message stringlengths 2 20.1k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 1.95k 109k | cluster float64 17 17 | __index_level_0__ int64 3.91k 217k |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Japan achieved the second straight victory in the national baseball competition WBC !! A baseball tournament was held at Aizu Gakuen High School as baseball became more popular. In this tournament, a round-robin league match will be held and the ranking will be decided in the following ways.
1. The team with the most wins is ranked high
2. If the number of wins is the same, the team with the fewest losses will be ranked higher.
Create a program that inputs the results of each team and outputs the team names in order from the top team. If there are teams with the same rank, output them in the order of input. However, the number of teams n is an integer from 2 to 10 and the team name t is a single-byte alphabetic character. , 2 for a draw. Also, the team name shall be unique.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by a single line of zeros. Each dataset is given in the following format:
n
score1
score2
::
scoren
The number of teams n (2 ≤ n ≤ 10) is given on the first line, and the scorei of the i-th team is given on the following n lines. Each grade is given in the following format.
t r1 r2 ... rn−1
The team name t (one-character half-width alphabetic character) and the result ri (0, 1, or 2) for each match of t are given separated by blanks.
The number of datasets does not exceed 50.
Output
For each dataset, the team name is output in order from the top team.
Example
Input
6
A 1 0 0 2 0
B 0 0 1 1 0
C 1 1 1 1 1
D 1 0 0 1 2
E 2 0 0 0 0
F 1 1 0 2 1
4
g 1 1 1
h 0 1 2
w 0 0 0
b 0 2 1
0
Output
E
A
B
D
F
C
w
h
b
g
Submitted Solution:
```
def solve():
N = int(input())
if N == 0: return -1
score = []
for i in range(N):
s = input().split()
win, draw = 0, 0
for j in range(1, len(s)):
if s[j] == '1': continue
elif s[j] == '0': win += 1
else : draw += 1
score.append((s[0], win, draw))
res = sorted(score, reverse=True, key=lambda x:(x[1], x[2]))
for s in res: print(s[0])
while True:
if solve() != None: break
``` | instruction | 0 | 81,627 | 17 | 163,254 |
Yes | output | 1 | 81,627 | 17 | 163,255 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Japan achieved the second straight victory in the national baseball competition WBC !! A baseball tournament was held at Aizu Gakuen High School as baseball became more popular. In this tournament, a round-robin league match will be held and the ranking will be decided in the following ways.
1. The team with the most wins is ranked high
2. If the number of wins is the same, the team with the fewest losses will be ranked higher.
Create a program that inputs the results of each team and outputs the team names in order from the top team. If there are teams with the same rank, output them in the order of input. However, the number of teams n is an integer from 2 to 10 and the team name t is a single-byte alphabetic character. , 2 for a draw. Also, the team name shall be unique.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by a single line of zeros. Each dataset is given in the following format:
n
score1
score2
::
scoren
The number of teams n (2 ≤ n ≤ 10) is given on the first line, and the scorei of the i-th team is given on the following n lines. Each grade is given in the following format.
t r1 r2 ... rn−1
The team name t (one-character half-width alphabetic character) and the result ri (0, 1, or 2) for each match of t are given separated by blanks.
The number of datasets does not exceed 50.
Output
For each dataset, the team name is output in order from the top team.
Example
Input
6
A 1 0 0 2 0
B 0 0 1 1 0
C 1 1 1 1 1
D 1 0 0 1 2
E 2 0 0 0 0
F 1 1 0 2 1
4
g 1 1 1
h 0 1 2
w 0 0 0
b 0 2 1
0
Output
E
A
B
D
F
C
w
h
b
g
Submitted Solution:
```
while True:
num = int(input())
if num==0:
break
L = []
for i in range(num):
n, *k = input().split()
L.append( (n, k.count("0"), k.count("1"), i) )
L.sort(key=lambda x:(-x[1],x[2],x[3]) )
for x in L:
print(x[0])
``` | instruction | 0 | 81,628 | 17 | 163,256 |
Yes | output | 1 | 81,628 | 17 | 163,257 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Japan achieved the second straight victory in the national baseball competition WBC !! A baseball tournament was held at Aizu Gakuen High School as baseball became more popular. In this tournament, a round-robin league match will be held and the ranking will be decided in the following ways.
1. The team with the most wins is ranked high
2. If the number of wins is the same, the team with the fewest losses will be ranked higher.
Create a program that inputs the results of each team and outputs the team names in order from the top team. If there are teams with the same rank, output them in the order of input. However, the number of teams n is an integer from 2 to 10 and the team name t is a single-byte alphabetic character. , 2 for a draw. Also, the team name shall be unique.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by a single line of zeros. Each dataset is given in the following format:
n
score1
score2
::
scoren
The number of teams n (2 ≤ n ≤ 10) is given on the first line, and the scorei of the i-th team is given on the following n lines. Each grade is given in the following format.
t r1 r2 ... rn−1
The team name t (one-character half-width alphabetic character) and the result ri (0, 1, or 2) for each match of t are given separated by blanks.
The number of datasets does not exceed 50.
Output
For each dataset, the team name is output in order from the top team.
Example
Input
6
A 1 0 0 2 0
B 0 0 1 1 0
C 1 1 1 1 1
D 1 0 0 1 2
E 2 0 0 0 0
F 1 1 0 2 1
4
g 1 1 1
h 0 1 2
w 0 0 0
b 0 2 1
0
Output
E
A
B
D
F
C
w
h
b
g
Submitted Solution:
```
import sys
import operator
f = sys.stdin
while True:
n = int(f.readline())
if n == 0:
break
teams = [f.readline().split() for i in range(n)]
teams = [(t[0], -t.count('0'), t.count('1'), i) for i, t in enumerate(teams)]
teams.sort(key=operator.itemgetter(1,2,3))
print('\n'.join([t[0] for t in teams]))
``` | instruction | 0 | 81,629 | 17 | 163,258 |
Yes | output | 1 | 81,629 | 17 | 163,259 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Japan achieved the second straight victory in the national baseball competition WBC !! A baseball tournament was held at Aizu Gakuen High School as baseball became more popular. In this tournament, a round-robin league match will be held and the ranking will be decided in the following ways.
1. The team with the most wins is ranked high
2. If the number of wins is the same, the team with the fewest losses will be ranked higher.
Create a program that inputs the results of each team and outputs the team names in order from the top team. If there are teams with the same rank, output them in the order of input. However, the number of teams n is an integer from 2 to 10 and the team name t is a single-byte alphabetic character. , 2 for a draw. Also, the team name shall be unique.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by a single line of zeros. Each dataset is given in the following format:
n
score1
score2
::
scoren
The number of teams n (2 ≤ n ≤ 10) is given on the first line, and the scorei of the i-th team is given on the following n lines. Each grade is given in the following format.
t r1 r2 ... rn−1
The team name t (one-character half-width alphabetic character) and the result ri (0, 1, or 2) for each match of t are given separated by blanks.
The number of datasets does not exceed 50.
Output
For each dataset, the team name is output in order from the top team.
Example
Input
6
A 1 0 0 2 0
B 0 0 1 1 0
C 1 1 1 1 1
D 1 0 0 1 2
E 2 0 0 0 0
F 1 1 0 2 1
4
g 1 1 1
h 0 1 2
w 0 0 0
b 0 2 1
0
Output
E
A
B
D
F
C
w
h
b
g
Submitted Solution:
```
import sys
f = sys.stdin
while True:
n = int(f.readline())
if n == 0:
break
teams = [f.readline().split() for i in range(n)]
teams = [(t[0], -t.count('0'), t.count('1'), i) for i, t in enumerate(teams)]
teams.sort(key=operator.itemgetter(1,2,3))
print('\n'.join([t[0] for t in teams]))
``` | instruction | 0 | 81,630 | 17 | 163,260 |
No | output | 1 | 81,630 | 17 | 163,261 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Japan achieved the second straight victory in the national baseball competition WBC !! A baseball tournament was held at Aizu Gakuen High School as baseball became more popular. In this tournament, a round-robin league match will be held and the ranking will be decided in the following ways.
1. The team with the most wins is ranked high
2. If the number of wins is the same, the team with the fewest losses will be ranked higher.
Create a program that inputs the results of each team and outputs the team names in order from the top team. If there are teams with the same rank, output them in the order of input. However, the number of teams n is an integer from 2 to 10 and the team name t is a single-byte alphabetic character. , 2 for a draw. Also, the team name shall be unique.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by a single line of zeros. Each dataset is given in the following format:
n
score1
score2
::
scoren
The number of teams n (2 ≤ n ≤ 10) is given on the first line, and the scorei of the i-th team is given on the following n lines. Each grade is given in the following format.
t r1 r2 ... rn−1
The team name t (one-character half-width alphabetic character) and the result ri (0, 1, or 2) for each match of t are given separated by blanks.
The number of datasets does not exceed 50.
Output
For each dataset, the team name is output in order from the top team.
Example
Input
6
A 1 0 0 2 0
B 0 0 1 1 0
C 1 1 1 1 1
D 1 0 0 1 2
E 2 0 0 0 0
F 1 1 0 2 1
4
g 1 1 1
h 0 1 2
w 0 0 0
b 0 2 1
0
Output
E
A
B
D
F
C
w
h
b
g
Submitted Solution:
```
while True:
n = int(input())
if n == 0: break
tmp = [list(input().split()) for _ in range(n)]
score = {'0': 2, '1': 0, '2': 1}
ans = []
for t in tmp:
tmpv = 0
for v in t[1:]:
tmpv += score[v]
ans.append(t[0] + str(tmpv))
for i in range(len(ans)):
for j in range(i+1, len(ans))[::-1]:
if ans[j][1] > ans[j-1][1]:
ans[j], ans[j-1] = ans[j-1], ans[j]
for a in ans: print(a[0])
``` | instruction | 0 | 81,631 | 17 | 163,262 |
No | output | 1 | 81,631 | 17 | 163,263 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Japan achieved the second straight victory in the national baseball competition WBC !! A baseball tournament was held at Aizu Gakuen High School as baseball became more popular. In this tournament, a round-robin league match will be held and the ranking will be decided in the following ways.
1. The team with the most wins is ranked high
2. If the number of wins is the same, the team with the fewest losses will be ranked higher.
Create a program that inputs the results of each team and outputs the team names in order from the top team. If there are teams with the same rank, output them in the order of input. However, the number of teams n is an integer from 2 to 10 and the team name t is a single-byte alphabetic character. , 2 for a draw. Also, the team name shall be unique.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by a single line of zeros. Each dataset is given in the following format:
n
score1
score2
::
scoren
The number of teams n (2 ≤ n ≤ 10) is given on the first line, and the scorei of the i-th team is given on the following n lines. Each grade is given in the following format.
t r1 r2 ... rn−1
The team name t (one-character half-width alphabetic character) and the result ri (0, 1, or 2) for each match of t are given separated by blanks.
The number of datasets does not exceed 50.
Output
For each dataset, the team name is output in order from the top team.
Example
Input
6
A 1 0 0 2 0
B 0 0 1 1 0
C 1 1 1 1 1
D 1 0 0 1 2
E 2 0 0 0 0
F 1 1 0 2 1
4
g 1 1 1
h 0 1 2
w 0 0 0
b 0 2 1
0
Output
E
A
B
D
F
C
w
h
b
g
Submitted Solution:
```
while True:
n = int(input())
if n == 0: break
tmp = [list(input().split()) for _ in range(n)]
score = {'0': 2, '1': 0, '2': 1}
ans = []
for t in tmp:
tmpv = 0
for v in t[1:]:
tmpv += score[v]
ans.append(t[0] + str(tmpv))
for i in range(len(ans)):
for j in range(i+1, len(ans))[::-1]:
if ans[j][1] > ans[j-1][1]:
ans[j], ans[j-1] = ans[j-1], ans[j]
print('\n'.join(str(a[0]) for a in ans))
``` | instruction | 0 | 81,632 | 17 | 163,264 |
No | output | 1 | 81,632 | 17 | 163,265 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Japan achieved the second straight victory in the national baseball competition WBC !! A baseball tournament was held at Aizu Gakuen High School as baseball became more popular. In this tournament, a round-robin league match will be held and the ranking will be decided in the following ways.
1. The team with the most wins is ranked high
2. If the number of wins is the same, the team with the fewest losses will be ranked higher.
Create a program that inputs the results of each team and outputs the team names in order from the top team. If there are teams with the same rank, output them in the order of input. However, the number of teams n is an integer from 2 to 10 and the team name t is a single-byte alphabetic character. , 2 for a draw. Also, the team name shall be unique.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by a single line of zeros. Each dataset is given in the following format:
n
score1
score2
::
scoren
The number of teams n (2 ≤ n ≤ 10) is given on the first line, and the scorei of the i-th team is given on the following n lines. Each grade is given in the following format.
t r1 r2 ... rn−1
The team name t (one-character half-width alphabetic character) and the result ri (0, 1, or 2) for each match of t are given separated by blanks.
The number of datasets does not exceed 50.
Output
For each dataset, the team name is output in order from the top team.
Example
Input
6
A 1 0 0 2 0
B 0 0 1 1 0
C 1 1 1 1 1
D 1 0 0 1 2
E 2 0 0 0 0
F 1 1 0 2 1
4
g 1 1 1
h 0 1 2
w 0 0 0
b 0 2 1
0
Output
E
A
B
D
F
C
w
h
b
g
Submitted Solution:
```
while True:
n = int(input())
if n == 0: break
tmp = [list(input().split()) for _ in range(n)]
score = {'0': 2, '1': 0, '2': 1}
ans = []
for t in tmp:
tmpv = 0
for v in t[1:]:
tmpv += score[v]
ans.append(t[0] + str(tmpv))
for i in range(len(ans)):
for j in range(i+1, len(ans))[::-1]:
if ans[j][1] > ans[j-1][1]:
ans[j], ans[j-1] = ans[j-1], ans[j]
print('\n'.join([a[0] for a in ans]))
``` | instruction | 0 | 81,633 | 17 | 163,266 |
No | output | 1 | 81,633 | 17 | 163,267 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As a German University in Cairo (GUC) student and a basketball player, Herr Wafa was delighted once he heard the news. GUC is finally participating in the Annual Basketball Competition (ABC).
A team is to be formed of n players, all of which are GUC students. However, the team might have players belonging to different departments. There are m departments in GUC, numbered from 1 to m. Herr Wafa's department has number h. For each department i, Herr Wafa knows number si — how many students who play basketball belong to this department.
Herr Wafa was also able to guarantee a spot on the team, using his special powers. But since he hates floating-point numbers, he needs your help at finding the probability that he will have at least one teammate belonging to his department.
Note that every possible team containing Herr Wafa is equally probable. Consider all the students different from each other.
Input
The first line contains three integers n, m and h (1 ≤ n ≤ 100, 1 ≤ m ≤ 1000, 1 ≤ h ≤ m) — the number of players on the team, the number of departments in GUC and Herr Wafa's department, correspondingly.
The second line contains a single-space-separated list of m integers si (1 ≤ si ≤ 100), denoting the number of students in the i-th department. Note that sh includes Herr Wafa.
Output
Print the probability that Herr Wafa will have at least one teammate from his department. If there is not enough basketball players in GUC to participate in ABC, print -1. The answer will be accepted if it has absolute or relative error not exceeding 10 - 6.
Examples
Input
3 2 1
2 1
Output
1
Input
3 2 1
1 1
Output
-1
Input
3 2 1
2 2
Output
0.666667
Note
In the first example all 3 players (2 from department 1 and 1 from department 2) must be chosen for the team. Both players from Wafa's departments will be chosen, so he's guaranteed to have a teammate from his department.
In the second example, there are not enough players.
In the third example, there are three possibilities to compose the team containing Herr Wafa. In two of them the other player from Herr Wafa's department is part of the team. | instruction | 0 | 81,737 | 17 | 163,474 |
Tags: combinatorics, dp, math, probabilities
Correct Solution:
```
n, m, h = map(int, input().split())
s = list(map(int, input().split()))
ans = 1
for i in range(sum(s) - 1, sum(s) - s[h - 1], -1):
ans *= (i - n + 1) / i
print (1 - ans if sum(s) >= n else -1)
``` | output | 1 | 81,737 | 17 | 163,475 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As a German University in Cairo (GUC) student and a basketball player, Herr Wafa was delighted once he heard the news. GUC is finally participating in the Annual Basketball Competition (ABC).
A team is to be formed of n players, all of which are GUC students. However, the team might have players belonging to different departments. There are m departments in GUC, numbered from 1 to m. Herr Wafa's department has number h. For each department i, Herr Wafa knows number si — how many students who play basketball belong to this department.
Herr Wafa was also able to guarantee a spot on the team, using his special powers. But since he hates floating-point numbers, he needs your help at finding the probability that he will have at least one teammate belonging to his department.
Note that every possible team containing Herr Wafa is equally probable. Consider all the students different from each other.
Input
The first line contains three integers n, m and h (1 ≤ n ≤ 100, 1 ≤ m ≤ 1000, 1 ≤ h ≤ m) — the number of players on the team, the number of departments in GUC and Herr Wafa's department, correspondingly.
The second line contains a single-space-separated list of m integers si (1 ≤ si ≤ 100), denoting the number of students in the i-th department. Note that sh includes Herr Wafa.
Output
Print the probability that Herr Wafa will have at least one teammate from his department. If there is not enough basketball players in GUC to participate in ABC, print -1. The answer will be accepted if it has absolute or relative error not exceeding 10 - 6.
Examples
Input
3 2 1
2 1
Output
1
Input
3 2 1
1 1
Output
-1
Input
3 2 1
2 2
Output
0.666667
Note
In the first example all 3 players (2 from department 1 and 1 from department 2) must be chosen for the team. Both players from Wafa's departments will be chosen, so he's guaranteed to have a teammate from his department.
In the second example, there are not enough players.
In the third example, there are three possibilities to compose the team containing Herr Wafa. In two of them the other player from Herr Wafa's department is part of the team. | instruction | 0 | 81,738 | 17 | 163,476 |
Tags: combinatorics, dp, math, probabilities
Correct Solution:
```
# Legends Always Come Up with Solution
# Author: Manvir Singh
import os
from io import BytesIO, IOBase
import sys
from collections import defaultdict, deque, Counter
from bisect import *
from math import sqrt, pi, ceil, log, inf
from itertools import permutations
from copy import deepcopy
from heapq import *
from sys import setrecursionlimit
def main():
n, m, h = map(int, input().split())
a = list(map(int, input().split()))
z = sum(a)
if n > z:
print(-1)
else:
y = z - a[h - 1]
p = 1
z = z - 1
for i in range(n - 1):
p = p * (y / z)
y -= 1
z -= 1
print(1 - p)
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
if __name__ == "__main__":
main()
``` | output | 1 | 81,738 | 17 | 163,477 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As a German University in Cairo (GUC) student and a basketball player, Herr Wafa was delighted once he heard the news. GUC is finally participating in the Annual Basketball Competition (ABC).
A team is to be formed of n players, all of which are GUC students. However, the team might have players belonging to different departments. There are m departments in GUC, numbered from 1 to m. Herr Wafa's department has number h. For each department i, Herr Wafa knows number si — how many students who play basketball belong to this department.
Herr Wafa was also able to guarantee a spot on the team, using his special powers. But since he hates floating-point numbers, he needs your help at finding the probability that he will have at least one teammate belonging to his department.
Note that every possible team containing Herr Wafa is equally probable. Consider all the students different from each other.
Input
The first line contains three integers n, m and h (1 ≤ n ≤ 100, 1 ≤ m ≤ 1000, 1 ≤ h ≤ m) — the number of players on the team, the number of departments in GUC and Herr Wafa's department, correspondingly.
The second line contains a single-space-separated list of m integers si (1 ≤ si ≤ 100), denoting the number of students in the i-th department. Note that sh includes Herr Wafa.
Output
Print the probability that Herr Wafa will have at least one teammate from his department. If there is not enough basketball players in GUC to participate in ABC, print -1. The answer will be accepted if it has absolute or relative error not exceeding 10 - 6.
Examples
Input
3 2 1
2 1
Output
1
Input
3 2 1
1 1
Output
-1
Input
3 2 1
2 2
Output
0.666667
Note
In the first example all 3 players (2 from department 1 and 1 from department 2) must be chosen for the team. Both players from Wafa's departments will be chosen, so he's guaranteed to have a teammate from his department.
In the second example, there are not enough players.
In the third example, there are three possibilities to compose the team containing Herr Wafa. In two of them the other player from Herr Wafa's department is part of the team. | instruction | 0 | 81,739 | 17 | 163,478 |
Tags: combinatorics, dp, math, probabilities
Correct Solution:
```
# by the authority of GOD author: manhar singh sachdev #
import os,sys
from io import BytesIO, IOBase
def main():
n,m,h = map(int,input().split())
s = list(map(int,input().split()))
x = sum(s)
if x < n:
print(-1)
else:
b,a = [i for i in range(x-s[h-1]+1,x)],[i for i in range(x-n-s[h-1]+2,x-n+1)]
ans = 1
for x,y in zip(a,b):
ans *= x
ans /= y
print(1-ans)
#Fast IO Region
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
if __name__ == '__main__':
main()
``` | output | 1 | 81,739 | 17 | 163,479 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As a German University in Cairo (GUC) student and a basketball player, Herr Wafa was delighted once he heard the news. GUC is finally participating in the Annual Basketball Competition (ABC).
A team is to be formed of n players, all of which are GUC students. However, the team might have players belonging to different departments. There are m departments in GUC, numbered from 1 to m. Herr Wafa's department has number h. For each department i, Herr Wafa knows number si — how many students who play basketball belong to this department.
Herr Wafa was also able to guarantee a spot on the team, using his special powers. But since he hates floating-point numbers, he needs your help at finding the probability that he will have at least one teammate belonging to his department.
Note that every possible team containing Herr Wafa is equally probable. Consider all the students different from each other.
Input
The first line contains three integers n, m and h (1 ≤ n ≤ 100, 1 ≤ m ≤ 1000, 1 ≤ h ≤ m) — the number of players on the team, the number of departments in GUC and Herr Wafa's department, correspondingly.
The second line contains a single-space-separated list of m integers si (1 ≤ si ≤ 100), denoting the number of students in the i-th department. Note that sh includes Herr Wafa.
Output
Print the probability that Herr Wafa will have at least one teammate from his department. If there is not enough basketball players in GUC to participate in ABC, print -1. The answer will be accepted if it has absolute or relative error not exceeding 10 - 6.
Examples
Input
3 2 1
2 1
Output
1
Input
3 2 1
1 1
Output
-1
Input
3 2 1
2 2
Output
0.666667
Note
In the first example all 3 players (2 from department 1 and 1 from department 2) must be chosen for the team. Both players from Wafa's departments will be chosen, so he's guaranteed to have a teammate from his department.
In the second example, there are not enough players.
In the third example, there are three possibilities to compose the team containing Herr Wafa. In two of them the other player from Herr Wafa's department is part of the team. | instruction | 0 | 81,740 | 17 | 163,480 |
Tags: combinatorics, dp, math, probabilities
Correct Solution:
```
def fact(x):
ans = 1
for i in range(1, x + 1):
ans *= i
return ans
def C(x, y):
return fact(x) // fact(y) // fact(x - y)
m, n, h = map(int, input().split())
a = list(map(int, input().split()))
s = sum(a)
h -= 1
if(s < m):
print(-1)
elif(s == m):
if(a[h] > 1):
print(1)
else:
print(0)
else:
ans = 1
for i in range(m - 1):
ans *= (s - a[h] - i)
ans /= (s - 1 - i)
print(1 - ans)
``` | output | 1 | 81,740 | 17 | 163,481 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As a German University in Cairo (GUC) student and a basketball player, Herr Wafa was delighted once he heard the news. GUC is finally participating in the Annual Basketball Competition (ABC).
A team is to be formed of n players, all of which are GUC students. However, the team might have players belonging to different departments. There are m departments in GUC, numbered from 1 to m. Herr Wafa's department has number h. For each department i, Herr Wafa knows number si — how many students who play basketball belong to this department.
Herr Wafa was also able to guarantee a spot on the team, using his special powers. But since he hates floating-point numbers, he needs your help at finding the probability that he will have at least one teammate belonging to his department.
Note that every possible team containing Herr Wafa is equally probable. Consider all the students different from each other.
Input
The first line contains three integers n, m and h (1 ≤ n ≤ 100, 1 ≤ m ≤ 1000, 1 ≤ h ≤ m) — the number of players on the team, the number of departments in GUC and Herr Wafa's department, correspondingly.
The second line contains a single-space-separated list of m integers si (1 ≤ si ≤ 100), denoting the number of students in the i-th department. Note that sh includes Herr Wafa.
Output
Print the probability that Herr Wafa will have at least one teammate from his department. If there is not enough basketball players in GUC to participate in ABC, print -1. The answer will be accepted if it has absolute or relative error not exceeding 10 - 6.
Examples
Input
3 2 1
2 1
Output
1
Input
3 2 1
1 1
Output
-1
Input
3 2 1
2 2
Output
0.666667
Note
In the first example all 3 players (2 from department 1 and 1 from department 2) must be chosen for the team. Both players from Wafa's departments will be chosen, so he's guaranteed to have a teammate from his department.
In the second example, there are not enough players.
In the third example, there are three possibilities to compose the team containing Herr Wafa. In two of them the other player from Herr Wafa's department is part of the team. | instruction | 0 | 81,741 | 17 | 163,482 |
Tags: combinatorics, dp, math, probabilities
Correct Solution:
```
def nck(n,k):
if(n<k):
return 0
num=1
den=1
if(k>n-k):
return nck(n,n-k)
for i in range(k):
num=num*(n-i)
den=den*(i+1)
return num//den
l=input().split()
n=int(l[0])
m=int(l[1])
h=int(l[2])
l=input().split()
li=[int(i) for i in l]
poss=sum(li)-1
if(poss<n-1):
print(-1)
else:
ki=li[h-1]-1
print(1-(nck(poss-ki,n-1)/nck(poss,n-1)))
``` | output | 1 | 81,741 | 17 | 163,483 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As a German University in Cairo (GUC) student and a basketball player, Herr Wafa was delighted once he heard the news. GUC is finally participating in the Annual Basketball Competition (ABC).
A team is to be formed of n players, all of which are GUC students. However, the team might have players belonging to different departments. There are m departments in GUC, numbered from 1 to m. Herr Wafa's department has number h. For each department i, Herr Wafa knows number si — how many students who play basketball belong to this department.
Herr Wafa was also able to guarantee a spot on the team, using his special powers. But since he hates floating-point numbers, he needs your help at finding the probability that he will have at least one teammate belonging to his department.
Note that every possible team containing Herr Wafa is equally probable. Consider all the students different from each other.
Input
The first line contains three integers n, m and h (1 ≤ n ≤ 100, 1 ≤ m ≤ 1000, 1 ≤ h ≤ m) — the number of players on the team, the number of departments in GUC and Herr Wafa's department, correspondingly.
The second line contains a single-space-separated list of m integers si (1 ≤ si ≤ 100), denoting the number of students in the i-th department. Note that sh includes Herr Wafa.
Output
Print the probability that Herr Wafa will have at least one teammate from his department. If there is not enough basketball players in GUC to participate in ABC, print -1. The answer will be accepted if it has absolute or relative error not exceeding 10 - 6.
Examples
Input
3 2 1
2 1
Output
1
Input
3 2 1
1 1
Output
-1
Input
3 2 1
2 2
Output
0.666667
Note
In the first example all 3 players (2 from department 1 and 1 from department 2) must be chosen for the team. Both players from Wafa's departments will be chosen, so he's guaranteed to have a teammate from his department.
In the second example, there are not enough players.
In the third example, there are three possibilities to compose the team containing Herr Wafa. In two of them the other player from Herr Wafa's department is part of the team. | instruction | 0 | 81,742 | 17 | 163,484 |
Tags: combinatorics, dp, math, probabilities
Correct Solution:
```
#!/usr/bin/env python3
from functools import reduce
from operator import mul
from fractions import Fraction
def nCk(n,k):
return int( reduce(mul, (Fraction(n-i, i+1) for i in range(k)), 1) )
def divide(a, b):
return float(Fraction(a, b))
def main():
n, _, h = list(map(int, input().split()))
s = list(map(int, input().split()))
s_sum = sum(s)
if s_sum < n:
print(-1)
else:
print(1 - divide(nCk(s_sum - s[h - 1], n - 1), nCk(s_sum - 1, n - 1)))
if __name__ == '__main__':
main()
``` | output | 1 | 81,742 | 17 | 163,485 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As a German University in Cairo (GUC) student and a basketball player, Herr Wafa was delighted once he heard the news. GUC is finally participating in the Annual Basketball Competition (ABC).
A team is to be formed of n players, all of which are GUC students. However, the team might have players belonging to different departments. There are m departments in GUC, numbered from 1 to m. Herr Wafa's department has number h. For each department i, Herr Wafa knows number si — how many students who play basketball belong to this department.
Herr Wafa was also able to guarantee a spot on the team, using his special powers. But since he hates floating-point numbers, he needs your help at finding the probability that he will have at least one teammate belonging to his department.
Note that every possible team containing Herr Wafa is equally probable. Consider all the students different from each other.
Input
The first line contains three integers n, m and h (1 ≤ n ≤ 100, 1 ≤ m ≤ 1000, 1 ≤ h ≤ m) — the number of players on the team, the number of departments in GUC and Herr Wafa's department, correspondingly.
The second line contains a single-space-separated list of m integers si (1 ≤ si ≤ 100), denoting the number of students in the i-th department. Note that sh includes Herr Wafa.
Output
Print the probability that Herr Wafa will have at least one teammate from his department. If there is not enough basketball players in GUC to participate in ABC, print -1. The answer will be accepted if it has absolute or relative error not exceeding 10 - 6.
Examples
Input
3 2 1
2 1
Output
1
Input
3 2 1
1 1
Output
-1
Input
3 2 1
2 2
Output
0.666667
Note
In the first example all 3 players (2 from department 1 and 1 from department 2) must be chosen for the team. Both players from Wafa's departments will be chosen, so he's guaranteed to have a teammate from his department.
In the second example, there are not enough players.
In the third example, there are three possibilities to compose the team containing Herr Wafa. In two of them the other player from Herr Wafa's department is part of the team. | instruction | 0 | 81,743 | 17 | 163,486 |
Tags: combinatorics, dp, math, probabilities
Correct Solution:
```
n, _, h = [int(x) for x in input().split()]
s = [int(x) for x in input().split()]
h -= 1
n -= 1
s[h] -= 1
st = sum(s) - s[h]
if st + s[h] < n:
print(-1)
elif n == 0:
print(0)
elif st < n:
print(1)
elif st == 0:
print(0 if s[h] == 0 else 1)
else:
def cmb(a, b):
if b > a:
return 0
if b == 0:
return 1
import math
# print( math.factorial(a), math.factorial(b), math.factorial(a - b), math.factorial(b) * math.factorial(a - b))
return math.factorial(a) // (math.factorial(b) * math.factorial(a - b))
ps = [0] * (n + 1)
ca = [0] * (n + 1)
cb = [0] * (n + 1)
ca[1] = st
cb[1] = s[h]
ca[0] = 1
cb[0] = 1
for k in range(2, n + 1):
ca[k] = ca[k - 1] * (st - k + 1) // k
cb[k] = cb[k - 1] * (s[h] - k + 1) // k
# print(ca)
# print(cb)
for k in range(n + 1):
ps[k] = ca[n - k] * cb[k]
# ps[k] = cmb(st, n - k) * cmb(s[h], k)
# print(st - k, n - k, s[h], k, ps[k])
pass
pst = sum(ps)
print((pst - ps[0]) / pst)
``` | output | 1 | 81,743 | 17 | 163,487 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As a German University in Cairo (GUC) student and a basketball player, Herr Wafa was delighted once he heard the news. GUC is finally participating in the Annual Basketball Competition (ABC).
A team is to be formed of n players, all of which are GUC students. However, the team might have players belonging to different departments. There are m departments in GUC, numbered from 1 to m. Herr Wafa's department has number h. For each department i, Herr Wafa knows number si — how many students who play basketball belong to this department.
Herr Wafa was also able to guarantee a spot on the team, using his special powers. But since he hates floating-point numbers, he needs your help at finding the probability that he will have at least one teammate belonging to his department.
Note that every possible team containing Herr Wafa is equally probable. Consider all the students different from each other.
Input
The first line contains three integers n, m and h (1 ≤ n ≤ 100, 1 ≤ m ≤ 1000, 1 ≤ h ≤ m) — the number of players on the team, the number of departments in GUC and Herr Wafa's department, correspondingly.
The second line contains a single-space-separated list of m integers si (1 ≤ si ≤ 100), denoting the number of students in the i-th department. Note that sh includes Herr Wafa.
Output
Print the probability that Herr Wafa will have at least one teammate from his department. If there is not enough basketball players in GUC to participate in ABC, print -1. The answer will be accepted if it has absolute or relative error not exceeding 10 - 6.
Examples
Input
3 2 1
2 1
Output
1
Input
3 2 1
1 1
Output
-1
Input
3 2 1
2 2
Output
0.666667
Note
In the first example all 3 players (2 from department 1 and 1 from department 2) must be chosen for the team. Both players from Wafa's departments will be chosen, so he's guaranteed to have a teammate from his department.
In the second example, there are not enough players.
In the third example, there are three possibilities to compose the team containing Herr Wafa. In two of them the other player from Herr Wafa's department is part of the team. | instruction | 0 | 81,744 | 17 | 163,488 |
Tags: combinatorics, dp, math, probabilities
Correct Solution:
```
import math
def binom(k, n):
ret = 1
for i in range(0, k):
ret = ret * (n-i) // (i+1)
return ret
[n, m, h] = map(int, input().split())
l = list(map(int, input().split()))
t = sum(l) - 1
n = n - 1
l[h-1] = l[h-1] - 1
p = t - l[h-1]
if n > t:
print("{}".format(-1))
else:
print("{}".format(1 - (binom(n, p) / binom(n, t))))
``` | output | 1 | 81,744 | 17 | 163,489 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As a German University in Cairo (GUC) student and a basketball player, Herr Wafa was delighted once he heard the news. GUC is finally participating in the Annual Basketball Competition (ABC).
A team is to be formed of n players, all of which are GUC students. However, the team might have players belonging to different departments. There are m departments in GUC, numbered from 1 to m. Herr Wafa's department has number h. For each department i, Herr Wafa knows number si — how many students who play basketball belong to this department.
Herr Wafa was also able to guarantee a spot on the team, using his special powers. But since he hates floating-point numbers, he needs your help at finding the probability that he will have at least one teammate belonging to his department.
Note that every possible team containing Herr Wafa is equally probable. Consider all the students different from each other.
Input
The first line contains three integers n, m and h (1 ≤ n ≤ 100, 1 ≤ m ≤ 1000, 1 ≤ h ≤ m) — the number of players on the team, the number of departments in GUC and Herr Wafa's department, correspondingly.
The second line contains a single-space-separated list of m integers si (1 ≤ si ≤ 100), denoting the number of students in the i-th department. Note that sh includes Herr Wafa.
Output
Print the probability that Herr Wafa will have at least one teammate from his department. If there is not enough basketball players in GUC to participate in ABC, print -1. The answer will be accepted if it has absolute or relative error not exceeding 10 - 6.
Examples
Input
3 2 1
2 1
Output
1
Input
3 2 1
1 1
Output
-1
Input
3 2 1
2 2
Output
0.666667
Note
In the first example all 3 players (2 from department 1 and 1 from department 2) must be chosen for the team. Both players from Wafa's departments will be chosen, so he's guaranteed to have a teammate from his department.
In the second example, there are not enough players.
In the third example, there are three possibilities to compose the team containing Herr Wafa. In two of them the other player from Herr Wafa's department is part of the team.
Submitted Solution:
```
n, m, h = map(int, input().split())
s = list(map(int, input().split()))
a = 1
S = sum(s)
for i in range(S - s[h - 1] + 1, S):
a *= (i - n + 1) / i
print (-1 if S < n else 1 - a)
``` | instruction | 0 | 81,745 | 17 | 163,490 |
Yes | output | 1 | 81,745 | 17 | 163,491 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As a German University in Cairo (GUC) student and a basketball player, Herr Wafa was delighted once he heard the news. GUC is finally participating in the Annual Basketball Competition (ABC).
A team is to be formed of n players, all of which are GUC students. However, the team might have players belonging to different departments. There are m departments in GUC, numbered from 1 to m. Herr Wafa's department has number h. For each department i, Herr Wafa knows number si — how many students who play basketball belong to this department.
Herr Wafa was also able to guarantee a spot on the team, using his special powers. But since he hates floating-point numbers, he needs your help at finding the probability that he will have at least one teammate belonging to his department.
Note that every possible team containing Herr Wafa is equally probable. Consider all the students different from each other.
Input
The first line contains three integers n, m and h (1 ≤ n ≤ 100, 1 ≤ m ≤ 1000, 1 ≤ h ≤ m) — the number of players on the team, the number of departments in GUC and Herr Wafa's department, correspondingly.
The second line contains a single-space-separated list of m integers si (1 ≤ si ≤ 100), denoting the number of students in the i-th department. Note that sh includes Herr Wafa.
Output
Print the probability that Herr Wafa will have at least one teammate from his department. If there is not enough basketball players in GUC to participate in ABC, print -1. The answer will be accepted if it has absolute or relative error not exceeding 10 - 6.
Examples
Input
3 2 1
2 1
Output
1
Input
3 2 1
1 1
Output
-1
Input
3 2 1
2 2
Output
0.666667
Note
In the first example all 3 players (2 from department 1 and 1 from department 2) must be chosen for the team. Both players from Wafa's departments will be chosen, so he's guaranteed to have a teammate from his department.
In the second example, there are not enough players.
In the third example, there are three possibilities to compose the team containing Herr Wafa. In two of them the other player from Herr Wafa's department is part of the team.
Submitted Solution:
```
import math
from sys import stdin, stdout
def ncr(n, r):
if (n < r):
return 0
r = min(r, n-r)
ans = 1
for i in range(r):
ans *= (n-i)
ans //= (i+1)
return ans
n, m, h = map(int , stdin.readline().split())
s = list(map(int, stdin.readline().split()))
tot = sum(s)
if ( n > tot ):
stdout.write(str(-1))
else:
players_from_other_dept = tot - s[h-1]
players_remaining = tot -1
players_remaining_to_be_sel = n - 1
ans = 1.0
ans -= ncr(players_from_other_dept, players_remaining_to_be_sel)/ncr(players_remaining, players_remaining_to_be_sel)
stdout.write("%.8f"%(ans))
``` | instruction | 0 | 81,746 | 17 | 163,492 |
Yes | output | 1 | 81,746 | 17 | 163,493 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As a German University in Cairo (GUC) student and a basketball player, Herr Wafa was delighted once he heard the news. GUC is finally participating in the Annual Basketball Competition (ABC).
A team is to be formed of n players, all of which are GUC students. However, the team might have players belonging to different departments. There are m departments in GUC, numbered from 1 to m. Herr Wafa's department has number h. For each department i, Herr Wafa knows number si — how many students who play basketball belong to this department.
Herr Wafa was also able to guarantee a spot on the team, using his special powers. But since he hates floating-point numbers, he needs your help at finding the probability that he will have at least one teammate belonging to his department.
Note that every possible team containing Herr Wafa is equally probable. Consider all the students different from each other.
Input
The first line contains three integers n, m and h (1 ≤ n ≤ 100, 1 ≤ m ≤ 1000, 1 ≤ h ≤ m) — the number of players on the team, the number of departments in GUC and Herr Wafa's department, correspondingly.
The second line contains a single-space-separated list of m integers si (1 ≤ si ≤ 100), denoting the number of students in the i-th department. Note that sh includes Herr Wafa.
Output
Print the probability that Herr Wafa will have at least one teammate from his department. If there is not enough basketball players in GUC to participate in ABC, print -1. The answer will be accepted if it has absolute or relative error not exceeding 10 - 6.
Examples
Input
3 2 1
2 1
Output
1
Input
3 2 1
1 1
Output
-1
Input
3 2 1
2 2
Output
0.666667
Note
In the first example all 3 players (2 from department 1 and 1 from department 2) must be chosen for the team. Both players from Wafa's departments will be chosen, so he's guaranteed to have a teammate from his department.
In the second example, there are not enough players.
In the third example, there are three possibilities to compose the team containing Herr Wafa. In two of them the other player from Herr Wafa's department is part of the team.
Submitted Solution:
```
from math import *
# Function to find the nCr
def printNcR(n, r):
# p holds the value of n*(n-1)*(n-2)...,
# k holds the value of r*(r-1)...
p = 1
k = 1
# C(n, r) == C(n, n-r),
# choosing the smaller value
if (n - r < r):
r = n - r
if (r != 0):
while (r):
p *= n
k *= r
# gcd of p, k
m = gcd(p, k)
# dividing by gcd, to simplify product
# division by their gcd saves from
# the overflow
p //= m
k //= m
n -= 1
r -= 1
# print(r)
# k should be simplified to 1
# as C(n, r) is a natural number
# (denominator should be 1 )
else:
p = 1
# if our approach is correct p = ans and k =1
return p
n,m,h = map(int,(input().split()))
a = list(map(int,input().strip().split()))
if sum(a)<n:
print(-1)
elif a[h-1] == 1:
print(0)
elif sum(a) == n or sum(a)-a[h-1]<n-1 or sum(a)-1<n-1:
print(1)
else:
p = printNcR(sum(a)-a[h-1],n-1)/printNcR(sum(a)-1,n-1)
print(1 - p)
``` | instruction | 0 | 81,747 | 17 | 163,494 |
Yes | output | 1 | 81,747 | 17 | 163,495 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As a German University in Cairo (GUC) student and a basketball player, Herr Wafa was delighted once he heard the news. GUC is finally participating in the Annual Basketball Competition (ABC).
A team is to be formed of n players, all of which are GUC students. However, the team might have players belonging to different departments. There are m departments in GUC, numbered from 1 to m. Herr Wafa's department has number h. For each department i, Herr Wafa knows number si — how many students who play basketball belong to this department.
Herr Wafa was also able to guarantee a spot on the team, using his special powers. But since he hates floating-point numbers, he needs your help at finding the probability that he will have at least one teammate belonging to his department.
Note that every possible team containing Herr Wafa is equally probable. Consider all the students different from each other.
Input
The first line contains three integers n, m and h (1 ≤ n ≤ 100, 1 ≤ m ≤ 1000, 1 ≤ h ≤ m) — the number of players on the team, the number of departments in GUC and Herr Wafa's department, correspondingly.
The second line contains a single-space-separated list of m integers si (1 ≤ si ≤ 100), denoting the number of students in the i-th department. Note that sh includes Herr Wafa.
Output
Print the probability that Herr Wafa will have at least one teammate from his department. If there is not enough basketball players in GUC to participate in ABC, print -1. The answer will be accepted if it has absolute or relative error not exceeding 10 - 6.
Examples
Input
3 2 1
2 1
Output
1
Input
3 2 1
1 1
Output
-1
Input
3 2 1
2 2
Output
0.666667
Note
In the first example all 3 players (2 from department 1 and 1 from department 2) must be chosen for the team. Both players from Wafa's departments will be chosen, so he's guaranteed to have a teammate from his department.
In the second example, there are not enough players.
In the third example, there are three possibilities to compose the team containing Herr Wafa. In two of them the other player from Herr Wafa's department is part of the team.
Submitted Solution:
```
R = lambda: map(int, input().split())
def cnt(n, k):
res = 1
for i in range(k):
res = res * (n - i) / (k + 1)
return res
n, m, h = R()
h -= 1
ps = list(R())
tot = sum(ps)
if tot < n:
print(-1)
exit(0)
if tot <= 1 or n <= 1 or ps[h] <= 1:
print(0)
exit(0)
if tot == n:
print(1)
exit(0)
ac = cnt(tot - 1, n - 1)
rem = cnt(tot - ps[h], n - 1)
print((ac - rem) / ac)
``` | instruction | 0 | 81,748 | 17 | 163,496 |
Yes | output | 1 | 81,748 | 17 | 163,497 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As a German University in Cairo (GUC) student and a basketball player, Herr Wafa was delighted once he heard the news. GUC is finally participating in the Annual Basketball Competition (ABC).
A team is to be formed of n players, all of which are GUC students. However, the team might have players belonging to different departments. There are m departments in GUC, numbered from 1 to m. Herr Wafa's department has number h. For each department i, Herr Wafa knows number si — how many students who play basketball belong to this department.
Herr Wafa was also able to guarantee a spot on the team, using his special powers. But since he hates floating-point numbers, he needs your help at finding the probability that he will have at least one teammate belonging to his department.
Note that every possible team containing Herr Wafa is equally probable. Consider all the students different from each other.
Input
The first line contains three integers n, m and h (1 ≤ n ≤ 100, 1 ≤ m ≤ 1000, 1 ≤ h ≤ m) — the number of players on the team, the number of departments in GUC and Herr Wafa's department, correspondingly.
The second line contains a single-space-separated list of m integers si (1 ≤ si ≤ 100), denoting the number of students in the i-th department. Note that sh includes Herr Wafa.
Output
Print the probability that Herr Wafa will have at least one teammate from his department. If there is not enough basketball players in GUC to participate in ABC, print -1. The answer will be accepted if it has absolute or relative error not exceeding 10 - 6.
Examples
Input
3 2 1
2 1
Output
1
Input
3 2 1
1 1
Output
-1
Input
3 2 1
2 2
Output
0.666667
Note
In the first example all 3 players (2 from department 1 and 1 from department 2) must be chosen for the team. Both players from Wafa's departments will be chosen, so he's guaranteed to have a teammate from his department.
In the second example, there are not enough players.
In the third example, there are three possibilities to compose the team containing Herr Wafa. In two of them the other player from Herr Wafa's department is part of the team.
Submitted Solution:
```
# Legends Always Come Up with Solution
# Author: Manvir Singh
import os
from io import BytesIO, IOBase
import sys
from collections import defaultdict, deque, Counter
from bisect import *
from math import sqrt, pi, ceil, log, inf
from itertools import permutations
from copy import deepcopy
from heapq import *
from sys import setrecursionlimit
def main():
n, m, h = map(int, input().split())
a = list(map(int, input().split()))
z = sum(a)
if n > z:
print(-1)
elif n == z:
print(1)
else:
y = z - a[h - 1]
p = 1
z = z - 1
for i in range(n - 1):
p = p * (y / z)
y -= 1
z -= 1
print(1 - p)
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
if __name__ == "__main__":
main()
``` | instruction | 0 | 81,749 | 17 | 163,498 |
No | output | 1 | 81,749 | 17 | 163,499 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As a German University in Cairo (GUC) student and a basketball player, Herr Wafa was delighted once he heard the news. GUC is finally participating in the Annual Basketball Competition (ABC).
A team is to be formed of n players, all of which are GUC students. However, the team might have players belonging to different departments. There are m departments in GUC, numbered from 1 to m. Herr Wafa's department has number h. For each department i, Herr Wafa knows number si — how many students who play basketball belong to this department.
Herr Wafa was also able to guarantee a spot on the team, using his special powers. But since he hates floating-point numbers, he needs your help at finding the probability that he will have at least one teammate belonging to his department.
Note that every possible team containing Herr Wafa is equally probable. Consider all the students different from each other.
Input
The first line contains three integers n, m and h (1 ≤ n ≤ 100, 1 ≤ m ≤ 1000, 1 ≤ h ≤ m) — the number of players on the team, the number of departments in GUC and Herr Wafa's department, correspondingly.
The second line contains a single-space-separated list of m integers si (1 ≤ si ≤ 100), denoting the number of students in the i-th department. Note that sh includes Herr Wafa.
Output
Print the probability that Herr Wafa will have at least one teammate from his department. If there is not enough basketball players in GUC to participate in ABC, print -1. The answer will be accepted if it has absolute or relative error not exceeding 10 - 6.
Examples
Input
3 2 1
2 1
Output
1
Input
3 2 1
1 1
Output
-1
Input
3 2 1
2 2
Output
0.666667
Note
In the first example all 3 players (2 from department 1 and 1 from department 2) must be chosen for the team. Both players from Wafa's departments will be chosen, so he's guaranteed to have a teammate from his department.
In the second example, there are not enough players.
In the third example, there are three possibilities to compose the team containing Herr Wafa. In two of them the other player from Herr Wafa's department is part of the team.
Submitted Solution:
```
def ncr(x,y):
m=1
n=1
for i in range(x-y+1,x+1):
m*=i
for i in range(1,y+1):
n*=i
return m//n
a,b,c=map(int,input().split())
d=[int(d) for d in input().split()]
e=0
for i in d:
e+=i
if(e==a):
print("1")
elif(e<a):
print("-1")
else:
f=d[c-1]-1
g=e-d[c-1]
p=ncr(g,a-2)
o=(f*p)/ncr(e-1,a-1)
print(o)
``` | instruction | 0 | 81,750 | 17 | 163,500 |
No | output | 1 | 81,750 | 17 | 163,501 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As a German University in Cairo (GUC) student and a basketball player, Herr Wafa was delighted once he heard the news. GUC is finally participating in the Annual Basketball Competition (ABC).
A team is to be formed of n players, all of which are GUC students. However, the team might have players belonging to different departments. There are m departments in GUC, numbered from 1 to m. Herr Wafa's department has number h. For each department i, Herr Wafa knows number si — how many students who play basketball belong to this department.
Herr Wafa was also able to guarantee a spot on the team, using his special powers. But since he hates floating-point numbers, he needs your help at finding the probability that he will have at least one teammate belonging to his department.
Note that every possible team containing Herr Wafa is equally probable. Consider all the students different from each other.
Input
The first line contains three integers n, m and h (1 ≤ n ≤ 100, 1 ≤ m ≤ 1000, 1 ≤ h ≤ m) — the number of players on the team, the number of departments in GUC and Herr Wafa's department, correspondingly.
The second line contains a single-space-separated list of m integers si (1 ≤ si ≤ 100), denoting the number of students in the i-th department. Note that sh includes Herr Wafa.
Output
Print the probability that Herr Wafa will have at least one teammate from his department. If there is not enough basketball players in GUC to participate in ABC, print -1. The answer will be accepted if it has absolute or relative error not exceeding 10 - 6.
Examples
Input
3 2 1
2 1
Output
1
Input
3 2 1
1 1
Output
-1
Input
3 2 1
2 2
Output
0.666667
Note
In the first example all 3 players (2 from department 1 and 1 from department 2) must be chosen for the team. Both players from Wafa's departments will be chosen, so he's guaranteed to have a teammate from his department.
In the second example, there are not enough players.
In the third example, there are three possibilities to compose the team containing Herr Wafa. In two of them the other player from Herr Wafa's department is part of the team.
Submitted Solution:
```
n, m, h= input().split()
a = []
nums = sorted([int(i) for i in input().split() if i.isdigit()])
soma = 0
for x in range(int(m)):
soma = soma + nums[x]
if(soma < int(n)):
print("-1.0")
else:
temp1 = 1
temp2 = 1
soma1 = soma - 1
soma2 = soma - nums[int(h)]
for x in range(int(n) - 1):
soma = soma + nums[x]
temp2 = temp2 * soma2 / soma1
soma2 = soma2 - 1
soma1 = soma1 - 1
temp1 = 1.0 - temp2
print ('%.6f' % temp1)
``` | instruction | 0 | 81,751 | 17 | 163,502 |
No | output | 1 | 81,751 | 17 | 163,503 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As a German University in Cairo (GUC) student and a basketball player, Herr Wafa was delighted once he heard the news. GUC is finally participating in the Annual Basketball Competition (ABC).
A team is to be formed of n players, all of which are GUC students. However, the team might have players belonging to different departments. There are m departments in GUC, numbered from 1 to m. Herr Wafa's department has number h. For each department i, Herr Wafa knows number si — how many students who play basketball belong to this department.
Herr Wafa was also able to guarantee a spot on the team, using his special powers. But since he hates floating-point numbers, he needs your help at finding the probability that he will have at least one teammate belonging to his department.
Note that every possible team containing Herr Wafa is equally probable. Consider all the students different from each other.
Input
The first line contains three integers n, m and h (1 ≤ n ≤ 100, 1 ≤ m ≤ 1000, 1 ≤ h ≤ m) — the number of players on the team, the number of departments in GUC and Herr Wafa's department, correspondingly.
The second line contains a single-space-separated list of m integers si (1 ≤ si ≤ 100), denoting the number of students in the i-th department. Note that sh includes Herr Wafa.
Output
Print the probability that Herr Wafa will have at least one teammate from his department. If there is not enough basketball players in GUC to participate in ABC, print -1. The answer will be accepted if it has absolute or relative error not exceeding 10 - 6.
Examples
Input
3 2 1
2 1
Output
1
Input
3 2 1
1 1
Output
-1
Input
3 2 1
2 2
Output
0.666667
Note
In the first example all 3 players (2 from department 1 and 1 from department 2) must be chosen for the team. Both players from Wafa's departments will be chosen, so he's guaranteed to have a teammate from his department.
In the second example, there are not enough players.
In the third example, there are three possibilities to compose the team containing Herr Wafa. In two of them the other player from Herr Wafa's department is part of the team.
Submitted Solution:
```
n, m, h = map(int, input().split())
h -= 1
S = list(map(int, input().split()))
def solve():
if sum(S) < n:
return -1
if sum(S) - S[h] < n - 1:
return 1
num1 = [k for k in range(1, sum(S) - S[h] + 1)]
num2 = [k for k in range(1, sum(S) - n + 1)]
den1 = [k for k in range(1, sum(S))]
den2 = [k for k in range(1, sum(S) - S[h] - n + 2)]
num1.sort(reverse=True)
num2.sort(reverse=True)
den1.sort(reverse=False)
den2.sort(reverse=False)
ans = 1
for i in range(max(map(len, [num1, num2, den1, den2]))):
ans *= num1[i] if i < len(num1) else 1
ans *= num2[i] if i < len(num2) else 1
ans /= den1[i] if i < len(den1) else 1
ans /= den2[i] if i < len(den2) else 1
return 1 - ans
print("%.12f\n" % solve())
``` | instruction | 0 | 81,752 | 17 | 163,504 |
No | output | 1 | 81,752 | 17 | 163,505 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A big football championship will occur soon! n teams will compete in it, and each pair of teams will play exactly one game against each other.
There are two possible outcomes of a game:
* the game may result in a tie, then both teams get 1 point;
* one team might win in a game, then the winning team gets 3 points and the losing team gets 0 points.
The score of a team is the number of points it gained during all games that it played.
You are interested in a hypothetical situation when all teams get the same score at the end of the championship. A simple example of that situation is when all games result in ties, but you want to minimize the number of ties as well.
Your task is to describe a situation (choose the result of each game) so that all teams get the same score, and the number of ties is the minimum possible.
Input
The first line contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
Then the test cases follow. Each test case is described by one line containing one integer n (2 ≤ n ≤ 100) — the number of teams.
Output
For each test case, print (n(n - 1))/(2) integers describing the results of the games in the following order: the first integer should correspond to the match between team 1 and team 2, the second — between team 1 and team 3, then 1 and 4, ..., 1 and n, 2 and 3, 2 and 4, ..., 2 and n, and so on, until the game between the team n - 1 and the team n.
The integer corresponding to the game between the team x and the team y should be 1 if x wins, -1 if y wins, or 0 if the game results in a tie.
All teams should get the same score, and the number of ties should be the minimum possible. If there are multiple optimal answers, print any of them. It can be shown that there always exists a way to make all teams have the same score.
Example
Input
2
2
3
Output
0
1 -1 1
Note
In the first test case of the example, both teams get 1 point since the game between them is a tie.
In the second test case of the example, team 1 defeats team 2 (team 1 gets 3 points), team 1 loses to team 3 (team 3 gets 3 points), and team 2 wins against team 3 (team 2 gets 3 points). | instruction | 0 | 81,950 | 17 | 163,900 |
Tags: brute force, constructive algorithms, dfs and similar, graphs, greedy, implementation, math
Correct Solution:
```
from sys import stdin, gettrace
if gettrace():
def inputi():
return input()
else:
def input():
return next(stdin)[:-1]
def inputi():
return stdin.buffer.readline()
def solve():
n = int(input())
res = []
for i in range(1,n):
for j in range(min((n-1)//2, n-i)):
res.append(1)
if n%2 == 0 and (n-1)//2 < n-i:
res.append(0)
for j in range((n-1)//2 - i+1):
res.append(-1)
print(' '.join(map(str, res)))
def main():
t = int(input())
for _ in range(t):
solve()
if __name__ == "__main__":
main()
``` | output | 1 | 81,950 | 17 | 163,901 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A big football championship will occur soon! n teams will compete in it, and each pair of teams will play exactly one game against each other.
There are two possible outcomes of a game:
* the game may result in a tie, then both teams get 1 point;
* one team might win in a game, then the winning team gets 3 points and the losing team gets 0 points.
The score of a team is the number of points it gained during all games that it played.
You are interested in a hypothetical situation when all teams get the same score at the end of the championship. A simple example of that situation is when all games result in ties, but you want to minimize the number of ties as well.
Your task is to describe a situation (choose the result of each game) so that all teams get the same score, and the number of ties is the minimum possible.
Input
The first line contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
Then the test cases follow. Each test case is described by one line containing one integer n (2 ≤ n ≤ 100) — the number of teams.
Output
For each test case, print (n(n - 1))/(2) integers describing the results of the games in the following order: the first integer should correspond to the match between team 1 and team 2, the second — between team 1 and team 3, then 1 and 4, ..., 1 and n, 2 and 3, 2 and 4, ..., 2 and n, and so on, until the game between the team n - 1 and the team n.
The integer corresponding to the game between the team x and the team y should be 1 if x wins, -1 if y wins, or 0 if the game results in a tie.
All teams should get the same score, and the number of ties should be the minimum possible. If there are multiple optimal answers, print any of them. It can be shown that there always exists a way to make all teams have the same score.
Example
Input
2
2
3
Output
0
1 -1 1
Note
In the first test case of the example, both teams get 1 point since the game between them is a tie.
In the second test case of the example, team 1 defeats team 2 (team 1 gets 3 points), team 1 loses to team 3 (team 3 gets 3 points), and team 2 wins against team 3 (team 2 gets 3 points). | instruction | 0 | 81,951 | 17 | 163,902 |
Tags: brute force, constructive algorithms, dfs and similar, graphs, greedy, implementation, math
Correct Solution:
```
####################################################
import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
#######################################
for t in range(int(input())):
n=int(input())
if n%2:
l=[]
for i in range((n*(n-1))//2):
if i%2:
l.append(-1)
else:
l.append(1)
else:
l1=[]
l=[]
for i in range(1,n):
for j in range(i+1,n+1):
l1.append([i,j])
c=1
d=1
y=1
for i in l1:
if i[0]==c and i[1]==c+1:
l.append(0)
c=i[1]+1
else:
if i[0]!=y:
y=i[0]
d^=1
if d:
l.append(1)
d^=1
else:
l.append(-1)
d^=1
print(*l)
``` | output | 1 | 81,951 | 17 | 163,903 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A big football championship will occur soon! n teams will compete in it, and each pair of teams will play exactly one game against each other.
There are two possible outcomes of a game:
* the game may result in a tie, then both teams get 1 point;
* one team might win in a game, then the winning team gets 3 points and the losing team gets 0 points.
The score of a team is the number of points it gained during all games that it played.
You are interested in a hypothetical situation when all teams get the same score at the end of the championship. A simple example of that situation is when all games result in ties, but you want to minimize the number of ties as well.
Your task is to describe a situation (choose the result of each game) so that all teams get the same score, and the number of ties is the minimum possible.
Input
The first line contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
Then the test cases follow. Each test case is described by one line containing one integer n (2 ≤ n ≤ 100) — the number of teams.
Output
For each test case, print (n(n - 1))/(2) integers describing the results of the games in the following order: the first integer should correspond to the match between team 1 and team 2, the second — between team 1 and team 3, then 1 and 4, ..., 1 and n, 2 and 3, 2 and 4, ..., 2 and n, and so on, until the game between the team n - 1 and the team n.
The integer corresponding to the game between the team x and the team y should be 1 if x wins, -1 if y wins, or 0 if the game results in a tie.
All teams should get the same score, and the number of ties should be the minimum possible. If there are multiple optimal answers, print any of them. It can be shown that there always exists a way to make all teams have the same score.
Example
Input
2
2
3
Output
0
1 -1 1
Note
In the first test case of the example, both teams get 1 point since the game between them is a tie.
In the second test case of the example, team 1 defeats team 2 (team 1 gets 3 points), team 1 loses to team 3 (team 3 gets 3 points), and team 2 wins against team 3 (team 2 gets 3 points). | instruction | 0 | 81,952 | 17 | 163,904 |
Tags: brute force, constructive algorithms, dfs and similar, graphs, greedy, implementation, math
Correct Solution:
```
# ANKIT BEHIND
import math
t=int(input())
for i in range(0,t):
n=int(input())
ans=[]
if(n%2!=0):
for i in range(0,(n*(n-1)//2)):
if(i%2==0):
ans.append(1)
else:
ans.append(-1)
else:
a=n//2
s=0
aa=n-1
index=0
i=1
while(i<=n*(n-1)//2):
i=i+1
s=s+1
if(s==a) and (s<aa):
ans.append(0)
index=index+1
for j in range(s,aa):
ans.append(-1)
index=index+j
i=index+1
aa=aa-1
s=0
elif(s==a) and (s==aa):
ans.append(0)
i=index+1
break
else:
ans.append(1)
for i in range(0,n*(n-2)//8):
ans.append(1)
print(*ans)
#c=list(map(int,input().split()))
#n,k,kk=map(int,input().split())
``` | output | 1 | 81,952 | 17 | 163,905 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A big football championship will occur soon! n teams will compete in it, and each pair of teams will play exactly one game against each other.
There are two possible outcomes of a game:
* the game may result in a tie, then both teams get 1 point;
* one team might win in a game, then the winning team gets 3 points and the losing team gets 0 points.
The score of a team is the number of points it gained during all games that it played.
You are interested in a hypothetical situation when all teams get the same score at the end of the championship. A simple example of that situation is when all games result in ties, but you want to minimize the number of ties as well.
Your task is to describe a situation (choose the result of each game) so that all teams get the same score, and the number of ties is the minimum possible.
Input
The first line contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
Then the test cases follow. Each test case is described by one line containing one integer n (2 ≤ n ≤ 100) — the number of teams.
Output
For each test case, print (n(n - 1))/(2) integers describing the results of the games in the following order: the first integer should correspond to the match between team 1 and team 2, the second — between team 1 and team 3, then 1 and 4, ..., 1 and n, 2 and 3, 2 and 4, ..., 2 and n, and so on, until the game between the team n - 1 and the team n.
The integer corresponding to the game between the team x and the team y should be 1 if x wins, -1 if y wins, or 0 if the game results in a tie.
All teams should get the same score, and the number of ties should be the minimum possible. If there are multiple optimal answers, print any of them. It can be shown that there always exists a way to make all teams have the same score.
Example
Input
2
2
3
Output
0
1 -1 1
Note
In the first test case of the example, both teams get 1 point since the game between them is a tie.
In the second test case of the example, team 1 defeats team 2 (team 1 gets 3 points), team 1 loses to team 3 (team 3 gets 3 points), and team 2 wins against team 3 (team 2 gets 3 points). | instruction | 0 | 81,953 | 17 | 163,906 |
Tags: brute force, constructive algorithms, dfs and similar, graphs, greedy, implementation, math
Correct Solution:
```
# by the authority of GOD author: manhar singh sachdev #
import os,sys
from io import BytesIO,IOBase
def some_random_function():
"""due to the fast IO template, my code gets caught in
plag check for no reason. That is why, I am writing
random functions"""
x = 10
x *= 100
i_dont_know = x
why_am_i_writing_this = x*x
print(i_dont_know)
print(why_am_i_writing_this)
def main():
for _ in range(int(input())):
n = int(input())
if n&1:
aa = []
x,y = n//2,n//2
for i in range(n-1):
aa.extend([1]*x)
aa.extend([-1]*y)
if not y:
x -= 1
else:
y -= 1
print(*aa)
else:
aa = []
x,y = (n-1)//2,(n-1)//2
for i in range(n-1):
if not i&1:
aa.append(0)
aa.extend([1]*x+[-1]*y)
else:
aa.extend([-1]*y+[1]*x)
if not i&1:
continue
x -= 1
y -= 1
print(*aa)
# Fast IO Region
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self,file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd,max(os.fstat(self._fd).st_size,BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0,2),self.buffer.write(b),self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines==0:
b = os.read(self._fd,max(os.fstat(self._fd).st_size,BUFSIZE))
self.newlines = b.count(b"\n")+(not b)
ptr = self.buffer.tell()
self.buffer.seek(0,2),self.buffer.write(b),self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd,self.buffer.getvalue())
self.buffer.truncate(0),self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self,file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s:self.buffer.write(s.encode("ascii"))
self.read = lambda:self.buffer.read().decode("ascii")
self.readline = lambda:self.buffer.readline().decode("ascii")
sys.stdin,sys.stdout = IOWrapper(sys.stdin),IOWrapper(sys.stdout)
input = lambda:sys.stdin.readline().rstrip("\r\n")
def some_random_function1():
"""due to the fast IO template, my code gets caught in
plag check for no reason. That is why, I am writing
random functions"""
x = 10
x *= 100
i_dont_know = x
why_am_i_writing_this = x*x
print(i_dont_know)
print(why_am_i_writing_this)
def some_random_function2():
"""due to the fast IO template, my code gets caught in
plag check for no reason. That is why, I am writing
random functions"""
x = 10
x *= 100
i_dont_know = x
why_am_i_writing_this = x*x
print(i_dont_know)
print(why_am_i_writing_this)
def some_random_function3():
"""due to the fast IO template, my code gets caught in
plag check for no reason. That is why, I am writing
random functions"""
x = 10
x *= 100
i_dont_know = x
why_am_i_writing_this = x*x
print(i_dont_know)
print(why_am_i_writing_this)
def some_random_function4():
"""due to the fast IO template, my code gets caught in
plag check for no reason. That is why, I am writing
random functions"""
x = 10
x *= 100
i_dont_know = x
why_am_i_writing_this = x*x
print(i_dont_know)
print(why_am_i_writing_this)
if __name__=='__main__':
main()
``` | output | 1 | 81,953 | 17 | 163,907 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A big football championship will occur soon! n teams will compete in it, and each pair of teams will play exactly one game against each other.
There are two possible outcomes of a game:
* the game may result in a tie, then both teams get 1 point;
* one team might win in a game, then the winning team gets 3 points and the losing team gets 0 points.
The score of a team is the number of points it gained during all games that it played.
You are interested in a hypothetical situation when all teams get the same score at the end of the championship. A simple example of that situation is when all games result in ties, but you want to minimize the number of ties as well.
Your task is to describe a situation (choose the result of each game) so that all teams get the same score, and the number of ties is the minimum possible.
Input
The first line contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
Then the test cases follow. Each test case is described by one line containing one integer n (2 ≤ n ≤ 100) — the number of teams.
Output
For each test case, print (n(n - 1))/(2) integers describing the results of the games in the following order: the first integer should correspond to the match between team 1 and team 2, the second — between team 1 and team 3, then 1 and 4, ..., 1 and n, 2 and 3, 2 and 4, ..., 2 and n, and so on, until the game between the team n - 1 and the team n.
The integer corresponding to the game between the team x and the team y should be 1 if x wins, -1 if y wins, or 0 if the game results in a tie.
All teams should get the same score, and the number of ties should be the minimum possible. If there are multiple optimal answers, print any of them. It can be shown that there always exists a way to make all teams have the same score.
Example
Input
2
2
3
Output
0
1 -1 1
Note
In the first test case of the example, both teams get 1 point since the game between them is a tie.
In the second test case of the example, team 1 defeats team 2 (team 1 gets 3 points), team 1 loses to team 3 (team 3 gets 3 points), and team 2 wins against team 3 (team 2 gets 3 points). | instruction | 0 | 81,954 | 17 | 163,908 |
Tags: brute force, constructive algorithms, dfs and similar, graphs, greedy, implementation, math
Correct Solution:
```
from math import ceil,floor,log
import sys
from heapq import heappush,heappop
from collections import Counter,defaultdict,deque
input=lambda : sys.stdin.readline().strip()
c=lambda x: 10**9 if(x=="?") else int(x)
class node:
def __init__(self,x,y):
self.a=[x,y]
def __lt__(self,b):
return b.a[0]<self.a[0]
def __repr__(self):
return str(self.a[0])+" "+str(self.a[1])
def main():
for _ in range(int(input())):
n=int(input())
l=[0]*((n*(n-1))//2)
t=1
k=0
if(n%2==1):
for i in range(n*(n-1)//2):
l[i]=t
t=-t
print(*l)
else:
t=-1
k=0
l1=[[0]*n for i in range(n)]
for i in range(n):
for j in range(i+1,n):
if(i+1==j and i%2==0):
pass
else:
l1[i][j]=t
t=-t
l[k]=l1[i][j]
k+=1
if(i%2==0):
t=1
else:
t=-1
print(*l)
main()
``` | output | 1 | 81,954 | 17 | 163,909 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A big football championship will occur soon! n teams will compete in it, and each pair of teams will play exactly one game against each other.
There are two possible outcomes of a game:
* the game may result in a tie, then both teams get 1 point;
* one team might win in a game, then the winning team gets 3 points and the losing team gets 0 points.
The score of a team is the number of points it gained during all games that it played.
You are interested in a hypothetical situation when all teams get the same score at the end of the championship. A simple example of that situation is when all games result in ties, but you want to minimize the number of ties as well.
Your task is to describe a situation (choose the result of each game) so that all teams get the same score, and the number of ties is the minimum possible.
Input
The first line contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
Then the test cases follow. Each test case is described by one line containing one integer n (2 ≤ n ≤ 100) — the number of teams.
Output
For each test case, print (n(n - 1))/(2) integers describing the results of the games in the following order: the first integer should correspond to the match between team 1 and team 2, the second — between team 1 and team 3, then 1 and 4, ..., 1 and n, 2 and 3, 2 and 4, ..., 2 and n, and so on, until the game between the team n - 1 and the team n.
The integer corresponding to the game between the team x and the team y should be 1 if x wins, -1 if y wins, or 0 if the game results in a tie.
All teams should get the same score, and the number of ties should be the minimum possible. If there are multiple optimal answers, print any of them. It can be shown that there always exists a way to make all teams have the same score.
Example
Input
2
2
3
Output
0
1 -1 1
Note
In the first test case of the example, both teams get 1 point since the game between them is a tie.
In the second test case of the example, team 1 defeats team 2 (team 1 gets 3 points), team 1 loses to team 3 (team 3 gets 3 points), and team 2 wins against team 3 (team 2 gets 3 points). | instruction | 0 | 81,955 | 17 | 163,910 |
Tags: brute force, constructive algorithms, dfs and similar, graphs, greedy, implementation, math
Correct Solution:
```
for _ in range(int(input())):
n=int(input())
x=(n*(n-1))//2
z=[0]*n
k=x//n
y=[]
for i in range(n):
for j in range(i+1,n):
if z[i]<k:
z[i]+=1
y.append(1)
elif n%2==0 and z[i]<k+1:
z[i]+=1
y.append(0)
else:
z[j]+=1
y.append(-1)
print(*y)
``` | output | 1 | 81,955 | 17 | 163,911 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A big football championship will occur soon! n teams will compete in it, and each pair of teams will play exactly one game against each other.
There are two possible outcomes of a game:
* the game may result in a tie, then both teams get 1 point;
* one team might win in a game, then the winning team gets 3 points and the losing team gets 0 points.
The score of a team is the number of points it gained during all games that it played.
You are interested in a hypothetical situation when all teams get the same score at the end of the championship. A simple example of that situation is when all games result in ties, but you want to minimize the number of ties as well.
Your task is to describe a situation (choose the result of each game) so that all teams get the same score, and the number of ties is the minimum possible.
Input
The first line contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
Then the test cases follow. Each test case is described by one line containing one integer n (2 ≤ n ≤ 100) — the number of teams.
Output
For each test case, print (n(n - 1))/(2) integers describing the results of the games in the following order: the first integer should correspond to the match between team 1 and team 2, the second — between team 1 and team 3, then 1 and 4, ..., 1 and n, 2 and 3, 2 and 4, ..., 2 and n, and so on, until the game between the team n - 1 and the team n.
The integer corresponding to the game between the team x and the team y should be 1 if x wins, -1 if y wins, or 0 if the game results in a tie.
All teams should get the same score, and the number of ties should be the minimum possible. If there are multiple optimal answers, print any of them. It can be shown that there always exists a way to make all teams have the same score.
Example
Input
2
2
3
Output
0
1 -1 1
Note
In the first test case of the example, both teams get 1 point since the game between them is a tie.
In the second test case of the example, team 1 defeats team 2 (team 1 gets 3 points), team 1 loses to team 3 (team 3 gets 3 points), and team 2 wins against team 3 (team 2 gets 3 points). | instruction | 0 | 81,956 | 17 | 163,912 |
Tags: brute force, constructive algorithms, dfs and similar, graphs, greedy, implementation, math
Correct Solution:
```
from sys import stdin, setrecursionlimit, stdout
#setrecursionlimit(100000) #use "python" instead of "pypy" to avoid MLE
from collections import deque
from math import sqrt, floor, ceil, log, log2, log10, pi, gcd, sin, cos, asin
from heapq import heapify, heappop, heappush, heapreplace, heappushpop
from bisect import bisect_right, bisect_left
def ii(): return int(stdin.readline())
def fi(): return float(stdin.readline())
def mi(): return map(int, stdin.readline().split())
def fmi(): return map(float, stdin.readline().split())
def li(): return list(mi())
def si(): return stdin.readline().rstrip()
def lsi(): return list(si())
mod=1000000007
res=['NO', 'YES']
#######################################################################################
########################### M Y F U N C T I O N S ###########################
#######################################################################################
#######################################################################################
########################### M A I N P R O G R A M ###########################
#######################################################################################
test, test_case=0, 1
test_case=ii()
#print('>>>>>>>>>>>>>>>>>>>>>>>>>>>')
while test<test_case:
test+=1
n=ii()
tot=(n*(n-1))//2
gap=tot//n
f=0
if tot%n:
f=1
a=[]
for i in range(1, n+1):
for j in range(i+1, n+1):
g=j-i
x=-1
if g<=gap:
x=1
elif f and g==gap+1:
x=0
a.append(x)
print(*a)
``` | output | 1 | 81,956 | 17 | 163,913 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A big football championship will occur soon! n teams will compete in it, and each pair of teams will play exactly one game against each other.
There are two possible outcomes of a game:
* the game may result in a tie, then both teams get 1 point;
* one team might win in a game, then the winning team gets 3 points and the losing team gets 0 points.
The score of a team is the number of points it gained during all games that it played.
You are interested in a hypothetical situation when all teams get the same score at the end of the championship. A simple example of that situation is when all games result in ties, but you want to minimize the number of ties as well.
Your task is to describe a situation (choose the result of each game) so that all teams get the same score, and the number of ties is the minimum possible.
Input
The first line contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
Then the test cases follow. Each test case is described by one line containing one integer n (2 ≤ n ≤ 100) — the number of teams.
Output
For each test case, print (n(n - 1))/(2) integers describing the results of the games in the following order: the first integer should correspond to the match between team 1 and team 2, the second — between team 1 and team 3, then 1 and 4, ..., 1 and n, 2 and 3, 2 and 4, ..., 2 and n, and so on, until the game between the team n - 1 and the team n.
The integer corresponding to the game between the team x and the team y should be 1 if x wins, -1 if y wins, or 0 if the game results in a tie.
All teams should get the same score, and the number of ties should be the minimum possible. If there are multiple optimal answers, print any of them. It can be shown that there always exists a way to make all teams have the same score.
Example
Input
2
2
3
Output
0
1 -1 1
Note
In the first test case of the example, both teams get 1 point since the game between them is a tie.
In the second test case of the example, team 1 defeats team 2 (team 1 gets 3 points), team 1 loses to team 3 (team 3 gets 3 points), and team 2 wins against team 3 (team 2 gets 3 points). | instruction | 0 | 81,957 | 17 | 163,914 |
Tags: brute force, constructive algorithms, dfs and similar, graphs, greedy, implementation, math
Correct Solution:
```
def solve(n):
t = [[None for i in range(n)] for j in range(n)]
for i in range(n):
for j in range(i+1,n):
if(j==i+1 and n&1==0 and i&1==0):
t[i][j] = 0
t[j][i] = 0
elif(i&1==0):
if(j&1==0):
t[i][j] = 1
t[j][i] = -1
else:
t[i][j] = -1
t[j][i] = 1
else:
if(j&1==1):
t[i][j] = 1
t[j][i] = -1
else:
t[i][j] = -1
t[j][i] = 1
res = []
for i in range(n):
for j in range(i+1,n):
res.append(t[i][j])
for x in res:
print(x,end=" ")
print()
t = int(input())
for _ in range(t):
n = int(input())
solve(n)
``` | output | 1 | 81,957 | 17 | 163,915 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A big football championship will occur soon! n teams will compete in it, and each pair of teams will play exactly one game against each other.
There are two possible outcomes of a game:
* the game may result in a tie, then both teams get 1 point;
* one team might win in a game, then the winning team gets 3 points and the losing team gets 0 points.
The score of a team is the number of points it gained during all games that it played.
You are interested in a hypothetical situation when all teams get the same score at the end of the championship. A simple example of that situation is when all games result in ties, but you want to minimize the number of ties as well.
Your task is to describe a situation (choose the result of each game) so that all teams get the same score, and the number of ties is the minimum possible.
Input
The first line contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
Then the test cases follow. Each test case is described by one line containing one integer n (2 ≤ n ≤ 100) — the number of teams.
Output
For each test case, print (n(n - 1))/(2) integers describing the results of the games in the following order: the first integer should correspond to the match between team 1 and team 2, the second — between team 1 and team 3, then 1 and 4, ..., 1 and n, 2 and 3, 2 and 4, ..., 2 and n, and so on, until the game between the team n - 1 and the team n.
The integer corresponding to the game between the team x and the team y should be 1 if x wins, -1 if y wins, or 0 if the game results in a tie.
All teams should get the same score, and the number of ties should be the minimum possible. If there are multiple optimal answers, print any of them. It can be shown that there always exists a way to make all teams have the same score.
Example
Input
2
2
3
Output
0
1 -1 1
Note
In the first test case of the example, both teams get 1 point since the game between them is a tie.
In the second test case of the example, team 1 defeats team 2 (team 1 gets 3 points), team 1 loses to team 3 (team 3 gets 3 points), and team 2 wins against team 3 (team 2 gets 3 points).
Submitted Solution:
```
answers = []
def solve(n):
final = []
limit = (n*(n-1)) // 2
if n == 2:
answers.append([0])
return
if n % 2 == 0:
wins = (n-1) // 2
loss = (n-1) // 2
draw = 1
for i in range(1, n+1):
temp = []
if wins >= 1:
temp.append([1]*wins)
if draw == 1:
temp.append([0])
if loss >= 1:
temp.append([-1]*loss)
# print(temp)
for ele in temp:
final.extend(ele[:])
loss -= 1
if loss < 0:
wins -= 1
draw = 0
# print(len(final), limit)
if len(final) < limit:
# print('ll')
rem = limit-len(final)
for i in range(rem):
final.append(1)
answers.append(final)
else:
wins = (n-1) // 2
loss = (n-1) // 2
# count = 0
for i in range(1, n+1):
temp = []
if wins >= 1:
temp.append([1]*wins)
if loss >= 1:
temp.append([-1]*loss)
# print(temp)
for ele in temp:
final.extend(ele[:])
loss -= 1
if loss < 0:
wins -= 1
answers.append(final)
T = int(input())
while T:
n = int(input())
solve(n)
T -= 1
for ans in answers:
print(*ans)
``` | instruction | 0 | 81,958 | 17 | 163,916 |
Yes | output | 1 | 81,958 | 17 | 163,917 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A big football championship will occur soon! n teams will compete in it, and each pair of teams will play exactly one game against each other.
There are two possible outcomes of a game:
* the game may result in a tie, then both teams get 1 point;
* one team might win in a game, then the winning team gets 3 points and the losing team gets 0 points.
The score of a team is the number of points it gained during all games that it played.
You are interested in a hypothetical situation when all teams get the same score at the end of the championship. A simple example of that situation is when all games result in ties, but you want to minimize the number of ties as well.
Your task is to describe a situation (choose the result of each game) so that all teams get the same score, and the number of ties is the minimum possible.
Input
The first line contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
Then the test cases follow. Each test case is described by one line containing one integer n (2 ≤ n ≤ 100) — the number of teams.
Output
For each test case, print (n(n - 1))/(2) integers describing the results of the games in the following order: the first integer should correspond to the match between team 1 and team 2, the second — between team 1 and team 3, then 1 and 4, ..., 1 and n, 2 and 3, 2 and 4, ..., 2 and n, and so on, until the game between the team n - 1 and the team n.
The integer corresponding to the game between the team x and the team y should be 1 if x wins, -1 if y wins, or 0 if the game results in a tie.
All teams should get the same score, and the number of ties should be the minimum possible. If there are multiple optimal answers, print any of them. It can be shown that there always exists a way to make all teams have the same score.
Example
Input
2
2
3
Output
0
1 -1 1
Note
In the first test case of the example, both teams get 1 point since the game between them is a tie.
In the second test case of the example, team 1 defeats team 2 (team 1 gets 3 points), team 1 loses to team 3 (team 3 gets 3 points), and team 2 wins against team 3 (team 2 gets 3 points).
Submitted Solution:
```
for _ in range(int(input())):
n = int(input())
if n%2==1:
for i in range(n):
for j in range(i+1,n):
print(1 if (i+j)%2==1 else -1,end = " ")
else:
grid = [[0 for i in range(n)]for j in range(n)]
for i in range(0,n,2):
for j in range(i+2,n,2):
grid[i][j] = 1
grid[i][j+1] = -1
grid[i+1][j] = -1
grid[i+1][j+1] = 1
for i in range(n):
for j in range(i+1,n):
print(grid[i][j],end = " ")
print()
``` | instruction | 0 | 81,959 | 17 | 163,918 |
Yes | output | 1 | 81,959 | 17 | 163,919 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A big football championship will occur soon! n teams will compete in it, and each pair of teams will play exactly one game against each other.
There are two possible outcomes of a game:
* the game may result in a tie, then both teams get 1 point;
* one team might win in a game, then the winning team gets 3 points and the losing team gets 0 points.
The score of a team is the number of points it gained during all games that it played.
You are interested in a hypothetical situation when all teams get the same score at the end of the championship. A simple example of that situation is when all games result in ties, but you want to minimize the number of ties as well.
Your task is to describe a situation (choose the result of each game) so that all teams get the same score, and the number of ties is the minimum possible.
Input
The first line contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
Then the test cases follow. Each test case is described by one line containing one integer n (2 ≤ n ≤ 100) — the number of teams.
Output
For each test case, print (n(n - 1))/(2) integers describing the results of the games in the following order: the first integer should correspond to the match between team 1 and team 2, the second — between team 1 and team 3, then 1 and 4, ..., 1 and n, 2 and 3, 2 and 4, ..., 2 and n, and so on, until the game between the team n - 1 and the team n.
The integer corresponding to the game between the team x and the team y should be 1 if x wins, -1 if y wins, or 0 if the game results in a tie.
All teams should get the same score, and the number of ties should be the minimum possible. If there are multiple optimal answers, print any of them. It can be shown that there always exists a way to make all teams have the same score.
Example
Input
2
2
3
Output
0
1 -1 1
Note
In the first test case of the example, both teams get 1 point since the game between them is a tie.
In the second test case of the example, team 1 defeats team 2 (team 1 gets 3 points), team 1 loses to team 3 (team 3 gets 3 points), and team 2 wins against team 3 (team 2 gets 3 points).
Submitted Solution:
```
t=int(input())
for q in range(t):
n=int(input())
z=-1
if n%2==0:
for i in range(1,n+1):
for j in range(i+1,n+1):
if i%2==1 and j==i+1:
print(0,end=' ')
elif i%2==0 and j==i+1:
print(z,end=' ')
else:
if z==1:
z=-1
else:
z=1
print(z,end=' ')
else:
for i in range(1,n+1):
for j in range(i+1,n+1):
if z==1:
z=-1
else:
z=1
print(z,end=' ')
``` | instruction | 0 | 81,960 | 17 | 163,920 |
Yes | output | 1 | 81,960 | 17 | 163,921 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A big football championship will occur soon! n teams will compete in it, and each pair of teams will play exactly one game against each other.
There are two possible outcomes of a game:
* the game may result in a tie, then both teams get 1 point;
* one team might win in a game, then the winning team gets 3 points and the losing team gets 0 points.
The score of a team is the number of points it gained during all games that it played.
You are interested in a hypothetical situation when all teams get the same score at the end of the championship. A simple example of that situation is when all games result in ties, but you want to minimize the number of ties as well.
Your task is to describe a situation (choose the result of each game) so that all teams get the same score, and the number of ties is the minimum possible.
Input
The first line contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
Then the test cases follow. Each test case is described by one line containing one integer n (2 ≤ n ≤ 100) — the number of teams.
Output
For each test case, print (n(n - 1))/(2) integers describing the results of the games in the following order: the first integer should correspond to the match between team 1 and team 2, the second — between team 1 and team 3, then 1 and 4, ..., 1 and n, 2 and 3, 2 and 4, ..., 2 and n, and so on, until the game between the team n - 1 and the team n.
The integer corresponding to the game between the team x and the team y should be 1 if x wins, -1 if y wins, or 0 if the game results in a tie.
All teams should get the same score, and the number of ties should be the minimum possible. If there are multiple optimal answers, print any of them. It can be shown that there always exists a way to make all teams have the same score.
Example
Input
2
2
3
Output
0
1 -1 1
Note
In the first test case of the example, both teams get 1 point since the game between them is a tie.
In the second test case of the example, team 1 defeats team 2 (team 1 gets 3 points), team 1 loses to team 3 (team 3 gets 3 points), and team 2 wins against team 3 (team 2 gets 3 points).
Submitted Solution:
```
def main():
num_cases = int(input())
for case_num in range(1, num_cases + 1):
solve(case_num)
def solve(case_num):
# Code here
n = int(input())
result = [[0] * n for _ in range(n)]
for i in range(n):
for j in range(1, n):
if (n % 2 == 0 and j == n // 2):
continue
if j <= (n - 1) // 2:
result[i][(i + j) % n] = 1
else:
result[i][(i + j) % n] = -1
# cycle = [0, 1]
# right = 2
# left = n - 1
# for i in range(n - 2):
# if i % 2 == 0:
# cycle.append(right)
# right += 1
# else:
# cycle.append(left)
# left -= 1
# cycle.append(0)
# for _ in range(n // 2):
# for i, j in zip(cycle[:-1], cycle[1:]):
# result[i][j] = 1
# result[j][i] = -1
# new_cycle = []
# for i in cycle:
# if i == 0:
# new_cycle.append(i)
# elif i == n - 1:
# new_cycle.append(1)
# else:
# new_cycle.append(i + 1)
# cycle = new_cycle
for i in range(n):
for j in range(i + 1, n):
print(result[i][j], end=' ')
print()
if __name__ == '__main__':
main()
``` | instruction | 0 | 81,961 | 17 | 163,922 |
Yes | output | 1 | 81,961 | 17 | 163,923 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A big football championship will occur soon! n teams will compete in it, and each pair of teams will play exactly one game against each other.
There are two possible outcomes of a game:
* the game may result in a tie, then both teams get 1 point;
* one team might win in a game, then the winning team gets 3 points and the losing team gets 0 points.
The score of a team is the number of points it gained during all games that it played.
You are interested in a hypothetical situation when all teams get the same score at the end of the championship. A simple example of that situation is when all games result in ties, but you want to minimize the number of ties as well.
Your task is to describe a situation (choose the result of each game) so that all teams get the same score, and the number of ties is the minimum possible.
Input
The first line contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
Then the test cases follow. Each test case is described by one line containing one integer n (2 ≤ n ≤ 100) — the number of teams.
Output
For each test case, print (n(n - 1))/(2) integers describing the results of the games in the following order: the first integer should correspond to the match between team 1 and team 2, the second — between team 1 and team 3, then 1 and 4, ..., 1 and n, 2 and 3, 2 and 4, ..., 2 and n, and so on, until the game between the team n - 1 and the team n.
The integer corresponding to the game between the team x and the team y should be 1 if x wins, -1 if y wins, or 0 if the game results in a tie.
All teams should get the same score, and the number of ties should be the minimum possible. If there are multiple optimal answers, print any of them. It can be shown that there always exists a way to make all teams have the same score.
Example
Input
2
2
3
Output
0
1 -1 1
Note
In the first test case of the example, both teams get 1 point since the game between them is a tie.
In the second test case of the example, team 1 defeats team 2 (team 1 gets 3 points), team 1 loses to team 3 (team 3 gets 3 points), and team 2 wins against team 3 (team 2 gets 3 points).
Submitted Solution:
```
t = int(input())
for _ in range(t):
n = int(input())
matches = n-1
maxi = matches//2
arr = [0]*n
out=[]
for i in range(n):
for j in range(i+1,n):
if arr[i]<maxi:
out.append(1)
arr[i]+=1
elif arr[j]<maxi:
out.append(-1)
arr[j]+=1
else:
out.append(0)
print(*out)
``` | instruction | 0 | 81,962 | 17 | 163,924 |
No | output | 1 | 81,962 | 17 | 163,925 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A big football championship will occur soon! n teams will compete in it, and each pair of teams will play exactly one game against each other.
There are two possible outcomes of a game:
* the game may result in a tie, then both teams get 1 point;
* one team might win in a game, then the winning team gets 3 points and the losing team gets 0 points.
The score of a team is the number of points it gained during all games that it played.
You are interested in a hypothetical situation when all teams get the same score at the end of the championship. A simple example of that situation is when all games result in ties, but you want to minimize the number of ties as well.
Your task is to describe a situation (choose the result of each game) so that all teams get the same score, and the number of ties is the minimum possible.
Input
The first line contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
Then the test cases follow. Each test case is described by one line containing one integer n (2 ≤ n ≤ 100) — the number of teams.
Output
For each test case, print (n(n - 1))/(2) integers describing the results of the games in the following order: the first integer should correspond to the match between team 1 and team 2, the second — between team 1 and team 3, then 1 and 4, ..., 1 and n, 2 and 3, 2 and 4, ..., 2 and n, and so on, until the game between the team n - 1 and the team n.
The integer corresponding to the game between the team x and the team y should be 1 if x wins, -1 if y wins, or 0 if the game results in a tie.
All teams should get the same score, and the number of ties should be the minimum possible. If there are multiple optimal answers, print any of them. It can be shown that there always exists a way to make all teams have the same score.
Example
Input
2
2
3
Output
0
1 -1 1
Note
In the first test case of the example, both teams get 1 point since the game between them is a tie.
In the second test case of the example, team 1 defeats team 2 (team 1 gets 3 points), team 1 loses to team 3 (team 3 gets 3 points), and team 2 wins against team 3 (team 2 gets 3 points).
Submitted Solution:
```
for _ in '.'*int(input()):
n = int(input())
if n == 2:
print(0)
continue
total = n*(n-1)//2
mw = total//n
l = [0 for i in range(n)] #will store the number of wins
draws = total%n
a = n-1
while a>0:
for i in range(a):
if l[n-a-1] < mw:
print("1", end = " ")
l[n-a-1] +=1
else:
if draws >0:
print("0", end = " ")
draws = draws - 1
else:
print("-1", end = " ")
l[n-a+i] += 1
a = a -1
``` | instruction | 0 | 81,963 | 17 | 163,926 |
No | output | 1 | 81,963 | 17 | 163,927 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A big football championship will occur soon! n teams will compete in it, and each pair of teams will play exactly one game against each other.
There are two possible outcomes of a game:
* the game may result in a tie, then both teams get 1 point;
* one team might win in a game, then the winning team gets 3 points and the losing team gets 0 points.
The score of a team is the number of points it gained during all games that it played.
You are interested in a hypothetical situation when all teams get the same score at the end of the championship. A simple example of that situation is when all games result in ties, but you want to minimize the number of ties as well.
Your task is to describe a situation (choose the result of each game) so that all teams get the same score, and the number of ties is the minimum possible.
Input
The first line contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
Then the test cases follow. Each test case is described by one line containing one integer n (2 ≤ n ≤ 100) — the number of teams.
Output
For each test case, print (n(n - 1))/(2) integers describing the results of the games in the following order: the first integer should correspond to the match between team 1 and team 2, the second — between team 1 and team 3, then 1 and 4, ..., 1 and n, 2 and 3, 2 and 4, ..., 2 and n, and so on, until the game between the team n - 1 and the team n.
The integer corresponding to the game between the team x and the team y should be 1 if x wins, -1 if y wins, or 0 if the game results in a tie.
All teams should get the same score, and the number of ties should be the minimum possible. If there are multiple optimal answers, print any of them. It can be shown that there always exists a way to make all teams have the same score.
Example
Input
2
2
3
Output
0
1 -1 1
Note
In the first test case of the example, both teams get 1 point since the game between them is a tie.
In the second test case of the example, team 1 defeats team 2 (team 1 gets 3 points), team 1 loses to team 3 (team 3 gets 3 points), and team 2 wins against team 3 (team 2 gets 3 points).
Submitted Solution:
```
# ------------------- fast io --------------------
import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# ------------------- fast io --------------------
def main():
for j in range(int(input())):
n=int(input());games=(n*(n-1))//2
ans=[];count=(n-1)//2
for s in range(n):
for i in range(s+1,n):
if i<=s+count:
ans.append(1)
else:
if i==s+count+1:
ans.append(0)
else:
ans.append(-1)
print(*ans)
main()
``` | instruction | 0 | 81,964 | 17 | 163,928 |
No | output | 1 | 81,964 | 17 | 163,929 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A big football championship will occur soon! n teams will compete in it, and each pair of teams will play exactly one game against each other.
There are two possible outcomes of a game:
* the game may result in a tie, then both teams get 1 point;
* one team might win in a game, then the winning team gets 3 points and the losing team gets 0 points.
The score of a team is the number of points it gained during all games that it played.
You are interested in a hypothetical situation when all teams get the same score at the end of the championship. A simple example of that situation is when all games result in ties, but you want to minimize the number of ties as well.
Your task is to describe a situation (choose the result of each game) so that all teams get the same score, and the number of ties is the minimum possible.
Input
The first line contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
Then the test cases follow. Each test case is described by one line containing one integer n (2 ≤ n ≤ 100) — the number of teams.
Output
For each test case, print (n(n - 1))/(2) integers describing the results of the games in the following order: the first integer should correspond to the match between team 1 and team 2, the second — between team 1 and team 3, then 1 and 4, ..., 1 and n, 2 and 3, 2 and 4, ..., 2 and n, and so on, until the game between the team n - 1 and the team n.
The integer corresponding to the game between the team x and the team y should be 1 if x wins, -1 if y wins, or 0 if the game results in a tie.
All teams should get the same score, and the number of ties should be the minimum possible. If there are multiple optimal answers, print any of them. It can be shown that there always exists a way to make all teams have the same score.
Example
Input
2
2
3
Output
0
1 -1 1
Note
In the first test case of the example, both teams get 1 point since the game between them is a tie.
In the second test case of the example, team 1 defeats team 2 (team 1 gets 3 points), team 1 loses to team 3 (team 3 gets 3 points), and team 2 wins against team 3 (team 2 gets 3 points).
Submitted Solution:
```
#input()
#int(input())
#[s for s in input().split()]
#[int(s) for s in input().split()]
#for t in range(t):
import math
import collections
import bisect
def arrPrint(a):
return " ".join([str(i) for i in a])
def gridPrint(a):
return "\n".join([" ".join([str(j) for j in a[i]]) for i in range(len(a))])
def isPalindrome(s):
for i in range(len(s)//2):
if not s[i] == s[-i-1]:
return False
return True
def solve(n):
ans = []
for i in range(1, n + 1):
for j in range(n - i):
if n%2 == 0 and j == 0:
ans.append(0)
else:
if j % 2 == 0:
ans.append(1)
else:
ans.append(-1)
return(arrPrint(ans))
t = int(input())
for t in range(t):
n = int(input())
result = solve(n)
print(result)
``` | instruction | 0 | 81,965 | 17 | 163,930 |
No | output | 1 | 81,965 | 17 | 163,931 |
Provide a correct Python 3 solution for this coding contest problem.
In 21XX, an annual programming contest, Japan Algorithmist GrandPrix (JAG) has become one of the most popular mind sports events.
JAG is conducted as a knockout tournament. This year, $N$ contestants will compete in JAG. A tournament chart is represented as a string. '[[a-b]-[c-d]]' is an easy example. In this case, there are 4 contestants named a, b, c, and d, and all matches are described as follows:
* Match 1 is the match between a and b.
* Match 2 is the match between c and d.
* Match 3 is the match between [the winner of match 1] and [the winner of match 2].
More precisely, the tournament chart satisfies the following BNF:
* <winner> ::= <person> | "[" <winner> "-" <winner> "]"
* <person> ::= "a" | "b" | "c" | ... | "z"
You, the chairperson of JAG, are planning to announce the results of this year's JAG competition. However, you made a mistake and lost the results of all the matches. Fortunately, you found the tournament chart that was printed before all of the matches of the tournament. Of course, it does not contains results at all. Therefore, you asked every contestant for the number of wins in the tournament, and got $N$ pieces of information in the form of "The contestant $a_i$ won $v_i$ times".
Now, your job is to determine whether all of these replies can be true.
Input
The input consists of a single test case in the format below.
$S$
$a_1$ $v_1$
:
$a_N$ $v_N$
$S$ represents the tournament chart. $S$ satisfies the above BNF. The following $N$ lines represent the information of the number of wins. The ($i+1$)-th line consists of a lowercase letter $a_i$ and a non-negative integer $v_i$ ($v_i \leq 26$) separated by a space, and this means that the contestant $a_i$ won $v_i$ times. Note that $N$ ($2 \leq N \leq 26$) means that the number of contestants and it can be identified by string $S$. You can assume that each letter $a_i$ is distinct. It is guaranteed that $S$ contains each $a_i$ exactly once and doesn't contain any other lowercase letters.
Output
Print 'Yes' in one line if replies are all valid for the tournament chart. Otherwise, print 'No' in one line.
Examples
Input
[[m-y]-[a-o]]
o 0
a 1
y 2
m 0
Output
Yes
Input
[[r-i]-[m-e]]
e 0
r 1
i 1
m 2
Output
No | instruction | 0 | 82,526 | 17 | 165,052 |
"Correct Solution:
```
#!usr/bin/env python3
from collections import defaultdict
from collections import deque
from heapq import heappush, heappop
import sys
import math
import bisect
import random
def LI(): return list(map(int, sys.stdin.readline().split()))
def I(): return int(sys.stdin.readline())
def LS():return list(map(list, sys.stdin.readline().split()))
def S(): return list(sys.stdin.readline())[:-1]
def IR(n):
l = [None for i in range(n)]
for i in range(n):l[i] = I()
return l
def LIR(n):
l = [None for i in range(n)]
for i in range(n):l[i] = LI()
return l
def SR(n):
l = [None for i in range(n)]
for i in range(n):l[i] = S()
return l
def LSR(n):
l = [None for i in range(n)]
for i in range(n):l[i] = LS()
return l
sys.setrecursionlimit(1000000)
mod = 1000000007
#A
def A():
def parse_winner(s,i):
if s[i] == "[":
i += 1
w1, i = parse_winner(s,i)
i += 1
w2, i = parse_winner(s,i)
return calc(w1, w2), i+1
else:
p, i = parse_person(s,i)
return p, i
def parse_person(s,i):
return s[i], i+1
def calc(w1,w2):
if f[w1] == 0:
if f[w2] == 0:
return "0"
else:
f[w2] -= 1
return w2
else:
if f[w2] != 0:
return "0"
else:
f[w1] -= 1
return w1
s = S()
k = 0
for i in s:
if i not in "[-]":
k += 1
f = defaultdict(int)
f["0"] = 100000
for i in range(k):
a,b = input().split()
f[a] = int(b)
w = parse_winner(s,0)[0]
if f[w]:
print("No")
else:
print("Yes")
return
#B
def B():
n,k = LI()
s = S()
t = S()
q = deque()
ans = 0
for i in range(n):
if s[i] == "B" and t[i] == "W":
if q:
x = q.popleft()
if i-x >= k:
ans += 1
while q:
q.popleft()
q.append(i)
if q:
ans += 1
for i in range(n):
if s[i] == "W" and t[i] == "B":
if q:
x = q.popleft()
if i-x >= k:
ans += 1
while q:
q.popleft()
q.append(i)
if q:
ans += 1
print(ans)
return
#C
def C():
n = I()
s = SR(n)
t = S()
return
#D
from operator import mul
def D():
def dot(a,b):
return sum(map(mul,a,b))
def mul_matrix(a,b,m):
tb = tuple(zip(*b))
return [[dot(a_i,b_j)%m for b_j in tb] for a_i in a]
def pow_matrix(a,n,m):
h = len(a)
b = [[1 if i == j else 0 for j in range(h)] for i in range(h)]
k = n
while k:
if (k&1):
b = mul_matrix(b,a,m)
a = mul_matrix(a,a,m)
k >>= 1
return b
while 1:
n,m,a,b,c,t = LI()
if n == 0:
break
s = LI()
s2 = [[s[i] for j in range(1)] for i in range(n)]
mat = [[0 for j in range(n)] for i in range(n)]
mat[0][0] = b
mat[0][1] = c
for i in range(1,n-1):
mat[i][i-1] = a
mat[i][i] = b
mat[i][i+1] = c
mat[n-1][-2] = a
mat[n-1][-1] = b
mat = pow_matrix(mat,t,m)
mat = mul_matrix(mat,s2,m)
for i in mat[:-1]:
print(i[0],end = " ")
print(mat[-1][0])
return
#E
def E():
def surface(x,y,z):
return ((x == 0)|(x == a-1))+((y == 0)|(y == b-1))+((z == 0)|(z == c-1))+k
d = [(1,0,0),(-1,0,0),(0,1,0),(0,-1,0),(0,0,1),(0,0,-1)]
a,b,c,n = LI()
s = [0 for i in range(7)]
k = (a==1)+(b==1)+(c==1)
if k == 0:
s[1] = 2*(max(0,a-2)*max(0,b-2)+max(0,c-2)*max(0,b-2)+max(0,a-2)*max(0,c-2))
s[2] = 4*(max(0,a-2)+max(0,b-2)+max(0,c-2))
s[3] = 8
elif k == 1:
s[2] = max(0,a-2)*max(0,b-2)+max(0,c-2)*max(0,b-2)+max(0,a-2)*max(0,c-2)
s[3] = 2*(max(0,a-2)+max(0,b-2)+max(0,c-2))
s[4] = 4
elif k == 2:
s[4] = max(0,a-2)+max(0,b-2)+max(0,c-2)
s[5] = 2
else:
s[6] = 1
f = defaultdict(int)
for i in range(n):
x,y,z = LI()
s[surface(x,y,z)] -= 1
f[(x,y,z)] = -1
for dx,dy,dz in d:
if f[(x+dx,y+dy,z+dz)] != -1:
f[(x+dx,y+dy,z+dz)] += 1
ans = 0
for i,j in f.items():
if j != -1:
x,y,z = i
if 0 <= x < a and 0 <= y < b and 0 <= z < c:
ans += j+surface(x,y,z)
s[surface(x,y,z)] -= 1
for i in range(1,7):
ans += i*s[i]
print(ans)
return
#F
def F():
return
#G
def G():
return
#H
def H():
return
#I
def I_():
return
#J
def J():
return
#Solve
if __name__ == "__main__":
A()
``` | output | 1 | 82,526 | 17 | 165,053 |
Provide a correct Python 3 solution for this coding contest problem.
In 21XX, an annual programming contest, Japan Algorithmist GrandPrix (JAG) has become one of the most popular mind sports events.
JAG is conducted as a knockout tournament. This year, $N$ contestants will compete in JAG. A tournament chart is represented as a string. '[[a-b]-[c-d]]' is an easy example. In this case, there are 4 contestants named a, b, c, and d, and all matches are described as follows:
* Match 1 is the match between a and b.
* Match 2 is the match between c and d.
* Match 3 is the match between [the winner of match 1] and [the winner of match 2].
More precisely, the tournament chart satisfies the following BNF:
* <winner> ::= <person> | "[" <winner> "-" <winner> "]"
* <person> ::= "a" | "b" | "c" | ... | "z"
You, the chairperson of JAG, are planning to announce the results of this year's JAG competition. However, you made a mistake and lost the results of all the matches. Fortunately, you found the tournament chart that was printed before all of the matches of the tournament. Of course, it does not contains results at all. Therefore, you asked every contestant for the number of wins in the tournament, and got $N$ pieces of information in the form of "The contestant $a_i$ won $v_i$ times".
Now, your job is to determine whether all of these replies can be true.
Input
The input consists of a single test case in the format below.
$S$
$a_1$ $v_1$
:
$a_N$ $v_N$
$S$ represents the tournament chart. $S$ satisfies the above BNF. The following $N$ lines represent the information of the number of wins. The ($i+1$)-th line consists of a lowercase letter $a_i$ and a non-negative integer $v_i$ ($v_i \leq 26$) separated by a space, and this means that the contestant $a_i$ won $v_i$ times. Note that $N$ ($2 \leq N \leq 26$) means that the number of contestants and it can be identified by string $S$. You can assume that each letter $a_i$ is distinct. It is guaranteed that $S$ contains each $a_i$ exactly once and doesn't contain any other lowercase letters.
Output
Print 'Yes' in one line if replies are all valid for the tournament chart. Otherwise, print 'No' in one line.
Examples
Input
[[m-y]-[a-o]]
o 0
a 1
y 2
m 0
Output
Yes
Input
[[r-i]-[m-e]]
e 0
r 1
i 1
m 2
Output
No | instruction | 0 | 82,527 | 17 | 165,054 |
"Correct Solution:
```
import re
def main():
s = input()
d = {}
n = len(re.findall(r'\w', s))
for i in range(n):
a, v = input().split()
v = int(v)
d[a] = v
prog = re.compile(r'\[(\w)-(\w)\]')
m = prog.search(s)
while m is not None:
c1, c2 = m.group(1, 2)
if d[c1] == d[c2]:
print('No')
return 0
if d[c1] < d[c2]:
d[c2] -= 1
s = prog.sub(c2, s, 1)
if d[c1] != 0:
print('No')
return 0
else:
d[c1] -= 1
s = prog.sub(c1, s, 1)
if d[c2] != 0:
print('No')
return 0
m = prog.search(s)
if len(s) != 1:
print('No')
elif [0 for k, v in d.items() if v]:
print('No')
else:
print('Yes')
return 0
if __name__ == '__main__':
main()
``` | output | 1 | 82,527 | 17 | 165,055 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A football league has recently begun in Beautiful land. There are n teams participating in the league. Let's enumerate them with integers from 1 to n.
There will be played exactly (n(n-1))/(2) matches: each team will play against all other teams exactly once. In each match, there is always a winner and loser and there is no draw.
After all matches are played, the organizers will count the number of beautiful triples. Let's call a triple of three teams (A, B, C) beautiful if a team A win against a team B, a team B win against a team C and a team C win against a team A. We look only to a triples of different teams and the order of teams in the triple is important.
The beauty of the league is the number of beautiful triples.
At the moment, m matches were played and their results are known.
What is the maximum beauty of the league that can be, after playing all remaining matches? Also find a possible results for all remaining (n(n-1))/(2) - m matches, so that the league has this maximum beauty.
Input
The first line contains two integers n, m (3 ≤ n ≤ 50, 0 ≤ m ≤ (n(n-1))/(2)) — the number of teams in the football league and the number of matches that were played.
Each of m following lines contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v) denoting that the u-th team won against the v-th team. It is guaranteed that each unordered pair of teams appears at most once.
Output
Print n lines, each line having a string of exactly n characters. Each character must be either 0 or 1.
Let a_{ij} be the j-th number in the i-th line. For all 1 ≤ i ≤ n it should be true, that a_{ii} = 0. For all pairs of teams i ≠ j the number a_{ij} indicates the result of the match between the i-th team and the j-th team:
* If a_{ij} is 1, the i-th team wins against the j-th team;
* Otherwise the j-th team wins against the i-th team;
* Also, it should be true, that a_{ij} + a_{ji} = 1.
Also note that the results of the m matches that were already played cannot be changed in your league.
The beauty of the league in the output should be maximum possible. If there are multiple possible answers with maximum beauty, you can print any of them.
Examples
Input
3 1
1 2
Output
010
001
100
Input
4 2
1 2
1 3
Output
0110
0001
0100
1010
Note
The beauty of league in the first test case is equal to 3 because there exists three beautiful triples: (1, 2, 3), (2, 3, 1), (3, 1, 2).
The beauty of league in the second test is equal to 6 because there exists six beautiful triples: (1, 2, 4), (2, 4, 1), (4, 1, 2), (2, 4, 3), (4, 3, 2), (3, 2, 4).
Submitted Solution:
```
import sys
input = sys.stdin.readline
n,m=map(int,input().split())
W=[0]*n
WIN=0
ANS=[[-1]*n for i in range(n)]
for i in range(n):
ANS[i][i]=0
for i in range(m):
x,y=map(int,input().split())
x-=1
y-=1
ANS[x][y]=1
ANS[y][x]=0
W[x]+=1
WIN+=1
last=n*(n-1)//2
while WIN<last:
WL=sorted([(sc,ind) for ind,sc in enumerate(W)])
flag=0
for _,ind in WL:
if -1 in ANS[ind]:
for _,indr in WL[::-1]:
if ANS[ind][indr]==-1:
ANS[ind][indr]=1
ANS[indr][ind]=0
W[ind]+=1
WIN+=1
flag=1
break
if flag:
break
for ans in ANS:
print("".join(map(str,ans)))
``` | instruction | 0 | 82,708 | 17 | 165,416 |
No | output | 1 | 82,708 | 17 | 165,417 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A programming coach has n students to teach. We know that n is divisible by 3. Let's assume that all students are numbered from 1 to n, inclusive.
Before the university programming championship the coach wants to split all students into groups of three. For some pairs of students we know that they want to be on the same team. Besides, if the i-th student wants to be on the same team with the j-th one, then the j-th student wants to be on the same team with the i-th one. The coach wants the teams to show good results, so he wants the following condition to hold: if the i-th student wants to be on the same team with the j-th, then the i-th and the j-th students must be on the same team. Also, it is obvious that each student must be on exactly one team.
Help the coach and divide the teams the way he wants.
Input
The first line of the input contains integers n and m (3 ≤ n ≤ 48, <image>. Then follow m lines, each contains a pair of integers ai, bi (1 ≤ ai < bi ≤ n) — the pair ai, bi means that students with numbers ai and bi want to be on the same team.
It is guaranteed that n is divisible by 3. It is guaranteed that each pair ai, bi occurs in the input at most once.
Output
If the required division into teams doesn't exist, print number -1. Otherwise, print <image> lines. In each line print three integers xi, yi, zi (1 ≤ xi, yi, zi ≤ n) — the i-th team.
If there are multiple answers, you are allowed to print any of them.
Examples
Input
3 0
Output
3 2 1
Input
6 4
1 2
2 3
3 4
5 6
Output
-1
Input
3 3
1 2
2 3
1 3
Output
3 2 1 | instruction | 0 | 82,874 | 17 | 165,748 |
Tags: brute force, dfs and similar, graphs
Correct Solution:
```
import sys
nm = list(map(int, input().split()))
done = []
visited = []
for i in range(nm[1]):
pair = input().split()
if pair[0] not in visited and pair[1] not in visited:
done.append([pair[0],pair[1]])
visited.append(pair[0])
visited.append("-")
visited.append(pair[1])
elif pair[0] not in visited:
temp = visited.index(pair[1])
if temp < len(visited)-1 and visited[temp+1] != "-":
if temp < len(visited)-1 and visited[temp+1] == "|":
print("-1")
sys.exit()
visited.insert(temp+1,"-")
visited.insert(temp+2,pair[0])
#visited.insert(temp+3,"|")
else:
if temp < len(visited)-3 and visited[temp+3] == "|":
print("-1")
sys.exit()
visited.insert(temp+4,"-")
visited.insert(temp+5,pair[0])
#visited.insert(temp+6,"|")
elif pair[1] not in visited:
temp = visited.index(pair[0])
if temp <len(visited)-1 and visited[temp+1] != "-":
if temp < len(visited)-1 and visited[temp+1] == "|":
print("-1")
sys.exit()
visited.insert(temp+1,"-")
visited.insert(temp+2,pair[1])
#visited.insert(temp+3,"|")
else:
if temp < len(visited)-3 and visited[temp+3] == "|":
print("-1")
sys.exit()
visited.insert(temp+4,"-")
visited.insert(temp+5,pair[1])
#visited.insert(temp+6,"|")
k = 0
lis = [[] for i in range((int(nm[0])//3)*2)]
i = 0
while i<len(visited)-1:
if visited[i+1] == "-":
lis[k].append(visited[i])
i+=2
else:
lis[k].append(visited[i])
k+=1
i+=1
i = 0
while i < len(lis):
if lis[i] == []:
lis.pop()
i-=1
i+=1
if len(visited) > 2:
lis[-1].append(visited[-1])
for i in range(nm[0]):
flag = False
if str(i+1) not in visited:
for e in range(len(lis)):
if len(lis[e]) < 3:
lis[e].append(str(i+1))
flag = True
break
if not flag:
lis.append([str(i+1)])
for i in lis:
if len(i) != 3:
print(-1)
sys.exit()
for i in lis:
print(i[0],i[1],i[2])
``` | output | 1 | 82,874 | 17 | 165,749 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A programming coach has n students to teach. We know that n is divisible by 3. Let's assume that all students are numbered from 1 to n, inclusive.
Before the university programming championship the coach wants to split all students into groups of three. For some pairs of students we know that they want to be on the same team. Besides, if the i-th student wants to be on the same team with the j-th one, then the j-th student wants to be on the same team with the i-th one. The coach wants the teams to show good results, so he wants the following condition to hold: if the i-th student wants to be on the same team with the j-th, then the i-th and the j-th students must be on the same team. Also, it is obvious that each student must be on exactly one team.
Help the coach and divide the teams the way he wants.
Input
The first line of the input contains integers n and m (3 ≤ n ≤ 48, <image>. Then follow m lines, each contains a pair of integers ai, bi (1 ≤ ai < bi ≤ n) — the pair ai, bi means that students with numbers ai and bi want to be on the same team.
It is guaranteed that n is divisible by 3. It is guaranteed that each pair ai, bi occurs in the input at most once.
Output
If the required division into teams doesn't exist, print number -1. Otherwise, print <image> lines. In each line print three integers xi, yi, zi (1 ≤ xi, yi, zi ≤ n) — the i-th team.
If there are multiple answers, you are allowed to print any of them.
Examples
Input
3 0
Output
3 2 1
Input
6 4
1 2
2 3
3 4
5 6
Output
-1
Input
3 3
1 2
2 3
1 3
Output
3 2 1 | instruction | 0 | 82,875 | 17 | 165,750 |
Tags: brute force, dfs and similar, graphs
Correct Solution:
```
#http://codeforces.com/problemset/problem/300/B
from collections import defaultdict
from itertools import islice
def finding_connected_component(graph, current_student):
stack = [current_student]
result = []
while stack:
student = stack.pop()
result.append(student)
for connected_student in graph[student]:
if not visited[connected_student]:
visited[connected_student] = True
stack.append(connected_student)
return result
number_of_students, number_of_pairs = map(int, input().rstrip().split(" "))
graph = defaultdict(list)
group = defaultdict(list)
visited = [False] * (number_of_students+1)
isHavingSolution = True
for i in range(number_of_pairs):
studentA, studentB = map(int, input().rstrip().split(" "))
graph[studentA].append(studentB)
graph[studentB].append(studentA)
for index in range(1, len(visited)):
if not visited[index]:
visited[index] = True
list_of_connected_students = finding_connected_component(graph, index)
if len(list_of_connected_students) > 3:
isHavingSolution = False
break
else:
group[len(list_of_connected_students)].append(list_of_connected_students)
if len(group[2]) > len(group[1]) or not isHavingSolution:
print(-1)
else:
while len(group[2]) > 0:
current_group_2 = group[2].pop()
current_group_1 = group[1].pop()
group[3].append(current_group_2 + current_group_1)
if len(group[1])%3 != 0: print(-1)
else:
index = 0
group_whateverleft = []
for element in group[1]:
group_whateverleft.append(element[0])
last_group_3 = [group_whateverleft[i:i+3] for i in range(0, len(group_whateverleft), 3)]
for element in last_group_3:
group[3].append(element)
for element in group[3]:
element = sorted(element, reverse=True)
print(' '.join(map(str, element)))
``` | output | 1 | 82,875 | 17 | 165,751 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A programming coach has n students to teach. We know that n is divisible by 3. Let's assume that all students are numbered from 1 to n, inclusive.
Before the university programming championship the coach wants to split all students into groups of three. For some pairs of students we know that they want to be on the same team. Besides, if the i-th student wants to be on the same team with the j-th one, then the j-th student wants to be on the same team with the i-th one. The coach wants the teams to show good results, so he wants the following condition to hold: if the i-th student wants to be on the same team with the j-th, then the i-th and the j-th students must be on the same team. Also, it is obvious that each student must be on exactly one team.
Help the coach and divide the teams the way he wants.
Input
The first line of the input contains integers n and m (3 ≤ n ≤ 48, <image>. Then follow m lines, each contains a pair of integers ai, bi (1 ≤ ai < bi ≤ n) — the pair ai, bi means that students with numbers ai and bi want to be on the same team.
It is guaranteed that n is divisible by 3. It is guaranteed that each pair ai, bi occurs in the input at most once.
Output
If the required division into teams doesn't exist, print number -1. Otherwise, print <image> lines. In each line print three integers xi, yi, zi (1 ≤ xi, yi, zi ≤ n) — the i-th team.
If there are multiple answers, you are allowed to print any of them.
Examples
Input
3 0
Output
3 2 1
Input
6 4
1 2
2 3
3 4
5 6
Output
-1
Input
3 3
1 2
2 3
1 3
Output
3 2 1 | instruction | 0 | 82,876 | 17 | 165,752 |
Tags: brute force, dfs and similar, graphs
Correct Solution:
```
import sys
from collections import defaultdict
def make_graph():
return defaultdict(make_graph)
def return_pair_sets(person, graph, used):
if person in used:
return []
used[person] = True
if person not in graph:
return [person]
sets = [person]
for friend in graph[person]:
sets += return_pair_sets(friend, graph, used)
return sets
values = [int(x) for x in sys.stdin.readline().split()]
n_kids, n_pairings = values[0], values[1]
graph = make_graph()
for i in range(n_pairings):
pair_values = [int(x) for x in sys.stdin.readline().split()]
first, second = pair_values[0], pair_values[1]
if first not in graph:
graph[first] = []
if second not in graph:
graph[second] = []
graph[first].append(second)
graph[second].append(first)
used = {}
sets_of_three = []
for i in range(1, n_kids+1):
if i not in used:
group = return_pair_sets(i, graph, used)
# print(group)
if len(group) < 3:
did_insert = False
for j in sets_of_three:
if len(j) + len(group) == 3:
j += group
did_insert = True
break
if not did_insert:
sets_of_three.append(group)
else:
sets_of_three.append(group)
ones = [x[0] for x in sets_of_three if len(x) == 1]
if len(ones) % 3 != 0:
print("-1")
sys.exit(0)
for i in range(len(ones) // 3):
cur_set = []
index = i * 3
cur_set.append(ones[index])
cur_set.append(ones[index+1])
cur_set.append(ones[index+2])
sets_of_three.append(cur_set)
for i in sets_of_three:
if len(i) != 3 and len(i) != 1:
print("-1")
sys.exit(0)
for i in sets_of_three:
if len(i) == 3:
sys.stdout.write(str(i[0]) + " " + str(i[1]) + " " + str(i[2]) + "\n")
``` | output | 1 | 82,876 | 17 | 165,753 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A programming coach has n students to teach. We know that n is divisible by 3. Let's assume that all students are numbered from 1 to n, inclusive.
Before the university programming championship the coach wants to split all students into groups of three. For some pairs of students we know that they want to be on the same team. Besides, if the i-th student wants to be on the same team with the j-th one, then the j-th student wants to be on the same team with the i-th one. The coach wants the teams to show good results, so he wants the following condition to hold: if the i-th student wants to be on the same team with the j-th, then the i-th and the j-th students must be on the same team. Also, it is obvious that each student must be on exactly one team.
Help the coach and divide the teams the way he wants.
Input
The first line of the input contains integers n and m (3 ≤ n ≤ 48, <image>. Then follow m lines, each contains a pair of integers ai, bi (1 ≤ ai < bi ≤ n) — the pair ai, bi means that students with numbers ai and bi want to be on the same team.
It is guaranteed that n is divisible by 3. It is guaranteed that each pair ai, bi occurs in the input at most once.
Output
If the required division into teams doesn't exist, print number -1. Otherwise, print <image> lines. In each line print three integers xi, yi, zi (1 ≤ xi, yi, zi ≤ n) — the i-th team.
If there are multiple answers, you are allowed to print any of them.
Examples
Input
3 0
Output
3 2 1
Input
6 4
1 2
2 3
3 4
5 6
Output
-1
Input
3 3
1 2
2 3
1 3
Output
3 2 1 | instruction | 0 | 82,877 | 17 | 165,754 |
Tags: brute force, dfs and similar, graphs
Correct Solution:
```
n,m=map(int,input().split())
from collections import defaultdict
e=defaultdict(list)
f,v=[False]*(n+1),[]
def dfs(i,s):
s.add(i)
f[i]=True
for k in e[i]:
if not f[k]:
dfs(k,s)
for j in range(m):
a,b=map(int,input().split())
e[a].append(b)
e[b].append(a)
for i in range(1,n+1):
if e[i] and not f[i]:
s=set()
dfs(i,s)
if len(s)>3:
print(-1)
exit()
v.append(list(s))
if len(v)>(n//3):
print(-1)
exit()
while len(v)<(n//3):
v.append([])
vi = 0
for i in range(1, n + 1):
if not f[i]:
while len(v[vi]) == 3:
vi += 1
v[vi].append(i)
for vi in v:
print(vi[0], vi[1], vi[2])
``` | output | 1 | 82,877 | 17 | 165,755 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A programming coach has n students to teach. We know that n is divisible by 3. Let's assume that all students are numbered from 1 to n, inclusive.
Before the university programming championship the coach wants to split all students into groups of three. For some pairs of students we know that they want to be on the same team. Besides, if the i-th student wants to be on the same team with the j-th one, then the j-th student wants to be on the same team with the i-th one. The coach wants the teams to show good results, so he wants the following condition to hold: if the i-th student wants to be on the same team with the j-th, then the i-th and the j-th students must be on the same team. Also, it is obvious that each student must be on exactly one team.
Help the coach and divide the teams the way he wants.
Input
The first line of the input contains integers n and m (3 ≤ n ≤ 48, <image>. Then follow m lines, each contains a pair of integers ai, bi (1 ≤ ai < bi ≤ n) — the pair ai, bi means that students with numbers ai and bi want to be on the same team.
It is guaranteed that n is divisible by 3. It is guaranteed that each pair ai, bi occurs in the input at most once.
Output
If the required division into teams doesn't exist, print number -1. Otherwise, print <image> lines. In each line print three integers xi, yi, zi (1 ≤ xi, yi, zi ≤ n) — the i-th team.
If there are multiple answers, you are allowed to print any of them.
Examples
Input
3 0
Output
3 2 1
Input
6 4
1 2
2 3
3 4
5 6
Output
-1
Input
3 3
1 2
2 3
1 3
Output
3 2 1 | instruction | 0 | 82,878 | 17 | 165,756 |
Tags: brute force, dfs and similar, graphs
Correct Solution:
```
n, m = map(int, input().split())
visited = [False for i in range(n)]
g = [[] for _ in range(n)]
for _ in range(m):
a, b = map(int, input().split())
g[a-1].append(b-1)
g[b-1].append(a-1)
def dfs(x, s):
s.add(x)
visited[x] = True
for adj in g[x]:
if not visited[adj]:
dfs(adj, s)
groups = []
for i in range(n):
if g[i] and not visited[i]:
s = set()
dfs(i, s)
if len(s) > 3:
print(-1)
exit()
groups.append(list(s))
if len(groups) > n//3:
print(-1)
exit()
while len(groups) < n // 3:
groups.append([])
current_v = 0
for i in range(n):
if not visited[i]:
while len(groups[current_v]) == 3:
current_v += 1
groups[current_v].append(i)
for g in groups:
print(" ".join(map(lambda x: str(x+1), g)))
``` | output | 1 | 82,878 | 17 | 165,757 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A programming coach has n students to teach. We know that n is divisible by 3. Let's assume that all students are numbered from 1 to n, inclusive.
Before the university programming championship the coach wants to split all students into groups of three. For some pairs of students we know that they want to be on the same team. Besides, if the i-th student wants to be on the same team with the j-th one, then the j-th student wants to be on the same team with the i-th one. The coach wants the teams to show good results, so he wants the following condition to hold: if the i-th student wants to be on the same team with the j-th, then the i-th and the j-th students must be on the same team. Also, it is obvious that each student must be on exactly one team.
Help the coach and divide the teams the way he wants.
Input
The first line of the input contains integers n and m (3 ≤ n ≤ 48, <image>. Then follow m lines, each contains a pair of integers ai, bi (1 ≤ ai < bi ≤ n) — the pair ai, bi means that students with numbers ai and bi want to be on the same team.
It is guaranteed that n is divisible by 3. It is guaranteed that each pair ai, bi occurs in the input at most once.
Output
If the required division into teams doesn't exist, print number -1. Otherwise, print <image> lines. In each line print three integers xi, yi, zi (1 ≤ xi, yi, zi ≤ n) — the i-th team.
If there are multiple answers, you are allowed to print any of them.
Examples
Input
3 0
Output
3 2 1
Input
6 4
1 2
2 3
3 4
5 6
Output
-1
Input
3 3
1 2
2 3
1 3
Output
3 2 1 | instruction | 0 | 82,879 | 17 | 165,758 |
Tags: brute force, dfs and similar, graphs
Correct Solution:
```
n, m = map(int, input().split())
edges = [[*map(int, input().split())] for _ in range(m)]
parent = [*range(n)]
for i, j in edges:
if parent[i - 1] != parent[j - 1]:
old, new = parent[i - 1], parent[j - 1]
for j in range(n):
if parent[j] == old:
parent[j] = new
d = {}
for i in range(n):
if parent[i] in d:
d[parent[i]].append(i + 1)
else:
d[parent[i]] = [i + 1]
d = [i for i in d.values()]
d.sort(key=lambda x: -len(x))
pl = [len(i) for i in d]
th, tw, o = [], [], []
for i in d:
if len(i) == 3:
th.append(i)
elif len(i) == 2:
tw.append(i)
elif len(i) == 1:
o.append(i)
elif len(i) > 3:
print(-1)
exit()
if len(tw) > len(o):
print(-1)
exit()
for i in th:
print(*i)
for i, j in zip(tw, o):
print(i[0], i[1], j[0])
for i in range(len(tw), len(o), 3):
print(o[i][0], o[i + 1][0], o[i + 2][0])
``` | output | 1 | 82,879 | 17 | 165,759 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A programming coach has n students to teach. We know that n is divisible by 3. Let's assume that all students are numbered from 1 to n, inclusive.
Before the university programming championship the coach wants to split all students into groups of three. For some pairs of students we know that they want to be on the same team. Besides, if the i-th student wants to be on the same team with the j-th one, then the j-th student wants to be on the same team with the i-th one. The coach wants the teams to show good results, so he wants the following condition to hold: if the i-th student wants to be on the same team with the j-th, then the i-th and the j-th students must be on the same team. Also, it is obvious that each student must be on exactly one team.
Help the coach and divide the teams the way he wants.
Input
The first line of the input contains integers n and m (3 ≤ n ≤ 48, <image>. Then follow m lines, each contains a pair of integers ai, bi (1 ≤ ai < bi ≤ n) — the pair ai, bi means that students with numbers ai and bi want to be on the same team.
It is guaranteed that n is divisible by 3. It is guaranteed that each pair ai, bi occurs in the input at most once.
Output
If the required division into teams doesn't exist, print number -1. Otherwise, print <image> lines. In each line print three integers xi, yi, zi (1 ≤ xi, yi, zi ≤ n) — the i-th team.
If there are multiple answers, you are allowed to print any of them.
Examples
Input
3 0
Output
3 2 1
Input
6 4
1 2
2 3
3 4
5 6
Output
-1
Input
3 3
1 2
2 3
1 3
Output
3 2 1 | instruction | 0 | 82,880 | 17 | 165,760 |
Tags: brute force, dfs and similar, graphs
Correct Solution:
```
n,m = map(int, input().split())
arr = [i for i in range(n+1)]
mem = {}
for i in range(m):
a,b = map(int, input().split())
if arr[a] in mem:
mem[arr[a]].append(arr[b])
arr[b] = arr[a]
elif arr[b] in mem:
mem[arr[b]].append(arr[a])
arr[a] = arr[b]
else:
mem[arr[a]] = [arr[a],arr[b]]
arr[b] = arr[a]
for i in mem.keys():
arr[i] = 0
brr = []
for i in range(1,n+1):
if arr[i]==i:
brr.append(i)
l1 = len(brr)
crr = list(mem.values())
l2 = len(crr)
for i in range(l2):
crr[i] = list(set(crr[i]))
flag = 0
for i in crr:
if len(i)>3:
flag = 1
break
elif len(i)==3:
continue
else:
try:
i.append(brr[0])
brr = brr[1:]
except:
flag = 1
break
if flag!=1:
for i in crr:
print(*i)
l1 = len(brr)
for j in range(0,l1,3):
print(brr[j],brr[j+1],brr[j+2])
else:
print(-1)
``` | output | 1 | 82,880 | 17 | 165,761 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A programming coach has n students to teach. We know that n is divisible by 3. Let's assume that all students are numbered from 1 to n, inclusive.
Before the university programming championship the coach wants to split all students into groups of three. For some pairs of students we know that they want to be on the same team. Besides, if the i-th student wants to be on the same team with the j-th one, then the j-th student wants to be on the same team with the i-th one. The coach wants the teams to show good results, so he wants the following condition to hold: if the i-th student wants to be on the same team with the j-th, then the i-th and the j-th students must be on the same team. Also, it is obvious that each student must be on exactly one team.
Help the coach and divide the teams the way he wants.
Input
The first line of the input contains integers n and m (3 ≤ n ≤ 48, <image>. Then follow m lines, each contains a pair of integers ai, bi (1 ≤ ai < bi ≤ n) — the pair ai, bi means that students with numbers ai and bi want to be on the same team.
It is guaranteed that n is divisible by 3. It is guaranteed that each pair ai, bi occurs in the input at most once.
Output
If the required division into teams doesn't exist, print number -1. Otherwise, print <image> lines. In each line print three integers xi, yi, zi (1 ≤ xi, yi, zi ≤ n) — the i-th team.
If there are multiple answers, you are allowed to print any of them.
Examples
Input
3 0
Output
3 2 1
Input
6 4
1 2
2 3
3 4
5 6
Output
-1
Input
3 3
1 2
2 3
1 3
Output
3 2 1 | instruction | 0 | 82,881 | 17 | 165,762 |
Tags: brute force, dfs and similar, graphs
Correct Solution:
```
import sys,math
from collections import deque,defaultdict
import operator as op
from functools import reduce
from itertools import permutations
import heapq
#sys.setrecursionlimit(10**7)
# OneDrive\Documents\codeforces
I=sys.stdin.readline
alpha="abcdefghijklmnopqrstuvwxyz"
mod=10**9 + 7
"""
x_move=[-1,0,1,0,-1,1,1,-1]
y_move=[0,1,0,-1,1,1,-1,-1]
"""
def ii():
return int(I().strip())
def li():
return list(map(int,I().strip().split()))
def mi():
return map(int,I().strip().split())
def ncr(n, r):
r = min(r, n-r)
numer = reduce(op.mul, range(n, n-r, -1), 1)
denom = reduce(op.mul, range(1, r+1), 1)
return numer // denom
def ispali(s):
i=0
j=len(s)-1
while i<j:
if s[i]!=s[j]:
return False
i+=1
j-=1
return True
def isPrime(n):
if n<=1:
return False
elif n<=2:
return True
else:
for i in range(2,int(n**.5)+1):
if n%i==0:
return False
return True
def main():
n,m=mi()
d=defaultdict(list)
for i in range(m):
u,v=mi()
d[u].append(v)
d[v].append(u)
vis=[-1]*(n+1)
cnt=1
groups=defaultdict(list)
def dfs(i):
vis[i]=cnt
con=1
for child in d[i]:
if vis[child]==-1:
con+=dfs(child)
return con
f=0
for i in range(1,n+1):
if vis[i]==-1:
grup=dfs(i)
tmp=[]
for j in range(1,n+1):
if vis[j]==cnt:
tmp.append(j)
if grup==3:
groups[3].append(tmp)
elif grup==2:
groups[2].append(tmp)
elif grup==1:
groups[1].append(tmp)
else:
f=1
break
cnt+=1
# print(groups)
if len(groups[2])>len(groups[1]):
f=1
else:
for i in range(len(groups[2])):
groups[3].append(groups[2].pop()+groups[1].pop())
for i in range(len(groups[1])//3):
tmp=[]
for _ in range(3):
tmp+=groups[1].pop()
groups[3].append(tmp)
if len(groups[1])>0 or len(groups[3])!=n//3:
f=1
if f:
print(-1)
else:
for i in groups[3]:
print(*i)
if __name__ == '__main__':
main()
``` | output | 1 | 82,881 | 17 | 165,763 |
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