+1, n-i-1))
break
seen[x] = True
print(len(ans))
for l1, l2 in ans:
print(l1, l2)
|
The sequence of $m$ integers is called the permutation if it contains all integers from $1$ to $m$ exactly once. The number $m$ is called the length of the permutation.
Dreamoon has two permutations $p_1$ and $p_2$ of non-zero lengths $l_1$ and $l_2$.
Now Dreamoon concatenates these two permutations into another sequence $a$ of length $l_1 + l_2$. First $l_1$ elements of $a$ is the permutation $p_1$ and next $l_2$ elements of $a$ is the permutation $p_2$.
You are given the sequence $a$, and you need to find two permutations $p_1$ and $p_2$. If there are several possible ways to restore them, you should find all of them. (Note that it is also possible that there will be no ways.)
-----Input-----
The first line contains an integer $t$ ($1 \le t \le 10\,000$) denoting the number of test cases in the input.
Each test case contains two lines. The first line contains one integer $n$ ($2 \leq n \leq 200\,000$): the length of $a$. The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \leq a_i \leq n-1$).
The total sum of $n$ is less than $200\,000$.
-----Output-----
For each test case, the first line of output should contain one integer $k$: the number of ways to divide $a$ into permutations $p_1$ and $p_2$.
Each of the next $k$ lines should contain two integers $l_1$ and $l_2$ ($1 \leq l_1, l_2 \leq n, l_1 + l_2 = n$), denoting, that it is possible to divide $a$ into two permutations of length $l_1$ and $l_2$ ($p_1$ is the first $l_1$ elements of $a$, and $p_2$ is the last $l_2$ elements of $a$). You can print solutions in any order.
-----Example-----
Input
6
5
1 4 3 2 1
6
2 4 1 3 2 1
4
2 1 1 3
4
1 3 3 1
12
2 1 3 4 5 6 7 8 9 1 10 2
3
1 1 1
Output
2
1 4
4 1
1
4 2
0
0
1
2 10
0
-----Note-----
In the first example, two possible ways to divide $a$ into permutations are $\{1\} + \{4, 3, 2, 1\}$ and $\{1,4,3,2\} + \{1\}$.
In the second example, the only way to divide $a$ into permutations is $\{2,4,1,3\} + \{2,1\}$.
In the third example, there are no possible ways.
|
import sys
input=sys.stdin.readline
t=int(input())
for _ in range(t):
n=int(input())
arr=list(map(int,input().split()))
d=dict()
demand=1
pre=[0]*n
post=[0]*n
for i in range(n):
d[arr[i]]=1
if(demand in d):
while(demand in d):
demand+=1
pre[i]=demand-1
d2=dict()
#print(pre)
demand=1
for i in range(n-1,-1,-1):
d2[arr[i]]=1
if(demand in d2):
while(demand in d2):
demand+=1
post[i]=demand-1
#print(post)
l=[]
for i in range(1,n):
if(post[i]+pre[i-1]==n):
l+=[[pre[i-1],post[i]]]
print(len(l))
for i in l:
print(*i)
|
The sequence of $m$ integers is called the permutation if it contains all integers from $1$ to $m$ exactly once. The number $m$ is called the length of the permutation.
Dreamoon has two permutations $p_1$ and $p_2$ of non-zero lengths $l_1$ and $l_2$.
Now Dreamoon concatenates these two permutations into another sequence $a$ of length $l_1 + l_2$. First $l_1$ elements of $a$ is the permutation $p_1$ and next $l_2$ elements of $a$ is the permutation $p_2$.
You are given the sequence $a$, and you need to find two permutations $p_1$ and $p_2$. If there are several possible ways to restore them, you should find all of them. (Note that it is also possible that there will be no ways.)
-----Input-----
The first line contains an integer $t$ ($1 \le t \le 10\,000$) denoting the number of test cases in the input.
Each test case contains two lines. The first line contains one integer $n$ ($2 \leq n \leq 200\,000$): the length of $a$. The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \leq a_i \leq n-1$).
The total sum of $n$ is less than $200\,000$.
-----Output-----
For each test case, the first line of output should contain one integer $k$: the number of ways to divide $a$ into permutations $p_1$ and $p_2$.
Each of the next $k$ lines should contain two integers $l_1$ and $l_2$ ($1 \leq l_1, l_2 \leq n, l_1 + l_2 = n$), denoting, that it is possible to divide $a$ into two permutations of length $l_1$ and $l_2$ ($p_1$ is the first $l_1$ elements of $a$, and $p_2$ is the last $l_2$ elements of $a$). You can print solutions in any order.
-----Example-----
Input
6
5
1 4 3 2 1
6
2 4 1 3 2 1
4
2 1 1 3
4
1 3 3 1
12
2 1 3 4 5 6 7 8 9 1 10 2
3
1 1 1
Output
2
1 4
4 1
1
4 2
0
0
1
2 10
0
-----Note-----
In the first example, two possible ways to divide $a$ into permutations are $\{1\} + \{4, 3, 2, 1\}$ and $\{1,4,3,2\} + \{1\}$.
In the second example, the only way to divide $a$ into permutations is $\{2,4,1,3\} + \{2,1\}$.
In the third example, there are no possible ways.
|
import heapq, sys
def ps(l):
n = len(l)
nxt = 1
heap = []
ans = []
for i in range(n):
heapq.heappush(heap, l[i])
while heap and heap[0] == nxt:
nxt += 1
heapq.heappop(heap)
if not heap:
ans.append(i)
return ans
for q in range(int(sys.stdin.readline())):
n = int(sys.stdin.readline())
d = [int(i) for i in sys.stdin.readline().split()]
st = set(ps(d))
# print(st)
d.reverse()
anss = []
ap = ps(d)
# print(ap)
for a in ap:
b = n-2-a
if b in st:
anss.append(str(b+1)+' '+ str(n - b - 1) + '\n')
sys.stdout.write(str(len(anss)) + '\n')
sys.stdout.write(''.join(anss))
|
Arthur owns a ski resort on a mountain. There are $n$ landing spots on the mountain numbered from $1$ to $n$ from the top to the foot of the mountain. The spots are connected with one-directional ski tracks. All tracks go towards the foot of the mountain, so there are no directed cycles formed by the tracks. There are at most two tracks leaving each spot, but many tracks may enter the same spot.
A skier can start skiing from one spot and stop in another spot if there is a sequence of tracks that lead from the starting spot and end in the ending spot. Unfortunately, recently there were many accidents, because the structure of the resort allows a skier to go through dangerous paths, by reaching high speed and endangering himself and the other customers. Here, a path is called dangerous, if it consists of at least two tracks.
Arthur wants to secure his customers by closing some of the spots in a way that there are no dangerous paths in the resort. When a spot is closed, all tracks entering and leaving that spot become unusable.
Formally, after closing some of the spots, there should not be a path that consists of two or more tracks.
Arthur doesn't want to close too many spots. He will be happy to find any way to close at most $\frac{4}{7}n$ spots so that the remaining part is safe. Help him find any suitable way to do so.
-----Input-----
The first line contains a single positive integer $T$ — the number of test cases. $T$ test case description follows.
The first line of each description contains two integers $n$ and $m$ ($1 \leq n \leq 2 \cdot 10^5$) — the number of landing spots and tracks respectively.
The following $m$ lines describe the tracks. Each of these lines contains two integers $x$ and $y$ ($1 \leq x < y \leq n$) — indices of the starting and finishing spots for the respective track. It is guaranteed that at most two tracks start at each spot. There may be tracks in which starting and finishing spots both coincide.
It is guaranteed that the sum of $n$ over all test cases does not exceed $2 \cdot 10^5$.
-----Output-----
For each test case, print a single integer $k$ ($0 \leq k \leq \frac{4}{7}n$) — the number of spots to be closed. In the next line, print $k$ distinct integers — indices of all spots to be closed, in any order.
If there are several answers, you may output any of them. Note that you don't have to minimize $k$. It can be shown that a suitable answer always exists.
-----Example-----
Input
2
4 6
1 2
1 3
2 3
2 4
3 4
3 4
7 6
1 2
1 3
2 4
2 5
3 6
3 7
Output
2
3 4
4
4 5 6 7
-----Note-----
In the first sample case, closing any two spots is suitable.
In the second sample case, closing only the spot $1$ is also suitable.
|
import sys
input = sys.stdin.readline
for f in range(int(input())):
n,m=list(map(int,input().split()))
neig=[0]*n
for i in range(n):
neig[i]=[0]
for i in range(m):
a,b=list(map(int,input().split()))
a-=1
b-=1
neig[a][0]+=1
neig[a].append(b)
lev=[1]*n
for i in range(n):
for j in range(1,neig[i][0]+1):
x=lev[i]+1
if x==4:
x=1
lev[neig[i][j]]=max(lev[neig[i][j]],x)
sol=0
s=[]
for i in range(n):
if lev[i]==3:
sol+=1
s.append(i+1)
print(sol)
print(*s)
|
Arthur owns a ski resort on a mountain. There are $n$ landing spots on the mountain numbered from $1$ to $n$ from the top to the foot of the mountain. The spots are connected with one-directional ski tracks. All tracks go towards the foot of the mountain, so there are no directed cycles formed by the tracks. There are at most two tracks leaving each spot, but many tracks may enter the same spot.
A skier can start skiing from one spot and stop in another spot if there is a sequence of tracks that lead from the starting spot and end in the ending spot. Unfortunately, recently there were many accidents, because the structure of the resort allows a skier to go through dangerous paths, by reaching high speed and endangering himself and the other customers. Here, a path is called dangerous, if it consists of at least two tracks.
Arthur wants to secure his customers by closing some of the spots in a way that there are no dangerous paths in the resort. When a spot is closed, all tracks entering and leaving that spot become unusable.
Formally, after closing some of the spots, there should not be a path that consists of two or more tracks.
Arthur doesn't want to close too many spots. He will be happy to find any way to close at most $\frac{4}{7}n$ spots so that the remaining part is safe. Help him find any suitable way to do so.
-----Input-----
The first line contains a single positive integer $T$ — the number of test cases. $T$ test case description follows.
The first line of each description contains two integers $n$ and $m$ ($1 \leq n \leq 2 \cdot 10^5$) — the number of landing spots and tracks respectively.
The following $m$ lines describe the tracks. Each of these lines contains two integers $x$ and $y$ ($1 \leq x < y \leq n$) — indices of the starting and finishing spots for the respective track. It is guaranteed that at most two tracks start at each spot. There may be tracks in which starting and finishing spots both coincide.
It is guaranteed that the sum of $n$ over all test cases does not exceed $2 \cdot 10^5$.
-----Output-----
For each test case, print a single integer $k$ ($0 \leq k \leq \frac{4}{7}n$) — the number of spots to be closed. In the next line, print $k$ distinct integers — indices of all spots to be closed, in any order.
If there are several answers, you may output any of them. Note that you don't have to minimize $k$. It can be shown that a suitable answer always exists.
-----Example-----
Input
2
4 6
1 2
1 3
2 3
2 4
3 4
3 4
7 6
1 2
1 3
2 4
2 5
3 6
3 7
Output
2
3 4
4
4 5 6 7
-----Note-----
In the first sample case, closing any two spots is suitable.
In the second sample case, closing only the spot $1$ is also suitable.
|
import sys
input = sys.stdin.readline
from heapq import heapify,heappush,heappop
t = int(input())
for _ in range(t):
n,m = map(int,input().split())
ab = [list(map(int,input().split())) for i in range(m)]
go = [[] for i in range(n+1)]
come = [[] for i in range(n+1)]
for a,b in ab:
go[a].append(b)
come[b].append(a)
exist = [1]*(n+1)
flg = [10]*(n+1)
for i in range(1,n+1):
if flg[i] == 10:
flg[i] = 2
if flg[i] == 0:
exist[i] = 0
if go[i]:
if flg[i] == 0:
for j in go[i]:
flg[j] = min(flg[j],2)
else:
for j in go[i]:
flg[j] = min(flg[j],flg[i]-1)
print(exist.count(0))
ansls = []
for i in range(1,n+1):
if exist[i] == 0:
ansls.append(i)
print(*ansls)
|
Arthur owns a ski resort on a mountain. There are $n$ landing spots on the mountain numbered from $1$ to $n$ from the top to the foot of the mountain. The spots are connected with one-directional ski tracks. All tracks go towards the foot of the mountain, so there are no directed cycles formed by the tracks. There are at most two tracks leaving each spot, but many tracks may enter the same spot.
A skier can start skiing from one spot and stop in another spot if there is a sequence of tracks that lead from the starting spot and end in the ending spot. Unfortunately, recently there were many accidents, because the structure of the resort allows a skier to go through dangerous paths, by reaching high speed and endangering himself and the other customers. Here, a path is called dangerous, if it consists of at least two tracks.
Arthur wants to secure his customers by closing some of the spots in a way that there are no dangerous paths in the resort. When a spot is closed, all tracks entering and leaving that spot become unusable.
Formally, after closing some of the spots, there should not be a path that consists of two or more tracks.
Arthur doesn't want to close too many spots. He will be happy to find any way to close at most $\frac{4}{7}n$ spots so that the remaining part is safe. Help him find any suitable way to do so.
-----Input-----
The first line contains a single positive integer $T$ — the number of test cases. $T$ test case description follows.
The first line of each description contains two integers $n$ and $m$ ($1 \leq n \leq 2 \cdot 10^5$) — the number of landing spots and tracks respectively.
The following $m$ lines describe the tracks. Each of these lines contains two integers $x$ and $y$ ($1 \leq x < y \leq n$) — indices of the starting and finishing spots for the respective track. It is guaranteed that at most two tracks start at each spot. There may be tracks in which starting and finishing spots both coincide.
It is guaranteed that the sum of $n$ over all test cases does not exceed $2 \cdot 10^5$.
-----Output-----
For each test case, print a single integer $k$ ($0 \leq k \leq \frac{4}{7}n$) — the number of spots to be closed. In the next line, print $k$ distinct integers — indices of all spots to be closed, in any order.
If there are several answers, you may output any of them. Note that you don't have to minimize $k$. It can be shown that a suitable answer always exists.
-----Example-----
Input
2
4 6
1 2
1 3
2 3
2 4
3 4
3 4
7 6
1 2
1 3
2 4
2 5
3 6
3 7
Output
2
3 4
4
4 5 6 7
-----Note-----
In the first sample case, closing any two spots is suitable.
In the second sample case, closing only the spot $1$ is also suitable.
|
import sys
T = int(sys.stdin.readline().strip())
for t in range (0, T):
n, m = list(map(int, sys.stdin.readline().strip().split()))
P = [[] for i in range (0, n)]
G = [0] * n
for i in range (0, m):
x, y = list(map(int, sys.stdin.readline().strip().split()))
x, y = x-1, y-1
P[y].append(x)
ans = []
for i in range (0, n):
for j in P[i]:
for k in P[j]:
if G[j] == 0 and G[k] == 0:
if G[i] == 0:
ans.append(str(i+1))
G[i] = 1
print(len(ans))
print(" ".join(ans))
|
Arthur owns a ski resort on a mountain. There are $n$ landing spots on the mountain numbered from $1$ to $n$ from the top to the foot of the mountain. The spots are connected with one-directional ski tracks. All tracks go towards the foot of the mountain, so there are no directed cycles formed by the tracks. There are at most two tracks leaving each spot, but many tracks may enter the same spot.
A skier can start skiing from one spot and stop in another spot if there is a sequence of tracks that lead from the starting spot and end in the ending spot. Unfortunately, recently there were many accidents, because the structure of the resort allows a skier to go through dangerous paths, by reaching high speed and endangering himself and the other customers. Here, a path is called dangerous, if it consists of at least two tracks.
Arthur wants to secure his customers by closing some of the spots in a way that there are no dangerous paths in the resort. When a spot is closed, all tracks entering and leaving that spot become unusable.
Formally, after closing some of the spots, there should not be a path that consists of two or more tracks.
Arthur doesn't want to close too many spots. He will be happy to find any way to close at most $\frac{4}{7}n$ spots so that the remaining part is safe. Help him find any suitable way to do so.
-----Input-----
The first line contains a single positive integer $T$ — the number of test cases. $T$ test case description follows.
The first line of each description contains two integers $n$ and $m$ ($1 \leq n \leq 2 \cdot 10^5$) — the number of landing spots and tracks respectively.
The following $m$ lines describe the tracks. Each of these lines contains two integers $x$ and $y$ ($1 \leq x < y \leq n$) — indices of the starting and finishing spots for the respective track. It is guaranteed that at most two tracks start at each spot. There may be tracks in which starting and finishing spots both coincide.
It is guaranteed that the sum of $n$ over all test cases does not exceed $2 \cdot 10^5$.
-----Output-----
For each test case, print a single integer $k$ ($0 \leq k \leq \frac{4}{7}n$) — the number of spots to be closed. In the next line, print $k$ distinct integers — indices of all spots to be closed, in any order.
If there are several answers, you may output any of them. Note that you don't have to minimize $k$. It can be shown that a suitable answer always exists.
-----Example-----
Input
2
4 6
1 2
1 3
2 3
2 4
3 4
3 4
7 6
1 2
1 3
2 4
2 5
3 6
3 7
Output
2
3 4
4
4 5 6 7
-----Note-----
In the first sample case, closing any two spots is suitable.
In the second sample case, closing only the spot $1$ is also suitable.
|
import sys
inputr = lambda: sys.stdin.readline().rstrip('\n')
input = sys.stdin.readline
for _ in range(int(input())):
n, m = list(map(int, input().split()))
adj = [[] for _ in range(n)]
for _ in range(m):
a, b = list(map(int, input().split()))
a -= 1
b -= 1
adj[a].append(b)
LP = [0] * n
for i in range(n):
if LP[i] < 2:
for j in adj[i]:
LP[j] = max(LP[j], LP[i] + 1)
r = [i+1 for i in range(n) if LP[i] >= 2]
print(len(r))
print(' '.join(map(str, r)))
assert 7 * len(r) <= 4 * n
|
Arthur owns a ski resort on a mountain. There are $n$ landing spots on the mountain numbered from $1$ to $n$ from the top to the foot of the mountain. The spots are connected with one-directional ski tracks. All tracks go towards the foot of the mountain, so there are no directed cycles formed by the tracks. There are at most two tracks leaving each spot, but many tracks may enter the same spot.
A skier can start skiing from one spot and stop in another spot if there is a sequence of tracks that lead from the starting spot and end in the ending spot. Unfortunately, recently there were many accidents, because the structure of the resort allows a skier to go through dangerous paths, by reaching high speed and endangering himself and the other customers. Here, a path is called dangerous, if it consists of at least two tracks.
Arthur wants to secure his customers by closing some of the spots in a way that there are no dangerous paths in the resort. When a spot is closed, all tracks entering and leaving that spot become unusable.
Formally, after closing some of the spots, there should not be a path that consists of two or more tracks.
Arthur doesn't want to close too many spots. He will be happy to find any way to close at most $\frac{4}{7}n$ spots so that the remaining part is safe. Help him find any suitable way to do so.
-----Input-----
The first line contains a single positive integer $T$ — the number of test cases. $T$ test case description follows.
The first line of each description contains two integers $n$ and $m$ ($1 \leq n \leq 2 \cdot 10^5$) — the number of landing spots and tracks respectively.
The following $m$ lines describe the tracks. Each of these lines contains two integers $x$ and $y$ ($1 \leq x < y \leq n$) — indices of the starting and finishing spots for the respective track. It is guaranteed that at most two tracks start at each spot. There may be tracks in which starting and finishing spots both coincide.
It is guaranteed that the sum of $n$ over all test cases does not exceed $2 \cdot 10^5$.
-----Output-----
For each test case, print a single integer $k$ ($0 \leq k \leq \frac{4}{7}n$) — the number of spots to be closed. In the next line, print $k$ distinct integers — indices of all spots to be closed, in any order.
If there are several answers, you may output any of them. Note that you don't have to minimize $k$. It can be shown that a suitable answer always exists.
-----Example-----
Input
2
4 6
1 2
1 3
2 3
2 4
3 4
3 4
7 6
1 2
1 3
2 4
2 5
3 6
3 7
Output
2
3 4
4
4 5 6 7
-----Note-----
In the first sample case, closing any two spots is suitable.
In the second sample case, closing only the spot $1$ is also suitable.
|
import sys
inputr = lambda: sys.stdin.readline().rstrip('\n')
input = sys.stdin.readline
for _ in range(int(input())):
n, m = list(map(int, input().split()))
adj = [[] for _ in range(n)]
for _ in range(m):
a, b = list(map(int, input().split()))
a -= 1
b -= 1
adj[a].append(b)
LP = [0] * n
r = []
for i in range(n):
if LP[i] < 2:
for j in adj[i]:
LP[j] = max(LP[j], LP[i] + 1)
else:
r.append(str(i+1))
print(len(r))
print(*r)
assert 7 * len(r) <= 4 * n
|
Arthur owns a ski resort on a mountain. There are $n$ landing spots on the mountain numbered from $1$ to $n$ from the top to the foot of the mountain. The spots are connected with one-directional ski tracks. All tracks go towards the foot of the mountain, so there are no directed cycles formed by the tracks. There are at most two tracks leaving each spot, but many tracks may enter the same spot.
A skier can start skiing from one spot and stop in another spot if there is a sequence of tracks that lead from the starting spot and end in the ending spot. Unfortunately, recently there were many accidents, because the structure of the resort allows a skier to go through dangerous paths, by reaching high speed and endangering himself and the other customers. Here, a path is called dangerous, if it consists of at least two tracks.
Arthur wants to secure his customers by closing some of the spots in a way that there are no dangerous paths in the resort. When a spot is closed, all tracks entering and leaving that spot become unusable.
Formally, after closing some of the spots, there should not be a path that consists of two or more tracks.
Arthur doesn't want to close too many spots. He will be happy to find any way to close at most $\frac{4}{7}n$ spots so that the remaining part is safe. Help him find any suitable way to do so.
-----Input-----
The first line contains a single positive integer $T$ — the number of test cases. $T$ test case description follows.
The first line of each description contains two integers $n$ and $m$ ($1 \leq n \leq 2 \cdot 10^5$) — the number of landing spots and tracks respectively.
The following $m$ lines describe the tracks. Each of these lines contains two integers $x$ and $y$ ($1 \leq x < y \leq n$) — indices of the starting and finishing spots for the respective track. It is guaranteed that at most two tracks start at each spot. There may be tracks in which starting and finishing spots both coincide.
It is guaranteed that the sum of $n$ over all test cases does not exceed $2 \cdot 10^5$.
-----Output-----
For each test case, print a single integer $k$ ($0 \leq k \leq \frac{4}{7}n$) — the number of spots to be closed. In the next line, print $k$ distinct integers — indices of all spots to be closed, in any order.
If there are several answers, you may output any of them. Note that you don't have to minimize $k$. It can be shown that a suitable answer always exists.
-----Example-----
Input
2
4 6
1 2
1 3
2 3
2 4
3 4
3 4
7 6
1 2
1 3
2 4
2 5
3 6
3 7
Output
2
3 4
4
4 5 6 7
-----Note-----
In the first sample case, closing any two spots is suitable.
In the second sample case, closing only the spot $1$ is also suitable.
|
#!/usr/bin/env python3
import sys
input = sys.stdin.readline
from collections import deque
class DirectedGraph:
def __init__(self, adj):
self.n = len(adj)
self.adj = adj
self.is_asyclic = False
self.max_path_len = None
def topological_sort(self):
indegree = [0] * self.n
for vs in self.adj:
for dest in vs:
indegree[dest] += 1
zero_v = []
for v, indeg in enumerate(indegree):
if indeg == 0:
zero_v.append(v)
max_path_len = 1
tp_sorted = []
to_be_added = []
while True:
while zero_v:
v = zero_v.pop()
tp_sorted.append(v)
for dest in self.adj[v]:
indegree[dest] -= 1
if indegree[dest] == 0:
to_be_added.append(dest)
if len(to_be_added) > 0:
zero_v.extend(to_be_added)
to_be_added = []
max_path_len += 1
else:
break
if len(tp_sorted) == self.n:
self.is_asyclic = True
self.max_path_len = max_path_len
return tp_sorted
else:
self.is_asyclic = False
return None
t = int(input())
for case in range(t):
n, m = map(int, input().split())
forward = [[] for _ in range(n)]
backward = [[] for _ in range(n)]
seen = set()
for _ in range(m):
u, v = map(int, input().split())
u -= 1; v -= 1
if (u, v) in seen:
continue
seen.add((u, v))
forward[u].append(v)
backward[v].append(u)
DG = DirectedGraph(forward)
tps = DG.topological_sort()
state = [-1] * n
state[0] = 0
for v in tps:
if len(backward[v]) == 0:
state[v] = 0
for pv in backward[v]:
state[v] = max(state[v], (state[pv] + 1) % 3)
ans = []
for i, color in enumerate(state):
if color == 2:
ans.append(i + 1)
print(len(ans))
print(*ans)
|
Arthur owns a ski resort on a mountain. There are $n$ landing spots on the mountain numbered from $1$ to $n$ from the top to the foot of the mountain. The spots are connected with one-directional ski tracks. All tracks go towards the foot of the mountain, so there are no directed cycles formed by the tracks. There are at most two tracks leaving each spot, but many tracks may enter the same spot.
A skier can start skiing from one spot and stop in another spot if there is a sequence of tracks that lead from the starting spot and end in the ending spot. Unfortunately, recently there were many accidents, because the structure of the resort allows a skier to go through dangerous paths, by reaching high speed and endangering himself and the other customers. Here, a path is called dangerous, if it consists of at least two tracks.
Arthur wants to secure his customers by closing some of the spots in a way that there are no dangerous paths in the resort. When a spot is closed, all tracks entering and leaving that spot become unusable.
Formally, after closing some of the spots, there should not be a path that consists of two or more tracks.
Arthur doesn't want to close too many spots. He will be happy to find any way to close at most $\frac{4}{7}n$ spots so that the remaining part is safe. Help him find any suitable way to do so.
-----Input-----
The first line contains a single positive integer $T$ — the number of test cases. $T$ test case description follows.
The first line of each description contains two integers $n$ and $m$ ($1 \leq n \leq 2 \cdot 10^5$) — the number of landing spots and tracks respectively.
The following $m$ lines describe the tracks. Each of these lines contains two integers $x$ and $y$ ($1 \leq x < y \leq n$) — indices of the starting and finishing spots for the respective track. It is guaranteed that at most two tracks start at each spot. There may be tracks in which starting and finishing spots both coincide.
It is guaranteed that the sum of $n$ over all test cases does not exceed $2 \cdot 10^5$.
-----Output-----
For each test case, print a single integer $k$ ($0 \leq k \leq \frac{4}{7}n$) — the number of spots to be closed. In the next line, print $k$ distinct integers — indices of all spots to be closed, in any order.
If there are several answers, you may output any of them. Note that you don't have to minimize $k$. It can be shown that a suitable answer always exists.
-----Example-----
Input
2
4 6
1 2
1 3
2 3
2 4
3 4
3 4
7 6
1 2
1 3
2 4
2 5
3 6
3 7
Output
2
3 4
4
4 5 6 7
-----Note-----
In the first sample case, closing any two spots is suitable.
In the second sample case, closing only the spot $1$ is also suitable.
|
import sys
def rs(): return sys.stdin.readline().rstrip()
def ri(): return int(sys.stdin.readline())
def ria(): return list(map(int, sys.stdin.readline().split()))
def ws(s): sys.stdout.write(s + '\n')
def wi(n): sys.stdout.write(str(n) + '\n')
def wia(a): sys.stdout.write(' '.join([str(x) for x in a]) + '\n')
def solve(n, m, g):
dp = [0] * n
ans = []
for i in range(n):
for w in g[i]:
dp[i] = max(dp[i], dp[w] + 1)
if dp[i] >= 2:
dp[i] = -1
ans.append(i+1)
wi(len(ans))
wia(ans)
def main():
for _ in range(ri()):
n, m = ria()
g = [[] for i in range(n)]
for __ in range(m):
u, v = ria()
g[v-1].append(u-1)
solve(n, m, g)
def __starting_point():
main()
__starting_point()
|
The only difference between easy and hard versions is constraints.
Now elections are held in Berland and you want to win them. More precisely, you want everyone to vote for you.
There are $n$ voters, and two ways to convince each of them to vote for you. The first way to convince the $i$-th voter is to pay him $p_i$ coins. The second way is to make $m_i$ other voters vote for you, and the $i$-th voter will vote for free.
Moreover, the process of such voting takes place in several steps. For example, if there are five voters with $m_1 = 1$, $m_2 = 2$, $m_3 = 2$, $m_4 = 4$, $m_5 = 5$, then you can buy the vote of the fifth voter, and eventually everyone will vote for you. Set of people voting for you will change as follows: ${5} \rightarrow {1, 5} \rightarrow {1, 2, 3, 5} \rightarrow {1, 2, 3, 4, 5}$.
Calculate the minimum number of coins you have to spend so that everyone votes for you.
-----Input-----
The first line contains one integer $t$ ($1 \le t \le 2 \cdot 10^5$) — the number of test cases.
The first line of each test case contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the number of voters.
The next $n$ lines contains the description of voters. $i$-th line contains two integers $m_i$ and $p_i$ ($1 \le p_i \le 10^9, 0 \le m_i < n$).
It is guaranteed that the sum of all $n$ over all test cases does not exceed $2 \cdot 10^5$.
-----Output-----
For each test case print one integer — the minimum number of coins you have to spend so that everyone votes for you.
-----Example-----
Input
3
3
1 5
2 10
2 8
7
0 1
3 1
1 1
6 1
1 1
4 1
4 1
6
2 6
2 3
2 8
2 7
4 4
5 5
Output
8
0
7
-----Note-----
In the first test case you have to buy vote of the third voter. Then the set of people voting for you will change as follows: ${3} \rightarrow {1, 3} \rightarrow {1, 2, 3}$.
In the second example you don't need to buy votes. The set of people voting for you will change as follows: ${1} \rightarrow {1, 3, 5} \rightarrow {1, 2, 3, 5} \rightarrow {1, 2, 3, 5, 6, 7} \rightarrow {1, 2, 3, 4, 5, 6, 7}$.
In the third test case you have to buy votes of the second and the fifth voters. Then the set of people voting for you will change as follows: ${2, 5} \rightarrow {1, 2, 3, 4, 5} \rightarrow {1, 2, 3, 4, 5, 6}$.
|
import sys
def I():
return sys.stdin.readline().rstrip()
class Heap:
def __init__( self ):
self.l = [ -1 ]
self.n = 0
def n( self ):
return self.n
def top( self ):
return self.l[ 1 ]
def ins( self, x ):
self.l.append( x )
n = len( self.l ) - 1
i = n
while i > 1:
j = i // 2
if self.l[ j ] > self.l[ i ]:
self.l[ j ], self.l[ i ] = self.l[ i ], self.l[ j ]
i = j
else:
break
def pop( self ):
r = self.l[ 1 ]
l = self.l.pop()
n = len( self.l ) - 1
if n:
self.l[ 1 ] = l
i = 1
while True:
j = i * 2
k = j + 1
if k < len( self.l ) and self.l[ i ] > max( self.l[ j ], self.l[ k ] ):
if self.l[ j ] == min( self.l[ j ], self.l[ k ] ):
self.l[ i ], self.l[ j ] = self.l[ j ], self.l[ i ]
i = j
else:
self.l[ i ], self.l[ k ] = self.l[ k ], self.l[ i ]
i = k
elif k < len( self.l ) and self.l[ i ] > self.l[ k ]:
self.l[ i ], self.l[ k ] = self.l[ k ], self.l[ i ]
i = k
elif j < len( self.l ) and self.l[ i ] > self.l[ j ]:
self.l[ i ], self.l[ j ] = self.l[ j ], self.l[ i ]
i = j
else:
break
return r
t = int( I() )
for _ in range( t ):
n = int( I() )
voter = [ list( map( int, I().split() ) ) for _ in range( n ) ]
h = Heap()
d = {}
for m, p in voter:
if m not in d:
d[ m ] = []
d[ m ].append( p )
need = {}
c = 0
sk = sorted( d.keys() )
for m in sk:
need[ m ] = max( 0, m - c )
c += len( d[ m ] )
c = 0
ans = 0
for m in sk[::-1]:
for p in d[ m ]:
h.ins( p )
while c < need[ m ]:
c += 1
ans += h.pop()
print( ans )
|
The only difference between easy and hard versions is constraints.
Now elections are held in Berland and you want to win them. More precisely, you want everyone to vote for you.
There are $n$ voters, and two ways to convince each of them to vote for you. The first way to convince the $i$-th voter is to pay him $p_i$ coins. The second way is to make $m_i$ other voters vote for you, and the $i$-th voter will vote for free.
Moreover, the process of such voting takes place in several steps. For example, if there are five voters with $m_1 = 1$, $m_2 = 2$, $m_3 = 2$, $m_4 = 4$, $m_5 = 5$, then you can buy the vote of the fifth voter, and eventually everyone will vote for you. Set of people voting for you will change as follows: ${5} \rightarrow {1, 5} \rightarrow {1, 2, 3, 5} \rightarrow {1, 2, 3, 4, 5}$.
Calculate the minimum number of coins you have to spend so that everyone votes for you.
-----Input-----
The first line contains one integer $t$ ($1 \le t \le 2 \cdot 10^5$) — the number of test cases.
The first line of each test case contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the number of voters.
The next $n$ lines contains the description of voters. $i$-th line contains two integers $m_i$ and $p_i$ ($1 \le p_i \le 10^9, 0 \le m_i < n$).
It is guaranteed that the sum of all $n$ over all test cases does not exceed $2 \cdot 10^5$.
-----Output-----
For each test case print one integer — the minimum number of coins you have to spend so that everyone votes for you.
-----Example-----
Input
3
3
1 5
2 10
2 8
7
0 1
3 1
1 1
6 1
1 1
4 1
4 1
6
2 6
2 3
2 8
2 7
4 4
5 5
Output
8
0
7
-----Note-----
In the first test case you have to buy vote of the third voter. Then the set of people voting for you will change as follows: ${3} \rightarrow {1, 3} \rightarrow {1, 2, 3}$.
In the second example you don't need to buy votes. The set of people voting for you will change as follows: ${1} \rightarrow {1, 3, 5} \rightarrow {1, 2, 3, 5} \rightarrow {1, 2, 3, 5, 6, 7} \rightarrow {1, 2, 3, 4, 5, 6, 7}$.
In the third test case you have to buy votes of the second and the fifth voters. Then the set of people voting for you will change as follows: ${2, 5} \rightarrow {1, 2, 3, 4, 5} \rightarrow {1, 2, 3, 4, 5, 6}$.
|
import heapq
import sys
input = sys.stdin.readline
t = int(input())
for _ in range(t):
n = int(input())
info = [list(map(int, input().split())) for i in range(n)]
info = sorted(info)
cnt = [0] * n
for i in range(n):
ind = info[i][0]
cnt[ind] += 1
ruiseki_cnt = [0] * (n+1)
for i in range(n):
ruiseki_cnt[i+1] = ruiseki_cnt[i] + cnt[i]
# print(cnt)
# print(ruiseki_cnt)
need = [0] * n
for i in range(1,n):
if cnt[i] != 0 and i > ruiseki_cnt[i]:
need[i] = min(i - ruiseki_cnt[i], i)
# print(need)
info = sorted(info, reverse = True)
#print(info)
num = n - 1
pos = 0
q = []
used_cnt = 0
ans = 0
while True:
if num == -1:
break
while True:
if pos < n and info[pos][0] >= num:
heapq.heappush(q, info[pos][1])
pos += 1
else:
break
if need[num] - used_cnt > 0:
tmp = need[num] - used_cnt
for _ in range(tmp):
ans += heapq.heappop(q)
used_cnt += tmp
num -= 1
print(ans)
|
The only difference between easy and hard versions is constraints.
Now elections are held in Berland and you want to win them. More precisely, you want everyone to vote for you.
There are $n$ voters, and two ways to convince each of them to vote for you. The first way to convince the $i$-th voter is to pay him $p_i$ coins. The second way is to make $m_i$ other voters vote for you, and the $i$-th voter will vote for free.
Moreover, the process of such voting takes place in several steps. For example, if there are five voters with $m_1 = 1$, $m_2 = 2$, $m_3 = 2$, $m_4 = 4$, $m_5 = 5$, then you can buy the vote of the fifth voter, and eventually everyone will vote for you. Set of people voting for you will change as follows: ${5} \rightarrow {1, 5} \rightarrow {1, 2, 3, 5} \rightarrow {1, 2, 3, 4, 5}$.
Calculate the minimum number of coins you have to spend so that everyone votes for you.
-----Input-----
The first line contains one integer $t$ ($1 \le t \le 2 \cdot 10^5$) — the number of test cases.
The first line of each test case contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the number of voters.
The next $n$ lines contains the description of voters. $i$-th line contains two integers $m_i$ and $p_i$ ($1 \le p_i \le 10^9, 0 \le m_i < n$).
It is guaranteed that the sum of all $n$ over all test cases does not exceed $2 \cdot 10^5$.
-----Output-----
For each test case print one integer — the minimum number of coins you have to spend so that everyone votes for you.
-----Example-----
Input
3
3
1 5
2 10
2 8
7
0 1
3 1
1 1
6 1
1 1
4 1
4 1
6
2 6
2 3
2 8
2 7
4 4
5 5
Output
8
0
7
-----Note-----
In the first test case you have to buy vote of the third voter. Then the set of people voting for you will change as follows: ${3} \rightarrow {1, 3} \rightarrow {1, 2, 3}$.
In the second example you don't need to buy votes. The set of people voting for you will change as follows: ${1} \rightarrow {1, 3, 5} \rightarrow {1, 2, 3, 5} \rightarrow {1, 2, 3, 5, 6, 7} \rightarrow {1, 2, 3, 4, 5, 6, 7}$.
In the third test case you have to buy votes of the second and the fifth voters. Then the set of people voting for you will change as follows: ${2, 5} \rightarrow {1, 2, 3, 4, 5} \rightarrow {1, 2, 3, 4, 5, 6}$.
|
import sys
input = sys.stdin.readline
import heapq
from itertools import accumulate
t=int(input())
for test in range(t):
n=int(input())
M=[[] for i in range(n)]
MCOUNT=[0]*(n)
for i in range(n):
m,p=list(map(int,input().split()))
M[m].append(p)
MCOUNT[m]+=1
#print(M)
#print(MCOUNT)
ACC=list(accumulate(MCOUNT))
#print(ACC)
HQ=[]
ANS=0
use=0
for i in range(n-1,-1,-1):
for j in M[i]:
heapq.heappush(HQ,j)
#print(HQ)
while ACC[i-1]+use<i:
x=heapq.heappop(HQ)
ANS+=x
use+=1
print(ANS)
|
The only difference between easy and hard versions is constraints.
Now elections are held in Berland and you want to win them. More precisely, you want everyone to vote for you.
There are $n$ voters, and two ways to convince each of them to vote for you. The first way to convince the $i$-th voter is to pay him $p_i$ coins. The second way is to make $m_i$ other voters vote for you, and the $i$-th voter will vote for free.
Moreover, the process of such voting takes place in several steps. For example, if there are five voters with $m_1 = 1$, $m_2 = 2$, $m_3 = 2$, $m_4 = 4$, $m_5 = 5$, then you can buy the vote of the fifth voter, and eventually everyone will vote for you. Set of people voting for you will change as follows: ${5} \rightarrow {1, 5} \rightarrow {1, 2, 3, 5} \rightarrow {1, 2, 3, 4, 5}$.
Calculate the minimum number of coins you have to spend so that everyone votes for you.
-----Input-----
The first line contains one integer $t$ ($1 \le t \le 2 \cdot 10^5$) — the number of test cases.
The first line of each test case contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the number of voters.
The next $n$ lines contains the description of voters. $i$-th line contains two integers $m_i$ and $p_i$ ($1 \le p_i \le 10^9, 0 \le m_i < n$).
It is guaranteed that the sum of all $n$ over all test cases does not exceed $2 \cdot 10^5$.
-----Output-----
For each test case print one integer — the minimum number of coins you have to spend so that everyone votes for you.
-----Example-----
Input
3
3
1 5
2 10
2 8
7
0 1
3 1
1 1
6 1
1 1
4 1
4 1
6
2 6
2 3
2 8
2 7
4 4
5 5
Output
8
0
7
-----Note-----
In the first test case you have to buy vote of the third voter. Then the set of people voting for you will change as follows: ${3} \rightarrow {1, 3} \rightarrow {1, 2, 3}$.
In the second example you don't need to buy votes. The set of people voting for you will change as follows: ${1} \rightarrow {1, 3, 5} \rightarrow {1, 2, 3, 5} \rightarrow {1, 2, 3, 5, 6, 7} \rightarrow {1, 2, 3, 4, 5, 6, 7}$.
In the third test case you have to buy votes of the second and the fifth voters. Then the set of people voting for you will change as follows: ${2, 5} \rightarrow {1, 2, 3, 4, 5} \rightarrow {1, 2, 3, 4, 5, 6}$.
|
import sys
from heapq import heappop, heappush
reader = (line.rstrip() for line in sys.stdin)
input = reader.__next__
t = int(input())
for _ in range(t):
n = int(input())
mp = []
for i in range(n):
mi, pi = list(map(int, input().split()))
mp.append((mi, pi))
mp.sort()
prices = []
cost = 0
bribed = 0
i = n - 1
while i >= 0:
currM = mp[i][0]
heappush(prices, mp[i][1])
while i >= 1 and mp[i-1][0] == currM:
i -= 1
heappush(prices, mp[i][1])
already = i + bribed
for k in range(max(0, currM - already)):
cost += heappop(prices)
bribed += 1
i -= 1
print(cost)
|
The only difference between easy and hard versions is constraints.
Now elections are held in Berland and you want to win them. More precisely, you want everyone to vote for you.
There are $n$ voters, and two ways to convince each of them to vote for you. The first way to convince the $i$-th voter is to pay him $p_i$ coins. The second way is to make $m_i$ other voters vote for you, and the $i$-th voter will vote for free.
Moreover, the process of such voting takes place in several steps. For example, if there are five voters with $m_1 = 1$, $m_2 = 2$, $m_3 = 2$, $m_4 = 4$, $m_5 = 5$, then you can buy the vote of the fifth voter, and eventually everyone will vote for you. Set of people voting for you will change as follows: ${5} \rightarrow {1, 5} \rightarrow {1, 2, 3, 5} \rightarrow {1, 2, 3, 4, 5}$.
Calculate the minimum number of coins you have to spend so that everyone votes for you.
-----Input-----
The first line contains one integer $t$ ($1 \le t \le 2 \cdot 10^5$) — the number of test cases.
The first line of each test case contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the number of voters.
The next $n$ lines contains the description of voters. $i$-th line contains two integers $m_i$ and $p_i$ ($1 \le p_i \le 10^9, 0 \le m_i < n$).
It is guaranteed that the sum of all $n$ over all test cases does not exceed $2 \cdot 10^5$.
-----Output-----
For each test case print one integer — the minimum number of coins you have to spend so that everyone votes for you.
-----Example-----
Input
3
3
1 5
2 10
2 8
7
0 1
3 1
1 1
6 1
1 1
4 1
4 1
6
2 6
2 3
2 8
2 7
4 4
5 5
Output
8
0
7
-----Note-----
In the first test case you have to buy vote of the third voter. Then the set of people voting for you will change as follows: ${3} \rightarrow {1, 3} \rightarrow {1, 2, 3}$.
In the second example you don't need to buy votes. The set of people voting for you will change as follows: ${1} \rightarrow {1, 3, 5} \rightarrow {1, 2, 3, 5} \rightarrow {1, 2, 3, 5, 6, 7} \rightarrow {1, 2, 3, 4, 5, 6, 7}$.
In the third test case you have to buy votes of the second and the fifth voters. Then the set of people voting for you will change as follows: ${2, 5} \rightarrow {1, 2, 3, 4, 5} \rightarrow {1, 2, 3, 4, 5, 6}$.
|
import sys
input = sys.stdin.readline
import heapq as hq
t = int(input())
for _ in range(t):
n = int(input())
vt = [list(map(int,input().split())) for i in range(n)]
vt.sort(reverse=True)
q = []
hq.heapify(q)
ans = 0
cnt = 0
for i in range(n):
hq.heappush(q,vt[i][1])
if vt[i][0] >= n-i+cnt:
ans += hq.heappop(q)
cnt += 1
print(ans)
|
The only difference between easy and hard versions is constraints.
Now elections are held in Berland and you want to win them. More precisely, you want everyone to vote for you.
There are $n$ voters, and two ways to convince each of them to vote for you. The first way to convince the $i$-th voter is to pay him $p_i$ coins. The second way is to make $m_i$ other voters vote for you, and the $i$-th voter will vote for free.
Moreover, the process of such voting takes place in several steps. For example, if there are five voters with $m_1 = 1$, $m_2 = 2$, $m_3 = 2$, $m_4 = 4$, $m_5 = 5$, then you can buy the vote of the fifth voter, and eventually everyone will vote for you. Set of people voting for you will change as follows: ${5} \rightarrow {1, 5} \rightarrow {1, 2, 3, 5} \rightarrow {1, 2, 3, 4, 5}$.
Calculate the minimum number of coins you have to spend so that everyone votes for you.
-----Input-----
The first line contains one integer $t$ ($1 \le t \le 2 \cdot 10^5$) — the number of test cases.
The first line of each test case contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the number of voters.
The next $n$ lines contains the description of voters. $i$-th line contains two integers $m_i$ and $p_i$ ($1 \le p_i \le 10^9, 0 \le m_i < n$).
It is guaranteed that the sum of all $n$ over all test cases does not exceed $2 \cdot 10^5$.
-----Output-----
For each test case print one integer — the minimum number of coins you have to spend so that everyone votes for you.
-----Example-----
Input
3
3
1 5
2 10
2 8
7
0 1
3 1
1 1
6 1
1 1
4 1
4 1
6
2 6
2 3
2 8
2 7
4 4
5 5
Output
8
0
7
-----Note-----
In the first test case you have to buy vote of the third voter. Then the set of people voting for you will change as follows: ${3} \rightarrow {1, 3} \rightarrow {1, 2, 3}$.
In the second example you don't need to buy votes. The set of people voting for you will change as follows: ${1} \rightarrow {1, 3, 5} \rightarrow {1, 2, 3, 5} \rightarrow {1, 2, 3, 5, 6, 7} \rightarrow {1, 2, 3, 4, 5, 6, 7}$.
In the third test case you have to buy votes of the second and the fifth voters. Then the set of people voting for you will change as follows: ${2, 5} \rightarrow {1, 2, 3, 4, 5} \rightarrow {1, 2, 3, 4, 5, 6}$.
|
import sys
import heapq as hq
readline = sys.stdin.readline
read = sys.stdin.read
ns = lambda: readline().rstrip()
ni = lambda: int(readline().rstrip())
nm = lambda: map(int, readline().split())
nl = lambda: list(map(int, readline().split()))
prn = lambda x: print(*x, sep='\n')
def solve():
n = ni()
vot = [tuple(nm()) for _ in range(n)]
vot.sort(key = lambda x: (-x[0], x[1]))
q = list()
c = 0
cost = 0
for i in range(n):
hq.heappush(q, vot[i][1])
while n - i - 1 + c < vot[i][0]:
cost += hq.heappop(q)
c += 1
print(cost)
return
# solve()
T = ni()
for _ in range(T):
solve()
|
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