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 |
|---|---|---|---|---|---|
Provide tags and a correct Python 3 solution for this coding contest problem.
At the school where Vasya is studying, preparations are underway for the graduation ceremony. One of the planned performances is a ball, which will be attended by pairs of boys and girls.
Each class must present two couples to the ball. In Vasya's class, a boys and b girls wish to participate. But not all boys and not all girls are ready to dance in pairs.
Formally, you know k possible one-boy-one-girl pairs. You need to choose two of these pairs so that no person is in more than one pair.
For example, if a=3, b=4, k=4 and the couples (1, 2), (1, 3), (2, 2), (3, 4) are ready to dance together (in each pair, the boy's number comes first, then the girl's number), then the following combinations of two pairs are possible (not all possible options are listed below):
* (1, 3) and (2, 2);
* (3, 4) and (1, 3);
But the following combinations are not possible:
* (1, 3) and (1, 2) β the first boy enters two pairs;
* (1, 2) and (2, 2) β the second girl enters two pairs;
Find the number of ways to select two pairs that match the condition above. Two ways are considered different if they consist of different pairs.
Input
The first line contains one integer t (1 β€ t β€ 10^4) β the number of test cases. Then t test cases follow.
The first line of each test case contains three integers a, b and k (1 β€ a, b, k β€ 2 β
10^5) β the number of boys and girls in the class and the number of couples ready to dance together.
The second line of each test case contains k integers a_1, a_2, β¦ a_k. (1 β€ a_i β€ a), where a_i is the number of the boy in the pair with the number i.
The third line of each test case contains k integers b_1, b_2, β¦ b_k. (1 β€ b_i β€ b), where b_i is the number of the girl in the pair with the number i.
It is guaranteed that the sums of a, b, and k over all test cases do not exceed 2 β
10^5.
It is guaranteed that each pair is specified at most once in one test case.
Output
For each test case, on a separate line print one integer β the number of ways to choose two pairs that match the condition above.
Example
Input
3
3 4 4
1 1 2 3
2 3 2 4
1 1 1
1
1
2 2 4
1 1 2 2
1 2 1 2
Output
4
0
2
Note
In the first test case, the following combinations of pairs fit:
* (1, 2) and (3, 4);
* (1, 3) and (2, 2);
* (1, 3) and (3, 4);
* (2, 2) and (3, 4).
There is only one pair in the second test case.
In the third test case, the following combinations of pairs fit:
* (1, 1) and (2, 2);
* (1, 2) and (2, 1). | instruction | 0 | 88,627 | 17 | 177,254 |
Tags: combinatorics, graphs, math
Correct Solution:
```
from math import sqrt
import operator
import sys
# inf = open('input.txt', 'r')
inf = sys.stdin
input = inf.readline
def read_one_int():
return int(input().rstrip('\n'))
def read_list_of_ints():
res = [int(val) for val in (input().rstrip('\n')).split(' ')]
return res
def read_str():
return input().rstrip()
def check_seq(k, a_lst, b_lst):
res = 0
a_to_b = {}
b_to_a = {}
for i in range(k):
if a_lst[i] not in a_to_b:
a_to_b[a_lst[i]] = []
a_to_b[a_lst[i]].append(b_lst[i])
if b_lst[i] not in b_to_a:
b_to_a[b_lst[i]] = []
b_to_a[b_lst[i]].append(a_lst[i])
for a_cur, a_mp in a_to_b.items():
for b_cur in a_mp:
res += k - len(a_mp) - len(b_to_a[b_cur]) + 1
return res // 2
def main():
samples = read_one_int()
for _ in range(samples):
a, b, k = read_list_of_ints()
a_lst = read_list_of_ints()
b_lst = read_list_of_ints()
cur_res = check_seq(k, a_lst, b_lst)
print(cur_res)
if __name__== '__main__':
main()
``` | output | 1 | 88,627 | 17 | 177,255 |
Provide tags and a correct Python 3 solution for this coding contest problem.
At the school where Vasya is studying, preparations are underway for the graduation ceremony. One of the planned performances is a ball, which will be attended by pairs of boys and girls.
Each class must present two couples to the ball. In Vasya's class, a boys and b girls wish to participate. But not all boys and not all girls are ready to dance in pairs.
Formally, you know k possible one-boy-one-girl pairs. You need to choose two of these pairs so that no person is in more than one pair.
For example, if a=3, b=4, k=4 and the couples (1, 2), (1, 3), (2, 2), (3, 4) are ready to dance together (in each pair, the boy's number comes first, then the girl's number), then the following combinations of two pairs are possible (not all possible options are listed below):
* (1, 3) and (2, 2);
* (3, 4) and (1, 3);
But the following combinations are not possible:
* (1, 3) and (1, 2) β the first boy enters two pairs;
* (1, 2) and (2, 2) β the second girl enters two pairs;
Find the number of ways to select two pairs that match the condition above. Two ways are considered different if they consist of different pairs.
Input
The first line contains one integer t (1 β€ t β€ 10^4) β the number of test cases. Then t test cases follow.
The first line of each test case contains three integers a, b and k (1 β€ a, b, k β€ 2 β
10^5) β the number of boys and girls in the class and the number of couples ready to dance together.
The second line of each test case contains k integers a_1, a_2, β¦ a_k. (1 β€ a_i β€ a), where a_i is the number of the boy in the pair with the number i.
The third line of each test case contains k integers b_1, b_2, β¦ b_k. (1 β€ b_i β€ b), where b_i is the number of the girl in the pair with the number i.
It is guaranteed that the sums of a, b, and k over all test cases do not exceed 2 β
10^5.
It is guaranteed that each pair is specified at most once in one test case.
Output
For each test case, on a separate line print one integer β the number of ways to choose two pairs that match the condition above.
Example
Input
3
3 4 4
1 1 2 3
2 3 2 4
1 1 1
1
1
2 2 4
1 1 2 2
1 2 1 2
Output
4
0
2
Note
In the first test case, the following combinations of pairs fit:
* (1, 2) and (3, 4);
* (1, 3) and (2, 2);
* (1, 3) and (3, 4);
* (2, 2) and (3, 4).
There is only one pair in the second test case.
In the third test case, the following combinations of pairs fit:
* (1, 1) and (2, 2);
* (1, 2) and (2, 1). | instruction | 0 | 88,628 | 17 | 177,256 |
Tags: combinatorics, graphs, math
Correct Solution:
```
from math import *
from collections import *
def read():
return list(map(int, input().split(' ')))
n, = read()
for _ in range(n):
a, b, k = read()
boy = read()
girl = read()
pair = [(boy[i], girl[i]) for i in range(k)]
if(k==1):
print(0)
continue
# # print(pair)
# ans = 0
# for i in range(k):
# for j in range(i+1,k):
# if pair[i][0]!=pair[j][0] and pair[i][1]!=pair[j][1]:
# ans += 1
# print(ans)
dic1, dic2 = defaultdict(int), defaultdict(int)
for (i, j) in pair:
dic1[i] += 1
dic2[j] += 1
# print(pair)
ans = 0
for (i, j) in pair:
# print(i,j,(dic1[i] + dic2[j] - 1))
ans += k - (dic1[i] + dic2[j] - 1)
print(ans//2)
``` | output | 1 | 88,628 | 17 | 177,257 |
Provide tags and a correct Python 3 solution for this coding contest problem.
At the school where Vasya is studying, preparations are underway for the graduation ceremony. One of the planned performances is a ball, which will be attended by pairs of boys and girls.
Each class must present two couples to the ball. In Vasya's class, a boys and b girls wish to participate. But not all boys and not all girls are ready to dance in pairs.
Formally, you know k possible one-boy-one-girl pairs. You need to choose two of these pairs so that no person is in more than one pair.
For example, if a=3, b=4, k=4 and the couples (1, 2), (1, 3), (2, 2), (3, 4) are ready to dance together (in each pair, the boy's number comes first, then the girl's number), then the following combinations of two pairs are possible (not all possible options are listed below):
* (1, 3) and (2, 2);
* (3, 4) and (1, 3);
But the following combinations are not possible:
* (1, 3) and (1, 2) β the first boy enters two pairs;
* (1, 2) and (2, 2) β the second girl enters two pairs;
Find the number of ways to select two pairs that match the condition above. Two ways are considered different if they consist of different pairs.
Input
The first line contains one integer t (1 β€ t β€ 10^4) β the number of test cases. Then t test cases follow.
The first line of each test case contains three integers a, b and k (1 β€ a, b, k β€ 2 β
10^5) β the number of boys and girls in the class and the number of couples ready to dance together.
The second line of each test case contains k integers a_1, a_2, β¦ a_k. (1 β€ a_i β€ a), where a_i is the number of the boy in the pair with the number i.
The third line of each test case contains k integers b_1, b_2, β¦ b_k. (1 β€ b_i β€ b), where b_i is the number of the girl in the pair with the number i.
It is guaranteed that the sums of a, b, and k over all test cases do not exceed 2 β
10^5.
It is guaranteed that each pair is specified at most once in one test case.
Output
For each test case, on a separate line print one integer β the number of ways to choose two pairs that match the condition above.
Example
Input
3
3 4 4
1 1 2 3
2 3 2 4
1 1 1
1
1
2 2 4
1 1 2 2
1 2 1 2
Output
4
0
2
Note
In the first test case, the following combinations of pairs fit:
* (1, 2) and (3, 4);
* (1, 3) and (2, 2);
* (1, 3) and (3, 4);
* (2, 2) and (3, 4).
There is only one pair in the second test case.
In the third test case, the following combinations of pairs fit:
* (1, 1) and (2, 2);
* (1, 2) and (2, 1). | instruction | 0 | 88,629 | 17 | 177,258 |
Tags: combinatorics, graphs, math
Correct Solution:
```
import math, sys
from collections import defaultdict, Counter, deque
from bisect import bisect_left, bisect_right
INF = float('inf')
MOD = int(1e9) + 7
MAX = int(1e6) + 1
def solve():
a, b, k = vars()
boys = array()
girls = array()
pair = []
for i in range(k):
pair.append([boys[i], girls[i]])
pair = sorted(pair)
index = defaultdict(list)
for i in range(k):
index[pair[i][1]].append(i)
# print(pair)
ans = 0
i = 0
while i < k:
d = 0
tmp = i
while i < k - 1 and pair[i + 1][0] == pair[i][0]:
i += 1
d += 1
db = i
i = tmp
# print('#', db, pair[i])
while i <= db:
girl = pair[i][1]
discard = len(index[girl]) - bisect_right(index[girl], db)
# print(i, discard, db + 1)
ans += k - db - discard - 1
i += 1
print(ans)
def main():
t = 1
t = int(input())
for _ in range(t):
solve()
def gcd(a, b):
while b:
a, b = b, a%b
return a
def input():
return sys.stdin.readline().rstrip('\n').strip()
def print(*args, sep=' ', end='\n'):
first = True
for arg in args:
if not first:
sys.stdout.write(sep)
sys.stdout.write(str(arg))
first = False
sys.stdout.write(end)
primes = [ 1 for i in range(MAX) ]
def sieve():
global primes
primes[0] = primes[1] = 0
i = 2
while i <= MAX ** 0.5:
j = i * i
while primes[i] and j < MAX:
if j % i == 0:
primes[j] = 0
j += i
i += 1
def vars():
return map(int, input().split())
def array():
return list(map(int, input().split()))
if __name__ == "__main__":
main()
``` | output | 1 | 88,629 | 17 | 177,259 |
Provide tags and a correct Python 3 solution for this coding contest problem.
At the school where Vasya is studying, preparations are underway for the graduation ceremony. One of the planned performances is a ball, which will be attended by pairs of boys and girls.
Each class must present two couples to the ball. In Vasya's class, a boys and b girls wish to participate. But not all boys and not all girls are ready to dance in pairs.
Formally, you know k possible one-boy-one-girl pairs. You need to choose two of these pairs so that no person is in more than one pair.
For example, if a=3, b=4, k=4 and the couples (1, 2), (1, 3), (2, 2), (3, 4) are ready to dance together (in each pair, the boy's number comes first, then the girl's number), then the following combinations of two pairs are possible (not all possible options are listed below):
* (1, 3) and (2, 2);
* (3, 4) and (1, 3);
But the following combinations are not possible:
* (1, 3) and (1, 2) β the first boy enters two pairs;
* (1, 2) and (2, 2) β the second girl enters two pairs;
Find the number of ways to select two pairs that match the condition above. Two ways are considered different if they consist of different pairs.
Input
The first line contains one integer t (1 β€ t β€ 10^4) β the number of test cases. Then t test cases follow.
The first line of each test case contains three integers a, b and k (1 β€ a, b, k β€ 2 β
10^5) β the number of boys and girls in the class and the number of couples ready to dance together.
The second line of each test case contains k integers a_1, a_2, β¦ a_k. (1 β€ a_i β€ a), where a_i is the number of the boy in the pair with the number i.
The third line of each test case contains k integers b_1, b_2, β¦ b_k. (1 β€ b_i β€ b), where b_i is the number of the girl in the pair with the number i.
It is guaranteed that the sums of a, b, and k over all test cases do not exceed 2 β
10^5.
It is guaranteed that each pair is specified at most once in one test case.
Output
For each test case, on a separate line print one integer β the number of ways to choose two pairs that match the condition above.
Example
Input
3
3 4 4
1 1 2 3
2 3 2 4
1 1 1
1
1
2 2 4
1 1 2 2
1 2 1 2
Output
4
0
2
Note
In the first test case, the following combinations of pairs fit:
* (1, 2) and (3, 4);
* (1, 3) and (2, 2);
* (1, 3) and (3, 4);
* (2, 2) and (3, 4).
There is only one pair in the second test case.
In the third test case, the following combinations of pairs fit:
* (1, 1) and (2, 2);
* (1, 2) and (2, 1). | instruction | 0 | 88,630 | 17 | 177,260 |
Tags: combinatorics, graphs, math
Correct Solution:
```
for _ in range(int(input())):
a, b, k = map(int, input().split())
A = list(map(int, input().split()))
B = list(map(int, input().split()))
deg_a = [0]*(a+1)
deg_b = [0]*(b+1)
for i in range(k):
deg_a[A[i]] += 1
deg_b[B[i]] += 1
ans = 0
for i in range(k):
ans += k - deg_a[A[i]] - deg_b[B[i]] + 1
print(ans // 2)
``` | output | 1 | 88,630 | 17 | 177,261 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
At the school where Vasya is studying, preparations are underway for the graduation ceremony. One of the planned performances is a ball, which will be attended by pairs of boys and girls.
Each class must present two couples to the ball. In Vasya's class, a boys and b girls wish to participate. But not all boys and not all girls are ready to dance in pairs.
Formally, you know k possible one-boy-one-girl pairs. You need to choose two of these pairs so that no person is in more than one pair.
For example, if a=3, b=4, k=4 and the couples (1, 2), (1, 3), (2, 2), (3, 4) are ready to dance together (in each pair, the boy's number comes first, then the girl's number), then the following combinations of two pairs are possible (not all possible options are listed below):
* (1, 3) and (2, 2);
* (3, 4) and (1, 3);
But the following combinations are not possible:
* (1, 3) and (1, 2) β the first boy enters two pairs;
* (1, 2) and (2, 2) β the second girl enters two pairs;
Find the number of ways to select two pairs that match the condition above. Two ways are considered different if they consist of different pairs.
Input
The first line contains one integer t (1 β€ t β€ 10^4) β the number of test cases. Then t test cases follow.
The first line of each test case contains three integers a, b and k (1 β€ a, b, k β€ 2 β
10^5) β the number of boys and girls in the class and the number of couples ready to dance together.
The second line of each test case contains k integers a_1, a_2, β¦ a_k. (1 β€ a_i β€ a), where a_i is the number of the boy in the pair with the number i.
The third line of each test case contains k integers b_1, b_2, β¦ b_k. (1 β€ b_i β€ b), where b_i is the number of the girl in the pair with the number i.
It is guaranteed that the sums of a, b, and k over all test cases do not exceed 2 β
10^5.
It is guaranteed that each pair is specified at most once in one test case.
Output
For each test case, on a separate line print one integer β the number of ways to choose two pairs that match the condition above.
Example
Input
3
3 4 4
1 1 2 3
2 3 2 4
1 1 1
1
1
2 2 4
1 1 2 2
1 2 1 2
Output
4
0
2
Note
In the first test case, the following combinations of pairs fit:
* (1, 2) and (3, 4);
* (1, 3) and (2, 2);
* (1, 3) and (3, 4);
* (2, 2) and (3, 4).
There is only one pair in the second test case.
In the third test case, the following combinations of pairs fit:
* (1, 1) and (2, 2);
* (1, 2) and (2, 1).
Submitted Solution:
```
from collections import Counter
def read_int():
return int(input())
def read_ints():
return map(int, input().split(' '))
t = read_int()
for case_num in range(t):
a, b, k = read_ints()
pa = list(read_ints())
pb = list(read_ints())
ca = Counter()
cb = Counter()
for i in range(k):
ca[pa[i]] += 1
cb[pb[i]] += 1
ans = 0
for i in range(k - 1):
ca[pa[i]] -= 1
cb[pb[i]] -= 1
ans += k - i - 1 - ca[pa[i]] - cb[pb[i]]
print(ans)
``` | instruction | 0 | 88,631 | 17 | 177,262 |
Yes | output | 1 | 88,631 | 17 | 177,263 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
At the school where Vasya is studying, preparations are underway for the graduation ceremony. One of the planned performances is a ball, which will be attended by pairs of boys and girls.
Each class must present two couples to the ball. In Vasya's class, a boys and b girls wish to participate. But not all boys and not all girls are ready to dance in pairs.
Formally, you know k possible one-boy-one-girl pairs. You need to choose two of these pairs so that no person is in more than one pair.
For example, if a=3, b=4, k=4 and the couples (1, 2), (1, 3), (2, 2), (3, 4) are ready to dance together (in each pair, the boy's number comes first, then the girl's number), then the following combinations of two pairs are possible (not all possible options are listed below):
* (1, 3) and (2, 2);
* (3, 4) and (1, 3);
But the following combinations are not possible:
* (1, 3) and (1, 2) β the first boy enters two pairs;
* (1, 2) and (2, 2) β the second girl enters two pairs;
Find the number of ways to select two pairs that match the condition above. Two ways are considered different if they consist of different pairs.
Input
The first line contains one integer t (1 β€ t β€ 10^4) β the number of test cases. Then t test cases follow.
The first line of each test case contains three integers a, b and k (1 β€ a, b, k β€ 2 β
10^5) β the number of boys and girls in the class and the number of couples ready to dance together.
The second line of each test case contains k integers a_1, a_2, β¦ a_k. (1 β€ a_i β€ a), where a_i is the number of the boy in the pair with the number i.
The third line of each test case contains k integers b_1, b_2, β¦ b_k. (1 β€ b_i β€ b), where b_i is the number of the girl in the pair with the number i.
It is guaranteed that the sums of a, b, and k over all test cases do not exceed 2 β
10^5.
It is guaranteed that each pair is specified at most once in one test case.
Output
For each test case, on a separate line print one integer β the number of ways to choose two pairs that match the condition above.
Example
Input
3
3 4 4
1 1 2 3
2 3 2 4
1 1 1
1
1
2 2 4
1 1 2 2
1 2 1 2
Output
4
0
2
Note
In the first test case, the following combinations of pairs fit:
* (1, 2) and (3, 4);
* (1, 3) and (2, 2);
* (1, 3) and (3, 4);
* (2, 2) and (3, 4).
There is only one pair in the second test case.
In the third test case, the following combinations of pairs fit:
* (1, 1) and (2, 2);
* (1, 2) and (2, 1).
Submitted Solution:
```
###pyrival template for fast IO
import os
import sys
from io import BytesIO, IOBase
# 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")
t=int(input())
while t>0:
t-=1
a,b,k=[int(x) for x in input().split()]
boys=[int(x) for x in input().split()]
girls=[int(x) for x in input().split()]
same=0
boys.sort()
girls.sort()
temp1=1;temp2=1
for i in range(1,k):
if boys[i]==boys[i-1]:
temp1+=1
else:
same+=temp1*(temp1-1)//2
#print(temp1*(temp1-1)//2)
temp1=1
if girls[i]==girls[i-1]:
temp2+=1
else:
same+=temp2*(temp2-1)//2
#print(temp2*(temp2-1)//2)
temp2=1
same+=temp2*(temp2-1)//2
same+=temp1*(temp1-1)//2
sys.stdout.write(str(k*(k-1)//2-same)+"\n")
``` | instruction | 0 | 88,632 | 17 | 177,264 |
Yes | output | 1 | 88,632 | 17 | 177,265 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
At the school where Vasya is studying, preparations are underway for the graduation ceremony. One of the planned performances is a ball, which will be attended by pairs of boys and girls.
Each class must present two couples to the ball. In Vasya's class, a boys and b girls wish to participate. But not all boys and not all girls are ready to dance in pairs.
Formally, you know k possible one-boy-one-girl pairs. You need to choose two of these pairs so that no person is in more than one pair.
For example, if a=3, b=4, k=4 and the couples (1, 2), (1, 3), (2, 2), (3, 4) are ready to dance together (in each pair, the boy's number comes first, then the girl's number), then the following combinations of two pairs are possible (not all possible options are listed below):
* (1, 3) and (2, 2);
* (3, 4) and (1, 3);
But the following combinations are not possible:
* (1, 3) and (1, 2) β the first boy enters two pairs;
* (1, 2) and (2, 2) β the second girl enters two pairs;
Find the number of ways to select two pairs that match the condition above. Two ways are considered different if they consist of different pairs.
Input
The first line contains one integer t (1 β€ t β€ 10^4) β the number of test cases. Then t test cases follow.
The first line of each test case contains three integers a, b and k (1 β€ a, b, k β€ 2 β
10^5) β the number of boys and girls in the class and the number of couples ready to dance together.
The second line of each test case contains k integers a_1, a_2, β¦ a_k. (1 β€ a_i β€ a), where a_i is the number of the boy in the pair with the number i.
The third line of each test case contains k integers b_1, b_2, β¦ b_k. (1 β€ b_i β€ b), where b_i is the number of the girl in the pair with the number i.
It is guaranteed that the sums of a, b, and k over all test cases do not exceed 2 β
10^5.
It is guaranteed that each pair is specified at most once in one test case.
Output
For each test case, on a separate line print one integer β the number of ways to choose two pairs that match the condition above.
Example
Input
3
3 4 4
1 1 2 3
2 3 2 4
1 1 1
1
1
2 2 4
1 1 2 2
1 2 1 2
Output
4
0
2
Note
In the first test case, the following combinations of pairs fit:
* (1, 2) and (3, 4);
* (1, 3) and (2, 2);
* (1, 3) and (3, 4);
* (2, 2) and (3, 4).
There is only one pair in the second test case.
In the third test case, the following combinations of pairs fit:
* (1, 1) and (2, 2);
* (1, 2) and (2, 1).
Submitted Solution:
```
from collections import Counter
t=int(input())
for z in range(t):
a,b,k=[int(q)for q in input().split()]
al=[int(q)for q in input().split()]
bl=[int(q)for q in input().split()]
alc=Counter(al)
blc=Counter(bl)
ans=0
for i in range(k):
da=alc[al[i]]
db=blc[bl[i]]
ans+=(k-da-db+1)
print(ans//2)
``` | instruction | 0 | 88,633 | 17 | 177,266 |
Yes | output | 1 | 88,633 | 17 | 177,267 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
At the school where Vasya is studying, preparations are underway for the graduation ceremony. One of the planned performances is a ball, which will be attended by pairs of boys and girls.
Each class must present two couples to the ball. In Vasya's class, a boys and b girls wish to participate. But not all boys and not all girls are ready to dance in pairs.
Formally, you know k possible one-boy-one-girl pairs. You need to choose two of these pairs so that no person is in more than one pair.
For example, if a=3, b=4, k=4 and the couples (1, 2), (1, 3), (2, 2), (3, 4) are ready to dance together (in each pair, the boy's number comes first, then the girl's number), then the following combinations of two pairs are possible (not all possible options are listed below):
* (1, 3) and (2, 2);
* (3, 4) and (1, 3);
But the following combinations are not possible:
* (1, 3) and (1, 2) β the first boy enters two pairs;
* (1, 2) and (2, 2) β the second girl enters two pairs;
Find the number of ways to select two pairs that match the condition above. Two ways are considered different if they consist of different pairs.
Input
The first line contains one integer t (1 β€ t β€ 10^4) β the number of test cases. Then t test cases follow.
The first line of each test case contains three integers a, b and k (1 β€ a, b, k β€ 2 β
10^5) β the number of boys and girls in the class and the number of couples ready to dance together.
The second line of each test case contains k integers a_1, a_2, β¦ a_k. (1 β€ a_i β€ a), where a_i is the number of the boy in the pair with the number i.
The third line of each test case contains k integers b_1, b_2, β¦ b_k. (1 β€ b_i β€ b), where b_i is the number of the girl in the pair with the number i.
It is guaranteed that the sums of a, b, and k over all test cases do not exceed 2 β
10^5.
It is guaranteed that each pair is specified at most once in one test case.
Output
For each test case, on a separate line print one integer β the number of ways to choose two pairs that match the condition above.
Example
Input
3
3 4 4
1 1 2 3
2 3 2 4
1 1 1
1
1
2 2 4
1 1 2 2
1 2 1 2
Output
4
0
2
Note
In the first test case, the following combinations of pairs fit:
* (1, 2) and (3, 4);
* (1, 3) and (2, 2);
* (1, 3) and (3, 4);
* (2, 2) and (3, 4).
There is only one pair in the second test case.
In the third test case, the following combinations of pairs fit:
* (1, 1) and (2, 2);
* (1, 2) and (2, 1).
Submitted Solution:
```
R=lambda:list(map(int,input().split()))
for _ in ' '*int(input()):
an,bn,k=R()
a=R()
b=R()
edges=[]
for i in range(k): edges.append([a[i], b[i]])
dega=[0]*(an+1)
degb=[0]*(bn+1)
for boy, girl in edges:
dega[boy] += 1
degb[girl] += 1
c=0
for b, g in edges:
c += (k + 1 - dega[b] - degb[g])
print(c//2)
``` | instruction | 0 | 88,634 | 17 | 177,268 |
Yes | output | 1 | 88,634 | 17 | 177,269 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
At the school where Vasya is studying, preparations are underway for the graduation ceremony. One of the planned performances is a ball, which will be attended by pairs of boys and girls.
Each class must present two couples to the ball. In Vasya's class, a boys and b girls wish to participate. But not all boys and not all girls are ready to dance in pairs.
Formally, you know k possible one-boy-one-girl pairs. You need to choose two of these pairs so that no person is in more than one pair.
For example, if a=3, b=4, k=4 and the couples (1, 2), (1, 3), (2, 2), (3, 4) are ready to dance together (in each pair, the boy's number comes first, then the girl's number), then the following combinations of two pairs are possible (not all possible options are listed below):
* (1, 3) and (2, 2);
* (3, 4) and (1, 3);
But the following combinations are not possible:
* (1, 3) and (1, 2) β the first boy enters two pairs;
* (1, 2) and (2, 2) β the second girl enters two pairs;
Find the number of ways to select two pairs that match the condition above. Two ways are considered different if they consist of different pairs.
Input
The first line contains one integer t (1 β€ t β€ 10^4) β the number of test cases. Then t test cases follow.
The first line of each test case contains three integers a, b and k (1 β€ a, b, k β€ 2 β
10^5) β the number of boys and girls in the class and the number of couples ready to dance together.
The second line of each test case contains k integers a_1, a_2, β¦ a_k. (1 β€ a_i β€ a), where a_i is the number of the boy in the pair with the number i.
The third line of each test case contains k integers b_1, b_2, β¦ b_k. (1 β€ b_i β€ b), where b_i is the number of the girl in the pair with the number i.
It is guaranteed that the sums of a, b, and k over all test cases do not exceed 2 β
10^5.
It is guaranteed that each pair is specified at most once in one test case.
Output
For each test case, on a separate line print one integer β the number of ways to choose two pairs that match the condition above.
Example
Input
3
3 4 4
1 1 2 3
2 3 2 4
1 1 1
1
1
2 2 4
1 1 2 2
1 2 1 2
Output
4
0
2
Note
In the first test case, the following combinations of pairs fit:
* (1, 2) and (3, 4);
* (1, 3) and (2, 2);
* (1, 3) and (3, 4);
* (2, 2) and (3, 4).
There is only one pair in the second test case.
In the third test case, the following combinations of pairs fit:
* (1, 1) and (2, 2);
* (1, 2) and (2, 1).
Submitted Solution:
```
t=int(input())
for i in range(t):
l=list(map(int,input().strip().split()))
l1=list(map(int,input().strip().split()))
l2=list(map(int,input().strip().split()))
s=0
if l[2]==1:
print("1")
continue
for i in range(l[2]-1):
count=0
for j in range(i+1,l[2]):
if l1[i]==l1[j]:
count+=1
if l2[i]==l2[j]:
count+=1
s+=(l[2]-i)-count-1
print(s)
``` | instruction | 0 | 88,635 | 17 | 177,270 |
No | output | 1 | 88,635 | 17 | 177,271 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
At the school where Vasya is studying, preparations are underway for the graduation ceremony. One of the planned performances is a ball, which will be attended by pairs of boys and girls.
Each class must present two couples to the ball. In Vasya's class, a boys and b girls wish to participate. But not all boys and not all girls are ready to dance in pairs.
Formally, you know k possible one-boy-one-girl pairs. You need to choose two of these pairs so that no person is in more than one pair.
For example, if a=3, b=4, k=4 and the couples (1, 2), (1, 3), (2, 2), (3, 4) are ready to dance together (in each pair, the boy's number comes first, then the girl's number), then the following combinations of two pairs are possible (not all possible options are listed below):
* (1, 3) and (2, 2);
* (3, 4) and (1, 3);
But the following combinations are not possible:
* (1, 3) and (1, 2) β the first boy enters two pairs;
* (1, 2) and (2, 2) β the second girl enters two pairs;
Find the number of ways to select two pairs that match the condition above. Two ways are considered different if they consist of different pairs.
Input
The first line contains one integer t (1 β€ t β€ 10^4) β the number of test cases. Then t test cases follow.
The first line of each test case contains three integers a, b and k (1 β€ a, b, k β€ 2 β
10^5) β the number of boys and girls in the class and the number of couples ready to dance together.
The second line of each test case contains k integers a_1, a_2, β¦ a_k. (1 β€ a_i β€ a), where a_i is the number of the boy in the pair with the number i.
The third line of each test case contains k integers b_1, b_2, β¦ b_k. (1 β€ b_i β€ b), where b_i is the number of the girl in the pair with the number i.
It is guaranteed that the sums of a, b, and k over all test cases do not exceed 2 β
10^5.
It is guaranteed that each pair is specified at most once in one test case.
Output
For each test case, on a separate line print one integer β the number of ways to choose two pairs that match the condition above.
Example
Input
3
3 4 4
1 1 2 3
2 3 2 4
1 1 1
1
1
2 2 4
1 1 2 2
1 2 1 2
Output
4
0
2
Note
In the first test case, the following combinations of pairs fit:
* (1, 2) and (3, 4);
* (1, 3) and (2, 2);
* (1, 3) and (3, 4);
* (2, 2) and (3, 4).
There is only one pair in the second test case.
In the third test case, the following combinations of pairs fit:
* (1, 1) and (2, 2);
* (1, 2) and (2, 1).
Submitted Solution:
```
t = int(input())
def func(a,b,k,boys,girls):
a = 0
for i in range(k-1):
for j in range(i+1,k):
if boys[i]!=boys[j] and girls[i]!=girls[j]:
a+=1
return a
for i in range(t):
a,b,k = list(map(int,input().split()))
boys = list(map(int,input().split()))
girls = list(map(int,input().split()))
print(a,b,k)
print(boys)
print(girls)
print(func(a,b,k,boys,girls))
# print(func(y))
``` | instruction | 0 | 88,636 | 17 | 177,272 |
No | output | 1 | 88,636 | 17 | 177,273 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
At the school where Vasya is studying, preparations are underway for the graduation ceremony. One of the planned performances is a ball, which will be attended by pairs of boys and girls.
Each class must present two couples to the ball. In Vasya's class, a boys and b girls wish to participate. But not all boys and not all girls are ready to dance in pairs.
Formally, you know k possible one-boy-one-girl pairs. You need to choose two of these pairs so that no person is in more than one pair.
For example, if a=3, b=4, k=4 and the couples (1, 2), (1, 3), (2, 2), (3, 4) are ready to dance together (in each pair, the boy's number comes first, then the girl's number), then the following combinations of two pairs are possible (not all possible options are listed below):
* (1, 3) and (2, 2);
* (3, 4) and (1, 3);
But the following combinations are not possible:
* (1, 3) and (1, 2) β the first boy enters two pairs;
* (1, 2) and (2, 2) β the second girl enters two pairs;
Find the number of ways to select two pairs that match the condition above. Two ways are considered different if they consist of different pairs.
Input
The first line contains one integer t (1 β€ t β€ 10^4) β the number of test cases. Then t test cases follow.
The first line of each test case contains three integers a, b and k (1 β€ a, b, k β€ 2 β
10^5) β the number of boys and girls in the class and the number of couples ready to dance together.
The second line of each test case contains k integers a_1, a_2, β¦ a_k. (1 β€ a_i β€ a), where a_i is the number of the boy in the pair with the number i.
The third line of each test case contains k integers b_1, b_2, β¦ b_k. (1 β€ b_i β€ b), where b_i is the number of the girl in the pair with the number i.
It is guaranteed that the sums of a, b, and k over all test cases do not exceed 2 β
10^5.
It is guaranteed that each pair is specified at most once in one test case.
Output
For each test case, on a separate line print one integer β the number of ways to choose two pairs that match the condition above.
Example
Input
3
3 4 4
1 1 2 3
2 3 2 4
1 1 1
1
1
2 2 4
1 1 2 2
1 2 1 2
Output
4
0
2
Note
In the first test case, the following combinations of pairs fit:
* (1, 2) and (3, 4);
* (1, 3) and (2, 2);
* (1, 3) and (3, 4);
* (2, 2) and (3, 4).
There is only one pair in the second test case.
In the third test case, the following combinations of pairs fit:
* (1, 1) and (2, 2);
* (1, 2) and (2, 1).
Submitted 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")
t=int(input())
for _ in range(t):
a,b,k=map(int,input().split())
l1=list(map(int,input().split()))
l2=list(map(int,input().split()))
ax,bx=set(l1),set(l2)
cx=set()
for i in range(k):
cx.add((l1[i],l2[i]))
t=(k*(k-1))//2
print(t-((k-len(ax))+(k-len(bx))))
``` | instruction | 0 | 88,637 | 17 | 177,274 |
No | output | 1 | 88,637 | 17 | 177,275 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
At the school where Vasya is studying, preparations are underway for the graduation ceremony. One of the planned performances is a ball, which will be attended by pairs of boys and girls.
Each class must present two couples to the ball. In Vasya's class, a boys and b girls wish to participate. But not all boys and not all girls are ready to dance in pairs.
Formally, you know k possible one-boy-one-girl pairs. You need to choose two of these pairs so that no person is in more than one pair.
For example, if a=3, b=4, k=4 and the couples (1, 2), (1, 3), (2, 2), (3, 4) are ready to dance together (in each pair, the boy's number comes first, then the girl's number), then the following combinations of two pairs are possible (not all possible options are listed below):
* (1, 3) and (2, 2);
* (3, 4) and (1, 3);
But the following combinations are not possible:
* (1, 3) and (1, 2) β the first boy enters two pairs;
* (1, 2) and (2, 2) β the second girl enters two pairs;
Find the number of ways to select two pairs that match the condition above. Two ways are considered different if they consist of different pairs.
Input
The first line contains one integer t (1 β€ t β€ 10^4) β the number of test cases. Then t test cases follow.
The first line of each test case contains three integers a, b and k (1 β€ a, b, k β€ 2 β
10^5) β the number of boys and girls in the class and the number of couples ready to dance together.
The second line of each test case contains k integers a_1, a_2, β¦ a_k. (1 β€ a_i β€ a), where a_i is the number of the boy in the pair with the number i.
The third line of each test case contains k integers b_1, b_2, β¦ b_k. (1 β€ b_i β€ b), where b_i is the number of the girl in the pair with the number i.
It is guaranteed that the sums of a, b, and k over all test cases do not exceed 2 β
10^5.
It is guaranteed that each pair is specified at most once in one test case.
Output
For each test case, on a separate line print one integer β the number of ways to choose two pairs that match the condition above.
Example
Input
3
3 4 4
1 1 2 3
2 3 2 4
1 1 1
1
1
2 2 4
1 1 2 2
1 2 1 2
Output
4
0
2
Note
In the first test case, the following combinations of pairs fit:
* (1, 2) and (3, 4);
* (1, 3) and (2, 2);
* (1, 3) and (3, 4);
* (2, 2) and (3, 4).
There is only one pair in the second test case.
In the third test case, the following combinations of pairs fit:
* (1, 1) and (2, 2);
* (1, 2) and (2, 1).
Submitted Solution:
```
from collections import Counter
for ad in range(int(input())):
a,b,k=list(map(int,input().split()))
x=list(map(int,input().split()))
y=list(map(int,input().split()))
ans=(k*(k-1))//2
p=Counter(x);q=Counter(y)
for i in range(k):
aa=p[i];bb=q[i]
ans-=(aa*(aa-1))//2
ans-=(bb*(bb-1))//2
print(ans)
``` | instruction | 0 | 88,638 | 17 | 177,276 |
No | output | 1 | 88,638 | 17 | 177,277 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Everyone knows that 2010 FIFA World Cup is being held in South Africa now. By the decision of BFA (Berland's Football Association) next World Cup will be held in Berland. BFA took the decision to change some World Cup regulations:
* the final tournament features n teams (n is always even)
* the first n / 2 teams (according to the standings) come through to the knockout stage
* the standings are made on the following principle: for a victory a team gets 3 points, for a draw β 1 point, for a defeat β 0 points. In the first place, teams are ordered in the standings in decreasing order of their points; in the second place β in decreasing order of the difference between scored and missed goals; in the third place β in the decreasing order of scored goals
* it's written in Berland's Constitution that the previous regulation helps to order the teams without ambiguity.
You are asked to write a program that, by the given list of the competing teams and the results of all the matches, will find the list of teams that managed to get through to the knockout stage.
Input
The first input line contains the only integer n (1 β€ n β€ 50) β amount of the teams, taking part in the final tournament of World Cup. The following n lines contain the names of these teams, a name is a string of lower-case and upper-case Latin letters, its length doesn't exceed 30 characters. The following nΒ·(n - 1) / 2 lines describe the held matches in the format name1-name2 num1:num2, where name1, name2 β names of the teams; num1, num2 (0 β€ num1, num2 β€ 100) β amount of the goals, scored by the corresponding teams. Accuracy of the descriptions is guaranteed: there are no two team names coinciding accurate to the letters' case; there is no match, where a team plays with itself; each match is met in the descriptions only once.
Output
Output n / 2 lines β names of the teams, which managed to get through to the knockout stage in lexicographical order. Output each name in a separate line. No odd characters (including spaces) are allowed. It's guaranteed that the described regulations help to order the teams without ambiguity.
Examples
Input
4
A
B
C
D
A-B 1:1
A-C 2:2
A-D 1:0
B-C 1:0
B-D 0:3
C-D 0:3
Output
A
D
Input
2
a
A
a-A 2:1
Output
a | instruction | 0 | 88,676 | 17 | 177,352 |
Tags: implementation
Correct Solution:
```
n = int(input())
res = {input(): [0, 0, 0] for i in range(n)}
for i in range(n * (n - 1) // 2):
teams, score = input().split()
team1, team2 = teams.split("-")
sc1, sc2 = map(int, score.split(":"))
if sc1 > sc2:
res[team1][0] += 3
elif sc2 > sc1:
res[team2][0] += 3
else:
res[team1][0] += 1
res[team2][0] += 1
res[team1][1] += sc1 - sc2
res[team2][1] += sc2 - sc1
res[team1][2] += sc1
res[team2][2] += sc2
table = sorted(((res[key], key) for key in res), reverse=True)
print("\n".join(sorted([row[1] for row in table[:n // 2]])))
``` | output | 1 | 88,676 | 17 | 177,353 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Everyone knows that 2010 FIFA World Cup is being held in South Africa now. By the decision of BFA (Berland's Football Association) next World Cup will be held in Berland. BFA took the decision to change some World Cup regulations:
* the final tournament features n teams (n is always even)
* the first n / 2 teams (according to the standings) come through to the knockout stage
* the standings are made on the following principle: for a victory a team gets 3 points, for a draw β 1 point, for a defeat β 0 points. In the first place, teams are ordered in the standings in decreasing order of their points; in the second place β in decreasing order of the difference between scored and missed goals; in the third place β in the decreasing order of scored goals
* it's written in Berland's Constitution that the previous regulation helps to order the teams without ambiguity.
You are asked to write a program that, by the given list of the competing teams and the results of all the matches, will find the list of teams that managed to get through to the knockout stage.
Input
The first input line contains the only integer n (1 β€ n β€ 50) β amount of the teams, taking part in the final tournament of World Cup. The following n lines contain the names of these teams, a name is a string of lower-case and upper-case Latin letters, its length doesn't exceed 30 characters. The following nΒ·(n - 1) / 2 lines describe the held matches in the format name1-name2 num1:num2, where name1, name2 β names of the teams; num1, num2 (0 β€ num1, num2 β€ 100) β amount of the goals, scored by the corresponding teams. Accuracy of the descriptions is guaranteed: there are no two team names coinciding accurate to the letters' case; there is no match, where a team plays with itself; each match is met in the descriptions only once.
Output
Output n / 2 lines β names of the teams, which managed to get through to the knockout stage in lexicographical order. Output each name in a separate line. No odd characters (including spaces) are allowed. It's guaranteed that the described regulations help to order the teams without ambiguity.
Examples
Input
4
A
B
C
D
A-B 1:1
A-C 2:2
A-D 1:0
B-C 1:0
B-D 0:3
C-D 0:3
Output
A
D
Input
2
a
A
a-A 2:1
Output
a | instruction | 0 | 88,677 | 17 | 177,354 |
Tags: implementation
Correct Solution:
```
#this function will update information about team after each match
def completeTeamInfo(team,scored,missed):
team.scoredGoals += scored
team.missedGoals += missed
team.differenceScoredAndMissed += scored - missed
#if the team scored more goals than missed, than it gains 3 points
if scored > missed:
team.points += 3
#if they scored the same amount of goals as they missed, team gains one point
elif scored == missed:
team.points += 1
#this function will sort the data according to required conditions
def arrangeTeams(totals):
totals.sort(key=lambda team: (team.points, team.differenceScoredAndMissed, team.scoredGoals ), reverse=True)
def main():
#make a class to store all necessary data
class Team:
scoredGoals = 0
missedGoals = 0
differenceScoredAndMissed = 0
points = 0
def __init__(self, name):
self.name = name
#get teams number
teamsNr = int(input())
totals = []
#get team's names
for _ in range(teamsNr):
name = input()
totals.append(Team(name))
#for each match
for _ in range(teamsNr*(teamsNr-1)//2):
#get score and participant teams
playingTeams,score = input().split()
#because teams are given in A-C 2:2 form
#we have to be more accurate in gathering data
teamOne = playingTeams[:playingTeams.index('-')]
teamTwo = playingTeams[playingTeams.index('-') + 1:]
scoreTeamOne = int(score[:score.index(":")])
scoreTeamTwo = int(score[score.index(':') + 1:])
#for each match
for team in totals:
#look for team in totals and complete teams info
if team.name == teamOne:
completeTeamInfo(team,scoreTeamOne,scoreTeamTwo)
if team.name == teamTwo:
completeTeamInfo(team,scoreTeamTwo,scoreTeamOne)
#arrange teams by the given conditions
arrangeTeams(totals)
winners = []
#we are interested only top half of the list
for i in range(teamsNr//2):
winners.append(totals[i].name)
#but arranged in alphabetical order
winners.sort()
#print results
for winner in winners:
print(winner)
main()
``` | output | 1 | 88,677 | 17 | 177,355 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Everyone knows that 2010 FIFA World Cup is being held in South Africa now. By the decision of BFA (Berland's Football Association) next World Cup will be held in Berland. BFA took the decision to change some World Cup regulations:
* the final tournament features n teams (n is always even)
* the first n / 2 teams (according to the standings) come through to the knockout stage
* the standings are made on the following principle: for a victory a team gets 3 points, for a draw β 1 point, for a defeat β 0 points. In the first place, teams are ordered in the standings in decreasing order of their points; in the second place β in decreasing order of the difference between scored and missed goals; in the third place β in the decreasing order of scored goals
* it's written in Berland's Constitution that the previous regulation helps to order the teams without ambiguity.
You are asked to write a program that, by the given list of the competing teams and the results of all the matches, will find the list of teams that managed to get through to the knockout stage.
Input
The first input line contains the only integer n (1 β€ n β€ 50) β amount of the teams, taking part in the final tournament of World Cup. The following n lines contain the names of these teams, a name is a string of lower-case and upper-case Latin letters, its length doesn't exceed 30 characters. The following nΒ·(n - 1) / 2 lines describe the held matches in the format name1-name2 num1:num2, where name1, name2 β names of the teams; num1, num2 (0 β€ num1, num2 β€ 100) β amount of the goals, scored by the corresponding teams. Accuracy of the descriptions is guaranteed: there are no two team names coinciding accurate to the letters' case; there is no match, where a team plays with itself; each match is met in the descriptions only once.
Output
Output n / 2 lines β names of the teams, which managed to get through to the knockout stage in lexicographical order. Output each name in a separate line. No odd characters (including spaces) are allowed. It's guaranteed that the described regulations help to order the teams without ambiguity.
Examples
Input
4
A
B
C
D
A-B 1:1
A-C 2:2
A-D 1:0
B-C 1:0
B-D 0:3
C-D 0:3
Output
A
D
Input
2
a
A
a-A 2:1
Output
a | instruction | 0 | 88,678 | 17 | 177,356 |
Tags: implementation
Correct Solution:
```
'''Problem 19A - World Football Cup'''
score = {} # dict pentru a face totalurile pentru fiecare din jucatori
n = int(input()) # citim nr de echipe
for i in range(n):
team = input() # key = numele la echipe
score[team] = [0,0,0] # numele la echipa = [puncte pentru fiecare match,
# diferenta intre golurile marcate si primite,
# toate golurile marcate ]
# citim rezultatele meciurilor
for _ in range(n*(n-1)//2):
data = input().split()
teams = data[0].split('-') # teams = ['team_first', 'team_second']
first_team = teams[0] # first team
sec_team = teams[1] # second team
scores = data[1].split(':')
first_tscore = int(scores[0]) # score for first_team
sec_tscore = int(scores[1]) # score for sec_team
# in caz ca a doua echipa a chastigat
if first_tscore < sec_tscore:
score[sec_team][0] += 3 # 'sec_team_name' = [3,0,0]
score[sec_team][1] += sec_tscore - first_tscore # 'sec_team_name' = [3,diferenta intre goluri,0]
score[sec_team][2] += sec_tscore # 'sec_team_name' = [3,diferenta intre goluri,goluri marcate in meci]
# la fel si pentru echipa 1 actualizam scorurile in dictionarul score
score[first_team][1] += first_tscore - sec_tscore
score[first_team][2] += first_tscore
# in caz ca echipa 1 a biruit
if first_tscore > sec_tscore:
score[first_team][0] += 3
score[first_team][1] += first_tscore - sec_tscore
score[first_team][2] += first_tscore
score[sec_team][1] += sec_tscore - first_tscore
score[sec_team][2] += sec_tscore
# in caz de egalitate
elif first_tscore == sec_tscore:
score[first_team][0] += 1
score[first_team][1] += first_tscore - sec_tscore
score[first_team][2] += first_tscore
score[sec_team][0] += 1
score[sec_team][1] += sec_tscore - first_tscore
score[sec_team][2] += sec_tscore
# print(score)
res = [] # valorile din dictionar le salvam in lista res
# team = key , # score = list of scores for each team
for team, score in score.items():
res.append([score[0], score[1], score[2], team]) # : res = [[5, 1, 4, 'A'], [4, -2, 2, 'B'], [1, -4, 2, 'C'], [6, 5, 6, 'D']]
res = sorted(res) # sortam lista , dupa condtitie
res.reverse() # reversam lista , ca echipele cu cele mai mari scoruri sa fie primele
nw = [] # lista pt echipele care sau calificat
for i in range(n//2):
nw.append(res[i][3]) # adougam in lisa numele acestor echipe
nw = sorted(nw)
print(*nw, sep='\n') # afisam numele la echipe din rand nou
``` | output | 1 | 88,678 | 17 | 177,357 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Everyone knows that 2010 FIFA World Cup is being held in South Africa now. By the decision of BFA (Berland's Football Association) next World Cup will be held in Berland. BFA took the decision to change some World Cup regulations:
* the final tournament features n teams (n is always even)
* the first n / 2 teams (according to the standings) come through to the knockout stage
* the standings are made on the following principle: for a victory a team gets 3 points, for a draw β 1 point, for a defeat β 0 points. In the first place, teams are ordered in the standings in decreasing order of their points; in the second place β in decreasing order of the difference between scored and missed goals; in the third place β in the decreasing order of scored goals
* it's written in Berland's Constitution that the previous regulation helps to order the teams without ambiguity.
You are asked to write a program that, by the given list of the competing teams and the results of all the matches, will find the list of teams that managed to get through to the knockout stage.
Input
The first input line contains the only integer n (1 β€ n β€ 50) β amount of the teams, taking part in the final tournament of World Cup. The following n lines contain the names of these teams, a name is a string of lower-case and upper-case Latin letters, its length doesn't exceed 30 characters. The following nΒ·(n - 1) / 2 lines describe the held matches in the format name1-name2 num1:num2, where name1, name2 β names of the teams; num1, num2 (0 β€ num1, num2 β€ 100) β amount of the goals, scored by the corresponding teams. Accuracy of the descriptions is guaranteed: there are no two team names coinciding accurate to the letters' case; there is no match, where a team plays with itself; each match is met in the descriptions only once.
Output
Output n / 2 lines β names of the teams, which managed to get through to the knockout stage in lexicographical order. Output each name in a separate line. No odd characters (including spaces) are allowed. It's guaranteed that the described regulations help to order the teams without ambiguity.
Examples
Input
4
A
B
C
D
A-B 1:1
A-C 2:2
A-D 1:0
B-C 1:0
B-D 0:3
C-D 0:3
Output
A
D
Input
2
a
A
a-A 2:1
Output
a | instruction | 0 | 88,679 | 17 | 177,358 |
Tags: implementation
Correct Solution:
```
class Info:
def __init__(self, newTeamName, newPoints, newGoalDiff, newScoredGoals):
self.teamName = newTeamName
self.points = newPoints
self.goalDiff = newGoalDiff
self.scoredGoals = newScoredGoals
def __str__(self):
temp = '\n**************\n'
temp += f'teamName: {self.teamName} \n'
temp += f'points: {self.points} \n'
temp += f'goalDiff: {self.goalDiff} \n'
temp += f'scoredGoals: {self.scoredGoals} \n'
temp += '**************\n'
return temp
# end of class
def findIndexByName(cont: list, searchName: str) -> int:
ln = len(cont)
for i in range(ln):
if searchName == cont[i].teamName:
return i
return -1
def output(cont: list):
for item in cont:
print(item)
n, cont = int(input()), []
for i in range(n):
# obj = Info(input(), 0, 0, 0)
# cont.append(obj)
cont.append(Info(input(), 0, 0, 0))
for i in range(n * (n - 1) // 2):
line = input()
dashInd = line.index('-')
spaceInd = line.index(' ')
colonInd = line.index(':')
team1Name = line[:dashInd]
team2Name = line[dashInd + 1:spaceInd]
score1 = int(line[spaceInd + 1:colonInd])
score2 = int(line[colonInd + 1:])
team1Ind = findIndexByName(cont, team1Name)
team2Ind = findIndexByName(cont, team2Name)
# update points
if score1 > score2:
cont[team1Ind].points += 3
elif score1 < score2:
cont[team2Ind].points += 3
else:
cont[team1Ind].points += 1
cont[team2Ind].points += 1
# uptade goalDiff
cont[team1Ind].goalDiff += score1 - score2
cont[team2Ind].goalDiff += score2 - score1
# update scoredGoals
cont[team1Ind].scoredGoals += score1
cont[team2Ind].scoredGoals += score2
cont.sort(key=lambda it: (it.points, it.goalDiff, it.scoredGoals), reverse=True)
del cont[n//2:]
cont.sort(key=lambda it: it.teamName)
#output(cont)
for item in cont:
print(item.teamName)
'''
0123456789....
line = 'barsa-real 15:10'
-------------------------------
4
A
B
C
D
A-B 1:1
A-C 2:2
A-D 1:0
B-C 1:0
B-D 0:3
C-D 0:3
points goalDiff scoredGoals
A -> 1+1+3=5 A -> (1+2+1)-(1+2+0) = 4-3=1 A -> 4
B -> 1+3+0=4 B -> (1+1+0)-(1+0+3) = 2-4=-2 B -> 2
C -> 1+0+0=1 C -> (2+0+0)-(2+1+3) = 2-6=-4 C -> 2
D -> 0+3+3=6 D -> (0+3+3)-(1+0+0) = 6-1=5 D -> 6
A, B, C, D -> D, A, B, C -> D, A -> A, D
0
| teamName: 'A' |
| points: 5 |
| goalDiff: 1 |
| scoredGoals: 4|
'''
``` | output | 1 | 88,679 | 17 | 177,359 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Everyone knows that 2010 FIFA World Cup is being held in South Africa now. By the decision of BFA (Berland's Football Association) next World Cup will be held in Berland. BFA took the decision to change some World Cup regulations:
* the final tournament features n teams (n is always even)
* the first n / 2 teams (according to the standings) come through to the knockout stage
* the standings are made on the following principle: for a victory a team gets 3 points, for a draw β 1 point, for a defeat β 0 points. In the first place, teams are ordered in the standings in decreasing order of their points; in the second place β in decreasing order of the difference between scored and missed goals; in the third place β in the decreasing order of scored goals
* it's written in Berland's Constitution that the previous regulation helps to order the teams without ambiguity.
You are asked to write a program that, by the given list of the competing teams and the results of all the matches, will find the list of teams that managed to get through to the knockout stage.
Input
The first input line contains the only integer n (1 β€ n β€ 50) β amount of the teams, taking part in the final tournament of World Cup. The following n lines contain the names of these teams, a name is a string of lower-case and upper-case Latin letters, its length doesn't exceed 30 characters. The following nΒ·(n - 1) / 2 lines describe the held matches in the format name1-name2 num1:num2, where name1, name2 β names of the teams; num1, num2 (0 β€ num1, num2 β€ 100) β amount of the goals, scored by the corresponding teams. Accuracy of the descriptions is guaranteed: there are no two team names coinciding accurate to the letters' case; there is no match, where a team plays with itself; each match is met in the descriptions only once.
Output
Output n / 2 lines β names of the teams, which managed to get through to the knockout stage in lexicographical order. Output each name in a separate line. No odd characters (including spaces) are allowed. It's guaranteed that the described regulations help to order the teams without ambiguity.
Examples
Input
4
A
B
C
D
A-B 1:1
A-C 2:2
A-D 1:0
B-C 1:0
B-D 0:3
C-D 0:3
Output
A
D
Input
2
a
A
a-A 2:1
Output
a | instruction | 0 | 88,680 | 17 | 177,360 |
Tags: implementation
Correct Solution:
```
#RAVENS
#TEAM_2
#ESSI-DAYI_MOHSEN-LORENZO
n = int(input())
team = dict()
for i in range(n):
team[input()] = [0,0,0]
for i in range(n*(n-1)//2):
a,b = input().split()
t1,t2=a.split('-')
g1,g2=map(int,b.split(':'))
team[t1][2]+=g1
team[t2][2]+=g2
if g1 > g2:team[t1][0]+=3
elif g2 > g1:team[t2][0]+=3
else:team[t1][0]+=1;team[t2][0]+=1
team[t1][1]+=(g1-g2)
team[t2][1]+=(g2-g1)
te = []
for i in team.keys():
te.append([i,team[i][0],team[i][1],team[i][2]])
for j in range(n):
for i in range(n-1-j):
if te[i][1] > te[i+1][1]:
te[i],te[i+1] = te[i+1],te[i]
elif te[i][1] == te[i+1][1]:
if te[i][2] > te[i+1][2]:
te[i],te[i+1] = te[i+1],te[i]
elif te[i][2] == te[i+1][2]:
if te[i][3] > te[i+1][3]:
te[i],te[i+1] = te[i+1],te[i]
res = []
for i in range(n-1,n//2-1,-1):
res.append(te[i][0])
res.sort()
for i in range(n//2):
print(res[i])
``` | output | 1 | 88,680 | 17 | 177,361 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Everyone knows that 2010 FIFA World Cup is being held in South Africa now. By the decision of BFA (Berland's Football Association) next World Cup will be held in Berland. BFA took the decision to change some World Cup regulations:
* the final tournament features n teams (n is always even)
* the first n / 2 teams (according to the standings) come through to the knockout stage
* the standings are made on the following principle: for a victory a team gets 3 points, for a draw β 1 point, for a defeat β 0 points. In the first place, teams are ordered in the standings in decreasing order of their points; in the second place β in decreasing order of the difference between scored and missed goals; in the third place β in the decreasing order of scored goals
* it's written in Berland's Constitution that the previous regulation helps to order the teams without ambiguity.
You are asked to write a program that, by the given list of the competing teams and the results of all the matches, will find the list of teams that managed to get through to the knockout stage.
Input
The first input line contains the only integer n (1 β€ n β€ 50) β amount of the teams, taking part in the final tournament of World Cup. The following n lines contain the names of these teams, a name is a string of lower-case and upper-case Latin letters, its length doesn't exceed 30 characters. The following nΒ·(n - 1) / 2 lines describe the held matches in the format name1-name2 num1:num2, where name1, name2 β names of the teams; num1, num2 (0 β€ num1, num2 β€ 100) β amount of the goals, scored by the corresponding teams. Accuracy of the descriptions is guaranteed: there are no two team names coinciding accurate to the letters' case; there is no match, where a team plays with itself; each match is met in the descriptions only once.
Output
Output n / 2 lines β names of the teams, which managed to get through to the knockout stage in lexicographical order. Output each name in a separate line. No odd characters (including spaces) are allowed. It's guaranteed that the described regulations help to order the teams without ambiguity.
Examples
Input
4
A
B
C
D
A-B 1:1
A-C 2:2
A-D 1:0
B-C 1:0
B-D 0:3
C-D 0:3
Output
A
D
Input
2
a
A
a-A 2:1
Output
a | instruction | 0 | 88,681 | 17 | 177,362 |
Tags: implementation
Correct Solution:
```
import math
n = int(input())
d= {}
for i in range(n):
d[input()] = [0,0,0]
for i in range(math.factorial(n)//(2*math.factorial(n-2))):
t,s = input().split()
t1,t2 = t.split('-')
s1,s2 = map(int,s.split(':'))
if s1>s2 :
d[t1][0] += 3
d[t1][1] += abs(s1-s2)
d[t2][1] -= abs(s1-s2)
elif s1==s2 :
d[t1][0] += 1
d[t2][0] += 1
else:
d[t2][0] += 3
d[t1][1] -= abs(s1-s2)
d[t2][1] += abs(s1-s2)
d[t1][2] += s1
d[t1][2] += s2
print(*sorted(sorted(d.keys(),key = lambda xx:\
(-d[xx][0],-d[xx][1],-d[xx][2]))[:n//2]),sep = '\n')
``` | output | 1 | 88,681 | 17 | 177,363 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Everyone knows that 2010 FIFA World Cup is being held in South Africa now. By the decision of BFA (Berland's Football Association) next World Cup will be held in Berland. BFA took the decision to change some World Cup regulations:
* the final tournament features n teams (n is always even)
* the first n / 2 teams (according to the standings) come through to the knockout stage
* the standings are made on the following principle: for a victory a team gets 3 points, for a draw β 1 point, for a defeat β 0 points. In the first place, teams are ordered in the standings in decreasing order of their points; in the second place β in decreasing order of the difference between scored and missed goals; in the third place β in the decreasing order of scored goals
* it's written in Berland's Constitution that the previous regulation helps to order the teams without ambiguity.
You are asked to write a program that, by the given list of the competing teams and the results of all the matches, will find the list of teams that managed to get through to the knockout stage.
Input
The first input line contains the only integer n (1 β€ n β€ 50) β amount of the teams, taking part in the final tournament of World Cup. The following n lines contain the names of these teams, a name is a string of lower-case and upper-case Latin letters, its length doesn't exceed 30 characters. The following nΒ·(n - 1) / 2 lines describe the held matches in the format name1-name2 num1:num2, where name1, name2 β names of the teams; num1, num2 (0 β€ num1, num2 β€ 100) β amount of the goals, scored by the corresponding teams. Accuracy of the descriptions is guaranteed: there are no two team names coinciding accurate to the letters' case; there is no match, where a team plays with itself; each match is met in the descriptions only once.
Output
Output n / 2 lines β names of the teams, which managed to get through to the knockout stage in lexicographical order. Output each name in a separate line. No odd characters (including spaces) are allowed. It's guaranteed that the described regulations help to order the teams without ambiguity.
Examples
Input
4
A
B
C
D
A-B 1:1
A-C 2:2
A-D 1:0
B-C 1:0
B-D 0:3
C-D 0:3
Output
A
D
Input
2
a
A
a-A 2:1
Output
a | instruction | 0 | 88,682 | 17 | 177,364 |
Tags: implementation
Correct Solution:
```
n = int(input())
# Score [points, diff, goals]
score = {}
for i in range(n):
score[input()] = [0, 0, 0]
for i in range(int(n*(n-1)/2)):
s = input().split()
p1 = s[0].split('-')[0]
p2 = s[0].split('-')[1]
s1 = int(s[1].split(':')[0])
s2 = int(s[1].split(':')[1])
score[p1][1] += s1 - s2
score[p1][2] += s1
score[p2][1] += s2 - s1
score[p2][2] += s2
if s1 > s2:
score[p1][0] += 3
elif s1 == s2:
score[p1][0] += 1
score[p2][0] += 1
else:
score[p2][0] += 3
score = sorted(score.items(), key=lambda x: x[1][0], reverse = True)
for i in range(n):
for j in range(n - i - 1):
if score[j][1][0] == score[j+1][1][0]:
if score[j][1][1] < score[j+1][1][1]:
score[j], score[j+1] = score[j+1], score[j]
for i in range(n):
for j in range(n - i - 1):
if score[j][1][1] == score[j+1][1][1]:
if score[j][1][2] < score[j+1][1][2]:
score[j], score[j+1] = score[j+1], score[j]
for i in sorted(score[:int(n/2)], key = lambda x : x[0]):
print(i[0])
``` | output | 1 | 88,682 | 17 | 177,365 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Everyone knows that 2010 FIFA World Cup is being held in South Africa now. By the decision of BFA (Berland's Football Association) next World Cup will be held in Berland. BFA took the decision to change some World Cup regulations:
* the final tournament features n teams (n is always even)
* the first n / 2 teams (according to the standings) come through to the knockout stage
* the standings are made on the following principle: for a victory a team gets 3 points, for a draw β 1 point, for a defeat β 0 points. In the first place, teams are ordered in the standings in decreasing order of their points; in the second place β in decreasing order of the difference between scored and missed goals; in the third place β in the decreasing order of scored goals
* it's written in Berland's Constitution that the previous regulation helps to order the teams without ambiguity.
You are asked to write a program that, by the given list of the competing teams and the results of all the matches, will find the list of teams that managed to get through to the knockout stage.
Input
The first input line contains the only integer n (1 β€ n β€ 50) β amount of the teams, taking part in the final tournament of World Cup. The following n lines contain the names of these teams, a name is a string of lower-case and upper-case Latin letters, its length doesn't exceed 30 characters. The following nΒ·(n - 1) / 2 lines describe the held matches in the format name1-name2 num1:num2, where name1, name2 β names of the teams; num1, num2 (0 β€ num1, num2 β€ 100) β amount of the goals, scored by the corresponding teams. Accuracy of the descriptions is guaranteed: there are no two team names coinciding accurate to the letters' case; there is no match, where a team plays with itself; each match is met in the descriptions only once.
Output
Output n / 2 lines β names of the teams, which managed to get through to the knockout stage in lexicographical order. Output each name in a separate line. No odd characters (including spaces) are allowed. It's guaranteed that the described regulations help to order the teams without ambiguity.
Examples
Input
4
A
B
C
D
A-B 1:1
A-C 2:2
A-D 1:0
B-C 1:0
B-D 0:3
C-D 0:3
Output
A
D
Input
2
a
A
a-A 2:1
Output
a | instruction | 0 | 88,683 | 17 | 177,366 |
Tags: implementation
Correct Solution:
```
n = int(input())
class Command:
def __init__(self, name):
self.name = name
self.score = 0
self.z = 0
self.p = 0
def get_r(self):
return self.z - self.p
arr = {}
for i in range(n):
name = input()
arr[name] = Command(name)
for i in range((n * (n - 1)) // 2):
string = input()
first, second = string.split()
name1, name2 = first.split('-')
num1, num2 = second.split(":")
num1, num2 = int(num1), int(num2)
if num1 > num2:
arr[name1].score += 3
elif num1 < num2:
arr[name2].score += 3
else:
arr[name1].score += 1
arr[name2].score += 1
arr[name1].z += num1
arr[name1].p += num2
arr[name2].z += num2
arr[name2].p += num1
arr = list(arr.values())
for i in range(len(arr)):
for j in range(len(arr) - 1):
if arr[j].score < arr[j + 1].score:
arr[j], arr[j + 1] = arr[j + 1], arr[j]
elif arr[j].score == arr[j + 1].score:
if arr[j].get_r() < arr[j + 1].get_r():
arr[j], arr[j + 1] = arr[j + 1], arr[j]
elif arr[j].get_r() == arr[j + 1].get_r():
if arr[j].z < arr[j + 1].z:
arr[j], arr[j + 1] = arr[j + 1], arr[j]
ans = []
for i in range(len(arr) // 2):
ans.append(arr[i].name)
ans.sort()
for i in ans:
print(i)
``` | output | 1 | 88,683 | 17 | 177,367 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Everyone knows that 2010 FIFA World Cup is being held in South Africa now. By the decision of BFA (Berland's Football Association) next World Cup will be held in Berland. BFA took the decision to change some World Cup regulations:
* the final tournament features n teams (n is always even)
* the first n / 2 teams (according to the standings) come through to the knockout stage
* the standings are made on the following principle: for a victory a team gets 3 points, for a draw β 1 point, for a defeat β 0 points. In the first place, teams are ordered in the standings in decreasing order of their points; in the second place β in decreasing order of the difference between scored and missed goals; in the third place β in the decreasing order of scored goals
* it's written in Berland's Constitution that the previous regulation helps to order the teams without ambiguity.
You are asked to write a program that, by the given list of the competing teams and the results of all the matches, will find the list of teams that managed to get through to the knockout stage.
Input
The first input line contains the only integer n (1 β€ n β€ 50) β amount of the teams, taking part in the final tournament of World Cup. The following n lines contain the names of these teams, a name is a string of lower-case and upper-case Latin letters, its length doesn't exceed 30 characters. The following nΒ·(n - 1) / 2 lines describe the held matches in the format name1-name2 num1:num2, where name1, name2 β names of the teams; num1, num2 (0 β€ num1, num2 β€ 100) β amount of the goals, scored by the corresponding teams. Accuracy of the descriptions is guaranteed: there are no two team names coinciding accurate to the letters' case; there is no match, where a team plays with itself; each match is met in the descriptions only once.
Output
Output n / 2 lines β names of the teams, which managed to get through to the knockout stage in lexicographical order. Output each name in a separate line. No odd characters (including spaces) are allowed. It's guaranteed that the described regulations help to order the teams without ambiguity.
Examples
Input
4
A
B
C
D
A-B 1:1
A-C 2:2
A-D 1:0
B-C 1:0
B-D 0:3
C-D 0:3
Output
A
D
Input
2
a
A
a-A 2:1
Output
a
Submitted Solution:
```
team_rankings = {}
n = int(input())
for i in range(n):
team = input() #particpating teams
team_rankings[team] = [0,0,0] #[points, goal count difference, goals]
for _ in range(n*(n-1)//2): #gathering data from each game
data = input().split()
teams = data[0].split('-') #finding the teams and their match-ups
team_1 = teams[0] #team 1
team_2 = teams[1] #team 2
scores = data[1].split(':')
score_t1 = int(scores[0]) #score of team 1
score_t2 = int(scores[1]) #score of team 2
#case: team 2 has won the match
if score_t1 < score_t2:
team_rankings[team_2][0] += 3 #gains 3 points
team_rankings[team_2][1] += score_t2 - score_t1 #goal count difference
team_rankings[team_2][2] += score_t2 #goals
#updating for team 1 as well
team_rankings[team_1][1] += score_t1 - score_t2
team_rankings[team_1][2] += score_t1
#case: team 1 has won the macth
if score_t1 > score_t2:
team_rankings[team_1][0] += 3 #gains 3 points
team_rankings[team_1][1] += score_t1 - score_t2 #goal count difference
team_rankings[team_1][2] += score_t1 #goals
#updating for team 1 as well
team_rankings[team_2][1] += score_t2 - score_t1
team_rankings[team_2][2] += score_t2
#case: stalemate or a tie
elif score_t1 == score_t2:
team_rankings[team_1][0] += 1
team_rankings[team_1][1] += score_t1 - score_t2
team_rankings[team_1][2] += score_t1
team_rankings[team_2][0] += 1
team_rankings[team_2][1] += score_t2 - score_t1
team_rankings[team_2][2] += score_t2
results = []
for team, team_rankings in team_rankings.items():
results.append([team_rankings[0], team_rankings[1], team_rankings[2], team])
results = sorted(results) #sorting the list
results.reverse() #reverse the order so we get the hoghest scoring team first
qualified = [] #list of qualified teams
for i in range(n//2):
qualified.append(results[i][3]) #adding the name of the qualified teams
qualified = sorted(qualified)
print(*qualified, sep='\n')
``` | instruction | 0 | 88,684 | 17 | 177,368 |
Yes | output | 1 | 88,684 | 17 | 177,369 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Everyone knows that 2010 FIFA World Cup is being held in South Africa now. By the decision of BFA (Berland's Football Association) next World Cup will be held in Berland. BFA took the decision to change some World Cup regulations:
* the final tournament features n teams (n is always even)
* the first n / 2 teams (according to the standings) come through to the knockout stage
* the standings are made on the following principle: for a victory a team gets 3 points, for a draw β 1 point, for a defeat β 0 points. In the first place, teams are ordered in the standings in decreasing order of their points; in the second place β in decreasing order of the difference between scored and missed goals; in the third place β in the decreasing order of scored goals
* it's written in Berland's Constitution that the previous regulation helps to order the teams without ambiguity.
You are asked to write a program that, by the given list of the competing teams and the results of all the matches, will find the list of teams that managed to get through to the knockout stage.
Input
The first input line contains the only integer n (1 β€ n β€ 50) β amount of the teams, taking part in the final tournament of World Cup. The following n lines contain the names of these teams, a name is a string of lower-case and upper-case Latin letters, its length doesn't exceed 30 characters. The following nΒ·(n - 1) / 2 lines describe the held matches in the format name1-name2 num1:num2, where name1, name2 β names of the teams; num1, num2 (0 β€ num1, num2 β€ 100) β amount of the goals, scored by the corresponding teams. Accuracy of the descriptions is guaranteed: there are no two team names coinciding accurate to the letters' case; there is no match, where a team plays with itself; each match is met in the descriptions only once.
Output
Output n / 2 lines β names of the teams, which managed to get through to the knockout stage in lexicographical order. Output each name in a separate line. No odd characters (including spaces) are allowed. It's guaranteed that the described regulations help to order the teams without ambiguity.
Examples
Input
4
A
B
C
D
A-B 1:1
A-C 2:2
A-D 1:0
B-C 1:0
B-D 0:3
C-D 0:3
Output
A
D
Input
2
a
A
a-A 2:1
Output
a
Submitted Solution:
```
# -*- coding: utf-8 -*-
dic,lista,final = {},[],[]
n = int(input())
for i in range(n): dic[input()] = {'points': 0,'scored': 0,'missed': 0}
matches = [input().split() for j in range(int(n*(n-1)/2))]
for j in range(len(matches)):
goals1,goals2 = int(matches[j][1].split(':')[0]),int(matches[j][1].split(':')[1])
team1,team2 = matches[j][0].split('-')[0], matches[j][0].split('-')[1]
dic[team1]['scored'] += goals1; dic[team1]['missed'] += goals2
dic[team2]['scored'] += goals2; dic[team2]['missed'] += goals1
if goals1 > goals2: dic[team1]['points'] += 3
elif goals1 < goals2: dic[team2]['points'] += 3
else: dic[team1]['points'] += 1; dic[team2]['points'] += 1
for i in dic:
lista.append([dic[i]['points'],dic[i]['scored']-dic[i]['missed'],dic[i]['scored'],i])
lista = sorted(lista)[::-1]
for i in range(int(n/2)):
final.append(lista[i][3])
final = sorted(final)
for i in final: print(i)
``` | instruction | 0 | 88,685 | 17 | 177,370 |
Yes | output | 1 | 88,685 | 17 | 177,371 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Everyone knows that 2010 FIFA World Cup is being held in South Africa now. By the decision of BFA (Berland's Football Association) next World Cup will be held in Berland. BFA took the decision to change some World Cup regulations:
* the final tournament features n teams (n is always even)
* the first n / 2 teams (according to the standings) come through to the knockout stage
* the standings are made on the following principle: for a victory a team gets 3 points, for a draw β 1 point, for a defeat β 0 points. In the first place, teams are ordered in the standings in decreasing order of their points; in the second place β in decreasing order of the difference between scored and missed goals; in the third place β in the decreasing order of scored goals
* it's written in Berland's Constitution that the previous regulation helps to order the teams without ambiguity.
You are asked to write a program that, by the given list of the competing teams and the results of all the matches, will find the list of teams that managed to get through to the knockout stage.
Input
The first input line contains the only integer n (1 β€ n β€ 50) β amount of the teams, taking part in the final tournament of World Cup. The following n lines contain the names of these teams, a name is a string of lower-case and upper-case Latin letters, its length doesn't exceed 30 characters. The following nΒ·(n - 1) / 2 lines describe the held matches in the format name1-name2 num1:num2, where name1, name2 β names of the teams; num1, num2 (0 β€ num1, num2 β€ 100) β amount of the goals, scored by the corresponding teams. Accuracy of the descriptions is guaranteed: there are no two team names coinciding accurate to the letters' case; there is no match, where a team plays with itself; each match is met in the descriptions only once.
Output
Output n / 2 lines β names of the teams, which managed to get through to the knockout stage in lexicographical order. Output each name in a separate line. No odd characters (including spaces) are allowed. It's guaranteed that the described regulations help to order the teams without ambiguity.
Examples
Input
4
A
B
C
D
A-B 1:1
A-C 2:2
A-D 1:0
B-C 1:0
B-D 0:3
C-D 0:3
Output
A
D
Input
2
a
A
a-A 2:1
Output
a
Submitted Solution:
```
clasificados = []
equipos = []
puntos = []
dg = []
gf = []
n = int(input())
for i in range(n):
equipos.append(input())
puntos.append(0)
dg.append(0)
gf.append(0)
n_partidos = (n * (n-1)) // 2
for i in range(n_partidos):
partido = input()
equipos_partido, resultado = partido.split()
e1, e2 = equipos_partido.split('-')
goles = resultado.split(':')
g1, g2 = int(goles[0]) , int(goles[1])
if g1 > g2:
p1 = 3
p2 = 0
elif g1 == g2:
p1 = 1
p2 = 1
else:
p1 = 0
p2 = 3
a = equipos.index(e1)
b = equipos.index(e2)
puntos[a] = puntos[a] + p1
dg[a] = dg[a] + g1 - g2
gf[a] = gf[a] + g1
puntos[b] = puntos[b] + p2
dg[b] = dg[b] + g2 - g1
gf[b] = gf[b] + g2
x = 0
for i in range(n//2):
mayor = 0
for j in range(n):
if mayor != j:
if puntos[mayor] < puntos[j]:
mayor = j
elif puntos[mayor] == puntos[j]:
if dg[mayor] < dg[j]:
mayor = j
elif dg[mayor] == dg[j]:
if gf[mayor] < gf[j]:
mayor = j
clasificados.append(equipos[mayor])
puntos[mayor] = -1
clasificados.sort()
for i in range(n//2):
print(clasificados[i])
``` | instruction | 0 | 88,686 | 17 | 177,372 |
Yes | output | 1 | 88,686 | 17 | 177,373 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Everyone knows that 2010 FIFA World Cup is being held in South Africa now. By the decision of BFA (Berland's Football Association) next World Cup will be held in Berland. BFA took the decision to change some World Cup regulations:
* the final tournament features n teams (n is always even)
* the first n / 2 teams (according to the standings) come through to the knockout stage
* the standings are made on the following principle: for a victory a team gets 3 points, for a draw β 1 point, for a defeat β 0 points. In the first place, teams are ordered in the standings in decreasing order of their points; in the second place β in decreasing order of the difference between scored and missed goals; in the third place β in the decreasing order of scored goals
* it's written in Berland's Constitution that the previous regulation helps to order the teams without ambiguity.
You are asked to write a program that, by the given list of the competing teams and the results of all the matches, will find the list of teams that managed to get through to the knockout stage.
Input
The first input line contains the only integer n (1 β€ n β€ 50) β amount of the teams, taking part in the final tournament of World Cup. The following n lines contain the names of these teams, a name is a string of lower-case and upper-case Latin letters, its length doesn't exceed 30 characters. The following nΒ·(n - 1) / 2 lines describe the held matches in the format name1-name2 num1:num2, where name1, name2 β names of the teams; num1, num2 (0 β€ num1, num2 β€ 100) β amount of the goals, scored by the corresponding teams. Accuracy of the descriptions is guaranteed: there are no two team names coinciding accurate to the letters' case; there is no match, where a team plays with itself; each match is met in the descriptions only once.
Output
Output n / 2 lines β names of the teams, which managed to get through to the knockout stage in lexicographical order. Output each name in a separate line. No odd characters (including spaces) are allowed. It's guaranteed that the described regulations help to order the teams without ambiguity.
Examples
Input
4
A
B
C
D
A-B 1:1
A-C 2:2
A-D 1:0
B-C 1:0
B-D 0:3
C-D 0:3
Output
A
D
Input
2
a
A
a-A 2:1
Output
a
Submitted Solution:
```
score = {}
n = int(input())
for i in range(n):
team = input()
score[team] = [0,0,0]
for i in range(n*(n-1)//2):
s = input().split()
teams = s[0].split('-')
# get team names
first = teams[0]
second = teams[1]
# get scores
scores = s[1].split(':')
first_score = int(scores[0])
second_score = int(scores[1])
# when team winned
if first_score < second_score:
score[second][0] += 3
score[second][1] = score[second][1] + second_score - first_score
score[second][2] += second_score
score[first][1] = score[first][1] + first_score - second_score
score[first][2] += first_score
if first_score > second_score:
score[first][0] += 3
score[first][1] += first_score - second_score
score[first][2] += first_score
score[second][1] += second_score - first_score
score[second][2] += second_score
# when draw
elif first_score == second_score:
score[first][0] += 1
score[first][1] += first_score - second_score
score[first][2] += first_score
score[second][0] += 1
score[second][1] += second_score - first_score
score[second][2] += second_score
results = []
# list of teams with 3 criteria scores
for team, score in score.items():
results.append([score[0], score[1], score[2], team])
# sort results in ascending order
results = sorted(results)
results.reverse()
nw = []
for i in range(n//2):
nw.append(results[i][3])
nw = sorted(nw)
print(*nw, sep='\n')
``` | instruction | 0 | 88,687 | 17 | 177,374 |
Yes | output | 1 | 88,687 | 17 | 177,375 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Everyone knows that 2010 FIFA World Cup is being held in South Africa now. By the decision of BFA (Berland's Football Association) next World Cup will be held in Berland. BFA took the decision to change some World Cup regulations:
* the final tournament features n teams (n is always even)
* the first n / 2 teams (according to the standings) come through to the knockout stage
* the standings are made on the following principle: for a victory a team gets 3 points, for a draw β 1 point, for a defeat β 0 points. In the first place, teams are ordered in the standings in decreasing order of their points; in the second place β in decreasing order of the difference between scored and missed goals; in the third place β in the decreasing order of scored goals
* it's written in Berland's Constitution that the previous regulation helps to order the teams without ambiguity.
You are asked to write a program that, by the given list of the competing teams and the results of all the matches, will find the list of teams that managed to get through to the knockout stage.
Input
The first input line contains the only integer n (1 β€ n β€ 50) β amount of the teams, taking part in the final tournament of World Cup. The following n lines contain the names of these teams, a name is a string of lower-case and upper-case Latin letters, its length doesn't exceed 30 characters. The following nΒ·(n - 1) / 2 lines describe the held matches in the format name1-name2 num1:num2, where name1, name2 β names of the teams; num1, num2 (0 β€ num1, num2 β€ 100) β amount of the goals, scored by the corresponding teams. Accuracy of the descriptions is guaranteed: there are no two team names coinciding accurate to the letters' case; there is no match, where a team plays with itself; each match is met in the descriptions only once.
Output
Output n / 2 lines β names of the teams, which managed to get through to the knockout stage in lexicographical order. Output each name in a separate line. No odd characters (including spaces) are allowed. It's guaranteed that the described regulations help to order the teams without ambiguity.
Examples
Input
4
A
B
C
D
A-B 1:1
A-C 2:2
A-D 1:0
B-C 1:0
B-D 0:3
C-D 0:3
Output
A
D
Input
2
a
A
a-A 2:1
Output
a
Submitted Solution:
```
n = int(input())
teamScores = dict()
for _ in range(n):
name = input()
teamScores[name] = 0
matchCount = (n)*(n-1)//2
for _ in range(matchCount):
match = input().split(" ")
team = match[0].split("-")
score = list(map(int,match[1].split(":")))
if score[0] == score[1]:
teamScores[team[0]] += 1
teamScores[team[1]] += 1
elif score[0] > score[1] :
teamScores[team[0]] += 3
else:
teamScores[team[1]] += 3
teams = list(teamScores.keys())
for i in range(n):
for j in range(n-i-1):
if teamScores[teams[j]] < teamScores[teams[j+1]] :
teams[j+1],teams[j] = teams[j],teams[j+1]
elif teamScores[teams[j]] == teamScores[teams[j+1]] :
if teams[j+1] < teams[j] :
teams[j+1],teams[j] = teams[j],teams[j+1]
print(teams)
teams = sorted(teams)
for i in range(n//2):
print(teams[i])
``` | instruction | 0 | 88,688 | 17 | 177,376 |
No | output | 1 | 88,688 | 17 | 177,377 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Everyone knows that 2010 FIFA World Cup is being held in South Africa now. By the decision of BFA (Berland's Football Association) next World Cup will be held in Berland. BFA took the decision to change some World Cup regulations:
* the final tournament features n teams (n is always even)
* the first n / 2 teams (according to the standings) come through to the knockout stage
* the standings are made on the following principle: for a victory a team gets 3 points, for a draw β 1 point, for a defeat β 0 points. In the first place, teams are ordered in the standings in decreasing order of their points; in the second place β in decreasing order of the difference between scored and missed goals; in the third place β in the decreasing order of scored goals
* it's written in Berland's Constitution that the previous regulation helps to order the teams without ambiguity.
You are asked to write a program that, by the given list of the competing teams and the results of all the matches, will find the list of teams that managed to get through to the knockout stage.
Input
The first input line contains the only integer n (1 β€ n β€ 50) β amount of the teams, taking part in the final tournament of World Cup. The following n lines contain the names of these teams, a name is a string of lower-case and upper-case Latin letters, its length doesn't exceed 30 characters. The following nΒ·(n - 1) / 2 lines describe the held matches in the format name1-name2 num1:num2, where name1, name2 β names of the teams; num1, num2 (0 β€ num1, num2 β€ 100) β amount of the goals, scored by the corresponding teams. Accuracy of the descriptions is guaranteed: there are no two team names coinciding accurate to the letters' case; there is no match, where a team plays with itself; each match is met in the descriptions only once.
Output
Output n / 2 lines β names of the teams, which managed to get through to the knockout stage in lexicographical order. Output each name in a separate line. No odd characters (including spaces) are allowed. It's guaranteed that the described regulations help to order the teams without ambiguity.
Examples
Input
4
A
B
C
D
A-B 1:1
A-C 2:2
A-D 1:0
B-C 1:0
B-D 0:3
C-D 0:3
Output
A
D
Input
2
a
A
a-A 2:1
Output
a
Submitted Solution:
```
n=int(input())
a1=[]
a2=[]
a3=[]
a4=[]
a5=[0]*n
a6=[]
goles=0
puntos=0
mayor=0
for k in range (n):
x=input()
a1.append(x)
for l in range ((n-1)*(n)//2):
b=input()
a2.append(b)
for a in range (n):
for b in range (len(a2)):
for c in range (3):
if a1[a]==a2[b][c]:
if c==0:
goles=goles+int(a2[b][4])
if c==2:
goles=goles+int(a2[b][6])
a3.append(goles)
goles=0
for d in range (n):
for e in range (len(a2)):
for f in range (3):
if a1[d]==a2[e][f]:
if f==0:
if int(a2[e][4])>int(a2[e][6]):
puntos=puntos+3
if int(a2[e][4])==int(a2[e][6]):
puntos=puntos+1
if f==2:
if int(a2[e][6])>int(a2[e][4]):
puntos=puntos+3
if int(a2[e][6])==int(a2[e][4]):
puntos=puntos+1
a4.append(puntos)
puntos=0
for g in range (n):
for h in range (n):
if int(a3[g])>int(a3[h]):
mayor=mayor+1
if int(a3[g])==int(a3[h]):
if int(a4[g])>int(a4[h]):
mayor=mayor+1
po=n-mayor-1
mayor=0
a5[po]=a1[g]
for la in range (n//2):
a6.append(a5[la])
a6.sort()
for yonax in range (len(a6)):
print(a6[yonax])
``` | instruction | 0 | 88,689 | 17 | 177,378 |
No | output | 1 | 88,689 | 17 | 177,379 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Everyone knows that 2010 FIFA World Cup is being held in South Africa now. By the decision of BFA (Berland's Football Association) next World Cup will be held in Berland. BFA took the decision to change some World Cup regulations:
* the final tournament features n teams (n is always even)
* the first n / 2 teams (according to the standings) come through to the knockout stage
* the standings are made on the following principle: for a victory a team gets 3 points, for a draw β 1 point, for a defeat β 0 points. In the first place, teams are ordered in the standings in decreasing order of their points; in the second place β in decreasing order of the difference between scored and missed goals; in the third place β in the decreasing order of scored goals
* it's written in Berland's Constitution that the previous regulation helps to order the teams without ambiguity.
You are asked to write a program that, by the given list of the competing teams and the results of all the matches, will find the list of teams that managed to get through to the knockout stage.
Input
The first input line contains the only integer n (1 β€ n β€ 50) β amount of the teams, taking part in the final tournament of World Cup. The following n lines contain the names of these teams, a name is a string of lower-case and upper-case Latin letters, its length doesn't exceed 30 characters. The following nΒ·(n - 1) / 2 lines describe the held matches in the format name1-name2 num1:num2, where name1, name2 β names of the teams; num1, num2 (0 β€ num1, num2 β€ 100) β amount of the goals, scored by the corresponding teams. Accuracy of the descriptions is guaranteed: there are no two team names coinciding accurate to the letters' case; there is no match, where a team plays with itself; each match is met in the descriptions only once.
Output
Output n / 2 lines β names of the teams, which managed to get through to the knockout stage in lexicographical order. Output each name in a separate line. No odd characters (including spaces) are allowed. It's guaranteed that the described regulations help to order the teams without ambiguity.
Examples
Input
4
A
B
C
D
A-B 1:1
A-C 2:2
A-D 1:0
B-C 1:0
B-D 0:3
C-D 0:3
Output
A
D
Input
2
a
A
a-A 2:1
Output
a
Submitted Solution:
```
clasificados = []
equipos = []
puntos = []
dg = []
gf = []
n = int(input())
for i in range(n):
equipos.append(input())
puntos.append(0)
dg.append(0)
gf.append(0)
n_partidos = (n * (n-1)) // 2
for i in range(n_partidos):
partido = input()
equipos_partido, resultado = partido.split()
e1, e2 = equipos_partido.split('-')
goles = resultado.split(':')
g1, g2 = int(goles[0]) , int(goles[1])
if g1 > g2:
p1 = 3
p2 = 0
elif g1 == g2:
p1 = 1
p2 = 1
else:
p1 = 0
p2 = 3
a = equipos.index(e1)
b = equipos.index(e2)
puntos[a] = puntos[a] + p1
dg[a] = dg[a] + g1 - g2
gf[a] = gf[a] + g1
puntos[b] = puntos[b] + p2
dg[b] = dg[b] + g2 - g1
gf[b] = gf[b] + g2
x = 0
for i in range(n//2):
mayor = 0
for j in range(n):
if mayor != j:
if puntos[mayor] < puntos[j]:
mayor = j
elif puntos[mayor] == puntos[j]:
if dg[mayor] < dg[j]:
mayor = j
elif dg[mayor] == dg[j]:
if gf[mayor] < gf[mayor]:
mayor = j
clasificados.append(equipos[mayor])
puntos[mayor] = -1
clasificados.sort()
for i in range(n//2):
print(clasificados[i])
``` | instruction | 0 | 88,690 | 17 | 177,380 |
No | output | 1 | 88,690 | 17 | 177,381 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Everyone knows that 2010 FIFA World Cup is being held in South Africa now. By the decision of BFA (Berland's Football Association) next World Cup will be held in Berland. BFA took the decision to change some World Cup regulations:
* the final tournament features n teams (n is always even)
* the first n / 2 teams (according to the standings) come through to the knockout stage
* the standings are made on the following principle: for a victory a team gets 3 points, for a draw β 1 point, for a defeat β 0 points. In the first place, teams are ordered in the standings in decreasing order of their points; in the second place β in decreasing order of the difference between scored and missed goals; in the third place β in the decreasing order of scored goals
* it's written in Berland's Constitution that the previous regulation helps to order the teams without ambiguity.
You are asked to write a program that, by the given list of the competing teams and the results of all the matches, will find the list of teams that managed to get through to the knockout stage.
Input
The first input line contains the only integer n (1 β€ n β€ 50) β amount of the teams, taking part in the final tournament of World Cup. The following n lines contain the names of these teams, a name is a string of lower-case and upper-case Latin letters, its length doesn't exceed 30 characters. The following nΒ·(n - 1) / 2 lines describe the held matches in the format name1-name2 num1:num2, where name1, name2 β names of the teams; num1, num2 (0 β€ num1, num2 β€ 100) β amount of the goals, scored by the corresponding teams. Accuracy of the descriptions is guaranteed: there are no two team names coinciding accurate to the letters' case; there is no match, where a team plays with itself; each match is met in the descriptions only once.
Output
Output n / 2 lines β names of the teams, which managed to get through to the knockout stage in lexicographical order. Output each name in a separate line. No odd characters (including spaces) are allowed. It's guaranteed that the described regulations help to order the teams without ambiguity.
Examples
Input
4
A
B
C
D
A-B 1:1
A-C 2:2
A-D 1:0
B-C 1:0
B-D 0:3
C-D 0:3
Output
A
D
Input
2
a
A
a-A 2:1
Output
a
Submitted Solution:
```
n = int(input())
d = dict()
for i in range(n):
d[input()] = [0, 0, 0]
for i in range(n * (n - 1) // 2):
t, s = input().split()
t1, t2 = t.split('-')
s1, s2 = [int(x) for x in s.split(':')]
a = d[t1]
b = d[t2]
a[2] += s1
b[2] += s2
a[1] += s1 - s2
b[1] += s2 - s1
if s1 == s2:
a[0] += 1
b[0] += 1
elif s1 < s2:
b[0] += 3
else:
a[0] += 3
print('\n'.join(b for a, b in sorted((b, a) for a, b in d.items())[n // 2: ]))
``` | instruction | 0 | 88,691 | 17 | 177,382 |
No | output | 1 | 88,691 | 17 | 177,383 |
Provide a correct Python 3 solution for this coding contest problem.
A smelt fishing tournament was held at Lake Hibara. It seems that catch and release is recommended this time.
Create a program that reads the participant number and the number of fish caught or released in order as one event, and outputs the participant number and the number of animals that have acquired the most smelt immediately after each event. please. If there are multiple participants with the highest number of participants (or if all participants are 0), output the one with the lowest participant number.
input
The input is given in the following format.
n q
a1 v1
a2 v2
::
aq vq
n (1 β€ n β€ 1000000) represents the number of participants and q (1 β€ q β€ 100000) represents the number of events. ai (1 β€ ai β€ n) vi (-100 β€ vi β€ 100) indicates that participant ai acquired or released vi at the i-th event. For vi, a positive value indicates acquisition, a negative value indicates release, and 0 is never given.
output
For each event, the participant number and the number of participants who have acquired the most smelt at hand are output on one line separated by one blank.
Example
Input
3 5
1 4
2 5
1 3
3 6
2 7
Output
1 4
2 5
1 7
1 7
2 12 | instruction | 0 | 89,142 | 17 | 178,284 |
"Correct Solution:
```
import sys
def f():
w=m=0;n,q=map(int,input().split());s=[0]*-~n
for e in sys.stdin:
a,v=map(int,e.split());s[a]+=v
if v<0 and a==w:m=max(s);w=s.index(m)
elif s[a]>m:w,m=a,s[a]
elif s[a]==m:w=min(w,a)
print(w,m)
if'__main__'==__name__:f()
``` | output | 1 | 89,142 | 17 | 178,285 |
Provide a correct Python 3 solution for this coding contest problem.
A smelt fishing tournament was held at Lake Hibara. It seems that catch and release is recommended this time.
Create a program that reads the participant number and the number of fish caught or released in order as one event, and outputs the participant number and the number of animals that have acquired the most smelt immediately after each event. please. If there are multiple participants with the highest number of participants (or if all participants are 0), output the one with the lowest participant number.
input
The input is given in the following format.
n q
a1 v1
a2 v2
::
aq vq
n (1 β€ n β€ 1000000) represents the number of participants and q (1 β€ q β€ 100000) represents the number of events. ai (1 β€ ai β€ n) vi (-100 β€ vi β€ 100) indicates that participant ai acquired or released vi at the i-th event. For vi, a positive value indicates acquisition, a negative value indicates release, and 0 is never given.
output
For each event, the participant number and the number of participants who have acquired the most smelt at hand are output on one line separated by one blank.
Example
Input
3 5
1 4
2 5
1 3
3 6
2 7
Output
1 4
2 5
1 7
1 7
2 12 | instruction | 0 | 89,143 | 17 | 178,286 |
"Correct Solution:
```
def solve():
import heapq
from sys import stdin
lines = stdin
n, q = map(int, lines.readline().split())
# hq element: [number of caught fish, participant number]
hq = [(0, i) for i in range(1, n + 1)]
heapq.heapify(hq)
fish = [0] * (n + 1) # number of caught fish
ans = ""
for i in range(q):
a, v = map(int, lines.readline().split())
# Because of Python heapq specification, negative values are used for fishing.
v *= -1
fish[a] += v
heapq.heappush(hq, (fish[a], a))
while hq[0][0] != fish[hq[0][1]]:
heapq.heappop(hq)
ans += f"{hq[0][1]} {-hq[0][0]}\n"
print(ans, end="")
solve()
``` | output | 1 | 89,143 | 17 | 178,287 |
Provide a correct Python 3 solution for this coding contest problem.
A smelt fishing tournament was held at Lake Hibara. It seems that catch and release is recommended this time.
Create a program that reads the participant number and the number of fish caught or released in order as one event, and outputs the participant number and the number of animals that have acquired the most smelt immediately after each event. please. If there are multiple participants with the highest number of participants (or if all participants are 0), output the one with the lowest participant number.
input
The input is given in the following format.
n q
a1 v1
a2 v2
::
aq vq
n (1 β€ n β€ 1000000) represents the number of participants and q (1 β€ q β€ 100000) represents the number of events. ai (1 β€ ai β€ n) vi (-100 β€ vi β€ 100) indicates that participant ai acquired or released vi at the i-th event. For vi, a positive value indicates acquisition, a negative value indicates release, and 0 is never given.
output
For each event, the participant number and the number of participants who have acquired the most smelt at hand are output on one line separated by one blank.
Example
Input
3 5
1 4
2 5
1 3
3 6
2 7
Output
1 4
2 5
1 7
1 7
2 12 | instruction | 0 | 89,144 | 17 | 178,288 |
"Correct Solution:
```
"""
γ»γ°γ‘γ³γζ¨
"""
INF = 10 ** 20
n, q = map(int, input().split())
size = 1
while size < n:
size *= 2
size = size * 2 - 1
seg_tree = [(INF, 0) for _ in range(size)]
def update(a, v):
ind = (size - 1) // 2 + a
prea, prev = seg_tree[ind]
seg_tree[ind] = (a + 1, prev + v)
while ind > 0:
ind = (ind - 1) // 2
ca1, cv1 = seg_tree[ind * 2 + 1]
ca2, cv2 = seg_tree[ind * 2 + 2]
if cv1 >= cv2:
seg_tree[ind] = (ca1, cv1)
else:
seg_tree[ind] = (ca2, cv2)
for i in range(q):
a, v = map(int, input().split())
a -= 1
update(a, v)
print(seg_tree[0][0], seg_tree[0][1])
``` | output | 1 | 89,144 | 17 | 178,289 |
Provide a correct Python 3 solution for this coding contest problem.
A smelt fishing tournament was held at Lake Hibara. It seems that catch and release is recommended this time.
Create a program that reads the participant number and the number of fish caught or released in order as one event, and outputs the participant number and the number of animals that have acquired the most smelt immediately after each event. please. If there are multiple participants with the highest number of participants (or if all participants are 0), output the one with the lowest participant number.
input
The input is given in the following format.
n q
a1 v1
a2 v2
::
aq vq
n (1 β€ n β€ 1000000) represents the number of participants and q (1 β€ q β€ 100000) represents the number of events. ai (1 β€ ai β€ n) vi (-100 β€ vi β€ 100) indicates that participant ai acquired or released vi at the i-th event. For vi, a positive value indicates acquisition, a negative value indicates release, and 0 is never given.
output
For each event, the participant number and the number of participants who have acquired the most smelt at hand are output on one line separated by one blank.
Example
Input
3 5
1 4
2 5
1 3
3 6
2 7
Output
1 4
2 5
1 7
1 7
2 12 | instruction | 0 | 89,145 | 17 | 178,290 |
"Correct Solution:
```
n,q=map(int,input().split())
s=[0]*-~n
w=m=0
for _ in[0]*q:
a,v=map(int,input().split())
s[a]+=v
if v<0 and a==w:m=max(s);w=s.index(m)
elif s[a]>m:w,m=a,s[a]
elif s[a]==m and a<w:w=a
print(w,m)
``` | output | 1 | 89,145 | 17 | 178,291 |
Provide a correct Python 3 solution for this coding contest problem.
A smelt fishing tournament was held at Lake Hibara. It seems that catch and release is recommended this time.
Create a program that reads the participant number and the number of fish caught or released in order as one event, and outputs the participant number and the number of animals that have acquired the most smelt immediately after each event. please. If there are multiple participants with the highest number of participants (or if all participants are 0), output the one with the lowest participant number.
input
The input is given in the following format.
n q
a1 v1
a2 v2
::
aq vq
n (1 β€ n β€ 1000000) represents the number of participants and q (1 β€ q β€ 100000) represents the number of events. ai (1 β€ ai β€ n) vi (-100 β€ vi β€ 100) indicates that participant ai acquired or released vi at the i-th event. For vi, a positive value indicates acquisition, a negative value indicates release, and 0 is never given.
output
For each event, the participant number and the number of participants who have acquired the most smelt at hand are output on one line separated by one blank.
Example
Input
3 5
1 4
2 5
1 3
3 6
2 7
Output
1 4
2 5
1 7
1 7
2 12 | instruction | 0 | 89,146 | 17 | 178,292 |
"Correct Solution:
```
[n,q] = map(int,input().split())
t = [ 0 for _ in range(n) ]
max_idx = 0
max_val = 0
for _ in range(q):
[a,v] = map(int,input().split())
t[a-1] += v
if max_idx == a-1:
if v > 0:
max_val = t[a-1]
else:
max_val = max(t)
max_idx = t.index(max_val)
elif t[a-1] > max_val:
max_val = t[a-1]
max_idx = a-1
elif t[a-1] == max_val:
if a-1 < max_idx:
max_val = t[a-1]
max_idx = a-1
print("{} {}".format(max_idx+1,max_val))
``` | output | 1 | 89,146 | 17 | 178,293 |
Provide a correct Python 3 solution for this coding contest problem.
A smelt fishing tournament was held at Lake Hibara. It seems that catch and release is recommended this time.
Create a program that reads the participant number and the number of fish caught or released in order as one event, and outputs the participant number and the number of animals that have acquired the most smelt immediately after each event. please. If there are multiple participants with the highest number of participants (or if all participants are 0), output the one with the lowest participant number.
input
The input is given in the following format.
n q
a1 v1
a2 v2
::
aq vq
n (1 β€ n β€ 1000000) represents the number of participants and q (1 β€ q β€ 100000) represents the number of events. ai (1 β€ ai β€ n) vi (-100 β€ vi β€ 100) indicates that participant ai acquired or released vi at the i-th event. For vi, a positive value indicates acquisition, a negative value indicates release, and 0 is never given.
output
For each event, the participant number and the number of participants who have acquired the most smelt at hand are output on one line separated by one blank.
Example
Input
3 5
1 4
2 5
1 3
3 6
2 7
Output
1 4
2 5
1 7
1 7
2 12 | instruction | 0 | 89,147 | 17 | 178,294 |
"Correct Solution:
```
import sys
def f():
w=m=0;n,q=map(int,input().split());s=[0]*-~n
for e in sys.stdin:
a,v=map(int,e.split());s[a]+=v
if v<0 and a==w:m=max(s);w=s.index(m)
elif s[a]>m:w,m=a,s[a]
elif s[a]==m and w>a:w=a
print(w,m)
f()
``` | output | 1 | 89,147 | 17 | 178,295 |
Provide a correct Python 3 solution for this coding contest problem.
A smelt fishing tournament was held at Lake Hibara. It seems that catch and release is recommended this time.
Create a program that reads the participant number and the number of fish caught or released in order as one event, and outputs the participant number and the number of animals that have acquired the most smelt immediately after each event. please. If there are multiple participants with the highest number of participants (or if all participants are 0), output the one with the lowest participant number.
input
The input is given in the following format.
n q
a1 v1
a2 v2
::
aq vq
n (1 β€ n β€ 1000000) represents the number of participants and q (1 β€ q β€ 100000) represents the number of events. ai (1 β€ ai β€ n) vi (-100 β€ vi β€ 100) indicates that participant ai acquired or released vi at the i-th event. For vi, a positive value indicates acquisition, a negative value indicates release, and 0 is never given.
output
For each event, the participant number and the number of participants who have acquired the most smelt at hand are output on one line separated by one blank.
Example
Input
3 5
1 4
2 5
1 3
3 6
2 7
Output
1 4
2 5
1 7
1 7
2 12 | instruction | 0 | 89,148 | 17 | 178,296 |
"Correct Solution:
```
import sys
def f():
w=m=0;n,q=map(int,input().split());s=[0]*-~n
for e in sys.stdin:
a,v=map(int,e.split());s[a]+=v
if v<0 and a==w:m=max(s);w=s.index(m)
elif s[a]>m:w,m=a,s[a]
elif s[a]==m:w=min(w,a)
print(w,m)
f()
``` | output | 1 | 89,148 | 17 | 178,297 |
Provide a correct Python 3 solution for this coding contest problem.
A smelt fishing tournament was held at Lake Hibara. It seems that catch and release is recommended this time.
Create a program that reads the participant number and the number of fish caught or released in order as one event, and outputs the participant number and the number of animals that have acquired the most smelt immediately after each event. please. If there are multiple participants with the highest number of participants (or if all participants are 0), output the one with the lowest participant number.
input
The input is given in the following format.
n q
a1 v1
a2 v2
::
aq vq
n (1 β€ n β€ 1000000) represents the number of participants and q (1 β€ q β€ 100000) represents the number of events. ai (1 β€ ai β€ n) vi (-100 β€ vi β€ 100) indicates that participant ai acquired or released vi at the i-th event. For vi, a positive value indicates acquisition, a negative value indicates release, and 0 is never given.
output
For each event, the participant number and the number of participants who have acquired the most smelt at hand are output on one line separated by one blank.
Example
Input
3 5
1 4
2 5
1 3
3 6
2 7
Output
1 4
2 5
1 7
1 7
2 12 | instruction | 0 | 89,149 | 17 | 178,298 |
"Correct Solution:
```
"""
γ»γ°γ‘γ³γζ¨
"""
INF = 10 ** 20
def main():
n, q = map(int, input().split())
size = 1
while size < n:
size *= 2
size = size * 2 - 1
seg_tree = [(INF, 0) for _ in range(size)]
def update(a, v):
ind = (size - 1) // 2 + a
prea, prev = seg_tree[ind]
seg_tree[ind] = (a + 1, prev + v)
while ind > 0:
ind = (ind - 1) // 2
ca1, cv1 = seg_tree[ind * 2 + 1]
ca2, cv2 = seg_tree[ind * 2 + 2]
if cv1 >= cv2:
seg_tree[ind] = (ca1, cv1)
else:
seg_tree[ind] = (ca2, cv2)
for i in range(q):
a, v = map(int, input().split())
update(a - 1, v)
print(seg_tree[0][0], seg_tree[0][1])
main()
``` | output | 1 | 89,149 | 17 | 178,299 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A smelt fishing tournament was held at Lake Hibara. It seems that catch and release is recommended this time.
Create a program that reads the participant number and the number of fish caught or released in order as one event, and outputs the participant number and the number of animals that have acquired the most smelt immediately after each event. please. If there are multiple participants with the highest number of participants (or if all participants are 0), output the one with the lowest participant number.
input
The input is given in the following format.
n q
a1 v1
a2 v2
::
aq vq
n (1 β€ n β€ 1000000) represents the number of participants and q (1 β€ q β€ 100000) represents the number of events. ai (1 β€ ai β€ n) vi (-100 β€ vi β€ 100) indicates that participant ai acquired or released vi at the i-th event. For vi, a positive value indicates acquisition, a negative value indicates release, and 0 is never given.
output
For each event, the participant number and the number of participants who have acquired the most smelt at hand are output on one line separated by one blank.
Example
Input
3 5
1 4
2 5
1 3
3 6
2 7
Output
1 4
2 5
1 7
1 7
2 12
Submitted Solution:
```
n, q = map(int, input().split())
n = [0]*(n+1)
maxv = winner = 0
for _ in range(q):
a, v = map(int, input().split())
n[a] += v
if v < 0 and a == winner:
maxv = max(n)
winner = n.index(maxv)
elif n[a] > maxv:
maxv, winner = n[a], a
elif n[a] == maxv:
winner = min(winner, a)
print(winner, maxv)
``` | instruction | 0 | 89,150 | 17 | 178,300 |
Yes | output | 1 | 89,150 | 17 | 178,301 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A smelt fishing tournament was held at Lake Hibara. It seems that catch and release is recommended this time.
Create a program that reads the participant number and the number of fish caught or released in order as one event, and outputs the participant number and the number of animals that have acquired the most smelt immediately after each event. please. If there are multiple participants with the highest number of participants (or if all participants are 0), output the one with the lowest participant number.
input
The input is given in the following format.
n q
a1 v1
a2 v2
::
aq vq
n (1 β€ n β€ 1000000) represents the number of participants and q (1 β€ q β€ 100000) represents the number of events. ai (1 β€ ai β€ n) vi (-100 β€ vi β€ 100) indicates that participant ai acquired or released vi at the i-th event. For vi, a positive value indicates acquisition, a negative value indicates release, and 0 is never given.
output
For each event, the participant number and the number of participants who have acquired the most smelt at hand are output on one line separated by one blank.
Example
Input
3 5
1 4
2 5
1 3
3 6
2 7
Output
1 4
2 5
1 7
1 7
2 12
Submitted Solution:
```
import queue
n,q=map(int,input().split())
p=queue.PriorityQueue()
s=[0]*-~n
for _ in[0]*q:
a,v=map(int,input().split())
s[a]+=v;p.put((-s[a],a))
while 1:
t=p.get()
if -t[0]==s[t[1]]:print(t[1],-t[0]);p.put(t);break
``` | instruction | 0 | 89,151 | 17 | 178,302 |
Yes | output | 1 | 89,151 | 17 | 178,303 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A smelt fishing tournament was held at Lake Hibara. It seems that catch and release is recommended this time.
Create a program that reads the participant number and the number of fish caught or released in order as one event, and outputs the participant number and the number of animals that have acquired the most smelt immediately after each event. please. If there are multiple participants with the highest number of participants (or if all participants are 0), output the one with the lowest participant number.
input
The input is given in the following format.
n q
a1 v1
a2 v2
::
aq vq
n (1 β€ n β€ 1000000) represents the number of participants and q (1 β€ q β€ 100000) represents the number of events. ai (1 β€ ai β€ n) vi (-100 β€ vi β€ 100) indicates that participant ai acquired or released vi at the i-th event. For vi, a positive value indicates acquisition, a negative value indicates release, and 0 is never given.
output
For each event, the participant number and the number of participants who have acquired the most smelt at hand are output on one line separated by one blank.
Example
Input
3 5
1 4
2 5
1 3
3 6
2 7
Output
1 4
2 5
1 7
1 7
2 12
Submitted Solution:
```
n, q = map(int, input().split())
result = [0 for i in range(n + 1)]
max_a = -1
max_v = 0
for i in range(q):
a, v = map(int, input().split())
result[a] += v
if result[a] > max_v:
max_a = a
max_v = result[a]
elif result[a] == max_v:
if a < max_a:
max_a = a
elif a == max_a and result[a] <= max_v:
max_a = result.index(max(result))
max_v = result[max_a]
print(max_a, max_v)
``` | instruction | 0 | 89,152 | 17 | 178,304 |
Yes | output | 1 | 89,152 | 17 | 178,305 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A smelt fishing tournament was held at Lake Hibara. It seems that catch and release is recommended this time.
Create a program that reads the participant number and the number of fish caught or released in order as one event, and outputs the participant number and the number of animals that have acquired the most smelt immediately after each event. please. If there are multiple participants with the highest number of participants (or if all participants are 0), output the one with the lowest participant number.
input
The input is given in the following format.
n q
a1 v1
a2 v2
::
aq vq
n (1 β€ n β€ 1000000) represents the number of participants and q (1 β€ q β€ 100000) represents the number of events. ai (1 β€ ai β€ n) vi (-100 β€ vi β€ 100) indicates that participant ai acquired or released vi at the i-th event. For vi, a positive value indicates acquisition, a negative value indicates release, and 0 is never given.
output
For each event, the participant number and the number of participants who have acquired the most smelt at hand are output on one line separated by one blank.
Example
Input
3 5
1 4
2 5
1 3
3 6
2 7
Output
1 4
2 5
1 7
1 7
2 12
Submitted Solution:
```
n, q = [int(el) for el in input().split(' ')]
data = [0] * n
result, index = 0, 1
for _ in range(q):
a, v = [int(el) for el in input().split(' ')]
data[a-1] += v
if v > 0:
if result < data[a-1]:
result, index = data[a-1], a
elif result == data[a-1]:
index = min(index, a)
else:
if index == a:
result = max(data)
index = data.index(result) + 1
print(index, result)
``` | instruction | 0 | 89,153 | 17 | 178,306 |
Yes | output | 1 | 89,153 | 17 | 178,307 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A smelt fishing tournament was held at Lake Hibara. It seems that catch and release is recommended this time.
Create a program that reads the participant number and the number of fish caught or released in order as one event, and outputs the participant number and the number of animals that have acquired the most smelt immediately after each event. please. If there are multiple participants with the highest number of participants (or if all participants are 0), output the one with the lowest participant number.
input
The input is given in the following format.
n q
a1 v1
a2 v2
::
aq vq
n (1 β€ n β€ 1000000) represents the number of participants and q (1 β€ q β€ 100000) represents the number of events. ai (1 β€ ai β€ n) vi (-100 β€ vi β€ 100) indicates that participant ai acquired or released vi at the i-th event. For vi, a positive value indicates acquisition, a negative value indicates release, and 0 is never given.
output
For each event, the participant number and the number of participants who have acquired the most smelt at hand are output on one line separated by one blank.
Example
Input
3 5
1 4
2 5
1 3
3 6
2 7
Output
1 4
2 5
1 7
1 7
2 12
Submitted Solution:
```
from sys import stdin
n, q = [int(_) for _ in input().split()]
book = [0] * (n+1)
max_per, max_val = q, 0
for _ in stdin.readlines() :
a, v = [int(_2) for _2 in _.split()]
book[a] += v
if v < 0 :
max_val = max(book)
max_per = book.index(max_val)
elif max_val < book[a] : max_per, max_val = a, book[a]
elif max_val == book[a] and a < max_per : max_per = a
print(max_per, max_val)
``` | instruction | 0 | 89,154 | 17 | 178,308 |
No | output | 1 | 89,154 | 17 | 178,309 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A smelt fishing tournament was held at Lake Hibara. It seems that catch and release is recommended this time.
Create a program that reads the participant number and the number of fish caught or released in order as one event, and outputs the participant number and the number of animals that have acquired the most smelt immediately after each event. please. If there are multiple participants with the highest number of participants (or if all participants are 0), output the one with the lowest participant number.
input
The input is given in the following format.
n q
a1 v1
a2 v2
::
aq vq
n (1 β€ n β€ 1000000) represents the number of participants and q (1 β€ q β€ 100000) represents the number of events. ai (1 β€ ai β€ n) vi (-100 β€ vi β€ 100) indicates that participant ai acquired or released vi at the i-th event. For vi, a positive value indicates acquisition, a negative value indicates release, and 0 is never given.
output
For each event, the participant number and the number of participants who have acquired the most smelt at hand are output on one line separated by one blank.
Example
Input
3 5
1 4
2 5
1 3
3 6
2 7
Output
1 4
2 5
1 7
1 7
2 12
Submitted Solution:
```
def liquidation(A):
A.sort()#??????????????????????????????
for i in range(len(A)):
for j in range(i+1,len(A)):
if A[i][1]==A[j][1]:
A[i][0]+=A[j][0]
del A[j]
break
return A
def rank(A):
Max=A[0][1]
champ=A[0][0]
for i in A:
if Max<i[1]:
champ=i[0]
Max=i[1]
print(champ,Max)
def Rank(A):
Max=A[0][0]
champ=A[0][1]
for i in range(1,len(A)):
if A[i][0]==Max:
champ=A[i][1]
else:
break
print(champ,Max)
n,q=map(int,input().split())
A=[]
for i in range(q):
a,v=map(int,input().split())
A.append([v,a])#a??????????????????, v?????Β£??Β£???????????Β°
A=liquidation(A)
A.sort(reverse=True)
#print(A)
Rank(A)
``` | instruction | 0 | 89,155 | 17 | 178,310 |
No | output | 1 | 89,155 | 17 | 178,311 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A smelt fishing tournament was held at Lake Hibara. It seems that catch and release is recommended this time.
Create a program that reads the participant number and the number of fish caught or released in order as one event, and outputs the participant number and the number of animals that have acquired the most smelt immediately after each event. please. If there are multiple participants with the highest number of participants (or if all participants are 0), output the one with the lowest participant number.
input
The input is given in the following format.
n q
a1 v1
a2 v2
::
aq vq
n (1 β€ n β€ 1000000) represents the number of participants and q (1 β€ q β€ 100000) represents the number of events. ai (1 β€ ai β€ n) vi (-100 β€ vi β€ 100) indicates that participant ai acquired or released vi at the i-th event. For vi, a positive value indicates acquisition, a negative value indicates release, and 0 is never given.
output
For each event, the participant number and the number of participants who have acquired the most smelt at hand are output on one line separated by one blank.
Example
Input
3 5
1 4
2 5
1 3
3 6
2 7
Output
1 4
2 5
1 7
1 7
2 12
Submitted Solution:
```
from collections import deque
n, q = [int(_) for _ in input().split()]
book = deque()
for _ in range(n+1) : book.append(0)
for _ in range(q) :
a, v = [int(_2) for _2 in input().split()]
book[a] += v
print(book.index(max(book)), max(book))
``` | instruction | 0 | 89,156 | 17 | 178,312 |
No | output | 1 | 89,156 | 17 | 178,313 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A smelt fishing tournament was held at Lake Hibara. It seems that catch and release is recommended this time.
Create a program that reads the participant number and the number of fish caught or released in order as one event, and outputs the participant number and the number of animals that have acquired the most smelt immediately after each event. please. If there are multiple participants with the highest number of participants (or if all participants are 0), output the one with the lowest participant number.
input
The input is given in the following format.
n q
a1 v1
a2 v2
::
aq vq
n (1 β€ n β€ 1000000) represents the number of participants and q (1 β€ q β€ 100000) represents the number of events. ai (1 β€ ai β€ n) vi (-100 β€ vi β€ 100) indicates that participant ai acquired or released vi at the i-th event. For vi, a positive value indicates acquisition, a negative value indicates release, and 0 is never given.
output
For each event, the participant number and the number of participants who have acquired the most smelt at hand are output on one line separated by one blank.
Example
Input
3 5
1 4
2 5
1 3
3 6
2 7
Output
1 4
2 5
1 7
1 7
2 12
Submitted Solution:
```
n,q=map(int,input().split())
s=[0]*-~n
w=m=0
for _ in[0]*q:
a,v=map(int,input().split())
s[a]+=v
if v<0 and a==w:m=max(s);w=s.index(m)
elif s[a]>maxv:w,m=a,s[a]
elif s[a]==maxv:w=min(w,a)
print(w,m)
``` | instruction | 0 | 89,157 | 17 | 178,314 |
No | output | 1 | 89,157 | 17 | 178,315 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Two semifinals have just been in the running tournament. Each semifinal had n participants. There are n participants advancing to the finals, they are chosen as follows: from each semifinal, we choose k people (0 β€ 2k β€ n) who showed the best result in their semifinals and all other places in the finals go to the people who haven't ranked in the top k in their semifinal but got to the n - 2k of the best among the others.
The tournament organizers hasn't yet determined the k value, so the participants want to know who else has any chance to get to the finals and who can go home.
Input
The first line contains a single integer n (1 β€ n β€ 105) β the number of participants in each semifinal.
Each of the next n lines contains two integers ai and bi (1 β€ ai, bi β€ 109) β the results of the i-th participant (the number of milliseconds he needs to cover the semifinals distance) of the first and second semifinals, correspondingly. All results are distinct. Sequences a1, a2, ..., an and b1, b2, ..., bn are sorted in ascending order, i.e. in the order the participants finished in the corresponding semifinal.
Output
Print two strings consisting of n characters, each equals either "0" or "1". The first line should correspond to the participants of the first semifinal, the second line should correspond to the participants of the second semifinal. The i-th character in the j-th line should equal "1" if the i-th participant of the j-th semifinal has any chances to advance to the finals, otherwise it should equal a "0".
Examples
Input
4
9840 9920
9860 9980
9930 10020
10040 10090
Output
1110
1100
Input
4
9900 9850
9940 9930
10000 10020
10060 10110
Output
1100
1100
Note
Consider the first sample. Each semifinal has 4 participants. The results of the first semifinal are 9840, 9860, 9930, 10040. The results of the second semifinal are 9920, 9980, 10020, 10090.
* If k = 0, the finalists are determined by the time only, so players 9840, 9860, 9920 and 9930 advance to the finals.
* If k = 1, the winners from both semifinals move to the finals (with results 9840 and 9920), and the other places are determined by the time (these places go to the sportsmen who run the distance in 9860 and 9930 milliseconds).
* If k = 2, then first and second places advance from each seminfial, these are participants with results 9840, 9860, 9920 and 9980 milliseconds. | instruction | 0 | 89,619 | 17 | 179,238 |
Tags: implementation, sortings
Correct Solution:
```
def semifinals(l1,l2):
output1=[0]*len(l1)
output2=[0]*len(l2)
n=len(l1)
k = len(l1)//2
output1[0:k]=[1]*k
output2[0:k]=[1]*k
i = k
while i < len(l1):
if l1[i]<l2[n-i-1]:
output1[i]=1
if l2[i]<l1[n-i-1]:
output2[i]=1
i+=1
output1=''.join([str(x) for x in output1])
output2=''.join([str(x) for x in output2])
return (output1,output2)
n=int(input())
l1=[]
l2=[]
for i in range(n):
a = [int(x) for x in input().split()]
l1.append(a[0])
l2.append(a[1])
print (semifinals(l1,l2)[0])
print (semifinals(l1,l2)[1])
``` | output | 1 | 89,619 | 17 | 179,239 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Two semifinals have just been in the running tournament. Each semifinal had n participants. There are n participants advancing to the finals, they are chosen as follows: from each semifinal, we choose k people (0 β€ 2k β€ n) who showed the best result in their semifinals and all other places in the finals go to the people who haven't ranked in the top k in their semifinal but got to the n - 2k of the best among the others.
The tournament organizers hasn't yet determined the k value, so the participants want to know who else has any chance to get to the finals and who can go home.
Input
The first line contains a single integer n (1 β€ n β€ 105) β the number of participants in each semifinal.
Each of the next n lines contains two integers ai and bi (1 β€ ai, bi β€ 109) β the results of the i-th participant (the number of milliseconds he needs to cover the semifinals distance) of the first and second semifinals, correspondingly. All results are distinct. Sequences a1, a2, ..., an and b1, b2, ..., bn are sorted in ascending order, i.e. in the order the participants finished in the corresponding semifinal.
Output
Print two strings consisting of n characters, each equals either "0" or "1". The first line should correspond to the participants of the first semifinal, the second line should correspond to the participants of the second semifinal. The i-th character in the j-th line should equal "1" if the i-th participant of the j-th semifinal has any chances to advance to the finals, otherwise it should equal a "0".
Examples
Input
4
9840 9920
9860 9980
9930 10020
10040 10090
Output
1110
1100
Input
4
9900 9850
9940 9930
10000 10020
10060 10110
Output
1100
1100
Note
Consider the first sample. Each semifinal has 4 participants. The results of the first semifinal are 9840, 9860, 9930, 10040. The results of the second semifinal are 9920, 9980, 10020, 10090.
* If k = 0, the finalists are determined by the time only, so players 9840, 9860, 9920 and 9930 advance to the finals.
* If k = 1, the winners from both semifinals move to the finals (with results 9840 and 9920), and the other places are determined by the time (these places go to the sportsmen who run the distance in 9860 and 9930 milliseconds).
* If k = 2, then first and second places advance from each seminfial, these are participants with results 9840, 9860, 9920 and 9980 milliseconds. | instruction | 0 | 89,620 | 17 | 179,240 |
Tags: implementation, sortings
Correct Solution:
```
n = int(input())
p1=['0' for _ in range(n)]
p2=['0' for _ in range(n)]
t1=[]
t2=[]
for i in range(n):
a, b = map(int, input().split())
t1.append((a,i))
t2.append((b,i))
t2.sort()
t1.sort()
for k in range((n//2)):
p1[t1[k][1]]='1'
p2[t2[k][1]]='1'
i=0
j=0
for k in range(n):
if t1[i][0]<t2[j][0]:
p1[t1[i][1]]='1'
i+=1
else:
p2[t2[j][1]]='1'
j+=1
print("".join(p1))
print("".join(p2))
``` | output | 1 | 89,620 | 17 | 179,241 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Two semifinals have just been in the running tournament. Each semifinal had n participants. There are n participants advancing to the finals, they are chosen as follows: from each semifinal, we choose k people (0 β€ 2k β€ n) who showed the best result in their semifinals and all other places in the finals go to the people who haven't ranked in the top k in their semifinal but got to the n - 2k of the best among the others.
The tournament organizers hasn't yet determined the k value, so the participants want to know who else has any chance to get to the finals and who can go home.
Input
The first line contains a single integer n (1 β€ n β€ 105) β the number of participants in each semifinal.
Each of the next n lines contains two integers ai and bi (1 β€ ai, bi β€ 109) β the results of the i-th participant (the number of milliseconds he needs to cover the semifinals distance) of the first and second semifinals, correspondingly. All results are distinct. Sequences a1, a2, ..., an and b1, b2, ..., bn are sorted in ascending order, i.e. in the order the participants finished in the corresponding semifinal.
Output
Print two strings consisting of n characters, each equals either "0" or "1". The first line should correspond to the participants of the first semifinal, the second line should correspond to the participants of the second semifinal. The i-th character in the j-th line should equal "1" if the i-th participant of the j-th semifinal has any chances to advance to the finals, otherwise it should equal a "0".
Examples
Input
4
9840 9920
9860 9980
9930 10020
10040 10090
Output
1110
1100
Input
4
9900 9850
9940 9930
10000 10020
10060 10110
Output
1100
1100
Note
Consider the first sample. Each semifinal has 4 participants. The results of the first semifinal are 9840, 9860, 9930, 10040. The results of the second semifinal are 9920, 9980, 10020, 10090.
* If k = 0, the finalists are determined by the time only, so players 9840, 9860, 9920 and 9930 advance to the finals.
* If k = 1, the winners from both semifinals move to the finals (with results 9840 and 9920), and the other places are determined by the time (these places go to the sportsmen who run the distance in 9860 and 9930 milliseconds).
* If k = 2, then first and second places advance from each seminfial, these are participants with results 9840, 9860, 9920 and 9980 milliseconds. | instruction | 0 | 89,621 | 17 | 179,242 |
Tags: implementation, sortings
Correct Solution:
```
def chose_p(arr, brr, k):
n = len(arr)
crr = []
for i in range(k):
crr.append(arr[i])
arr[i] = 0
for i in range(k):
crr.append(brr[i])
brr[i] = 0
drr = [x for x in arr if x] + [x for x in brr if x]
drr.sort()
for i in range(n - 2 * k):
crr.append(drr[i])
return crr
if __name__ == '__main__':
n = int(input())
arr = []
brr = []
for _ in range(n):
x, y = map(int, input().split())
arr.append(x)
brr.append(y)
s = set()
crr = chose_p(arr[:], brr[:], 0)
drr = chose_p(arr[:], brr[:], n // 2)
s.update(crr, drr)
for i, x in enumerate(arr):
arr[i] = str(int(x in s))
for i, x in enumerate(brr):
brr[i] = str(int(x in s))
print(''.join(arr))
print(''.join(brr))
``` | output | 1 | 89,621 | 17 | 179,243 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Two semifinals have just been in the running tournament. Each semifinal had n participants. There are n participants advancing to the finals, they are chosen as follows: from each semifinal, we choose k people (0 β€ 2k β€ n) who showed the best result in their semifinals and all other places in the finals go to the people who haven't ranked in the top k in their semifinal but got to the n - 2k of the best among the others.
The tournament organizers hasn't yet determined the k value, so the participants want to know who else has any chance to get to the finals and who can go home.
Input
The first line contains a single integer n (1 β€ n β€ 105) β the number of participants in each semifinal.
Each of the next n lines contains two integers ai and bi (1 β€ ai, bi β€ 109) β the results of the i-th participant (the number of milliseconds he needs to cover the semifinals distance) of the first and second semifinals, correspondingly. All results are distinct. Sequences a1, a2, ..., an and b1, b2, ..., bn are sorted in ascending order, i.e. in the order the participants finished in the corresponding semifinal.
Output
Print two strings consisting of n characters, each equals either "0" or "1". The first line should correspond to the participants of the first semifinal, the second line should correspond to the participants of the second semifinal. The i-th character in the j-th line should equal "1" if the i-th participant of the j-th semifinal has any chances to advance to the finals, otherwise it should equal a "0".
Examples
Input
4
9840 9920
9860 9980
9930 10020
10040 10090
Output
1110
1100
Input
4
9900 9850
9940 9930
10000 10020
10060 10110
Output
1100
1100
Note
Consider the first sample. Each semifinal has 4 participants. The results of the first semifinal are 9840, 9860, 9930, 10040. The results of the second semifinal are 9920, 9980, 10020, 10090.
* If k = 0, the finalists are determined by the time only, so players 9840, 9860, 9920 and 9930 advance to the finals.
* If k = 1, the winners from both semifinals move to the finals (with results 9840 and 9920), and the other places are determined by the time (these places go to the sportsmen who run the distance in 9860 and 9930 milliseconds).
* If k = 2, then first and second places advance from each seminfial, these are participants with results 9840, 9860, 9920 and 9980 milliseconds. | instruction | 0 | 89,622 | 17 | 179,244 |
Tags: implementation, sortings
Correct Solution:
```
import sys
from functools import lru_cache, cmp_to_key
from heapq import merge, heapify, heappop, heappush
from math import *
from collections import defaultdict as dd, deque, Counter as C
from itertools import combinations as comb, permutations as perm
from bisect import bisect_left as bl, bisect_right as br, bisect
from time import perf_counter
from fractions import Fraction
import copy
import time
starttime = time.time()
mod = int(pow(10, 9) + 7)
mod2 = 998244353
# from sys import stdin
# input = stdin.readline
def data(): return sys.stdin.readline().strip()
def out(*var, end="\n"): sys.stdout.write(' '.join(map(str, var))+end)
def L(): return list(sp())
def sl(): return list(ssp())
def sp(): return map(int, data().split())
def ssp(): return map(str, data().split())
def l1d(n, val=0): return [val for i in range(n)]
def l2d(n, m, val=0): return [l1d(n, val) for j in range(m)]
try:
# sys.setrecursionlimit(int(pow(10,6)))
sys.stdin = open("input.txt", "r")
# sys.stdout = open("../output.txt", "w")
except:
pass
def pmat(A):
for ele in A:
print(ele,end="\n")
n=L()[0]
A=[]
B=[]
for i in range(n):
a,b=L()
A.append(a)
B.append(b)
Z=sorted(A+B)[n-1]
for i in range(n):
if i<n//2 or A[i]<=Z:
print(1,end="")
else:
print(0,end="")
print()
for i in range(n):
if i<n//2 or B[i]<=Z:
print(1,end="")
else:
print(0,end="")
print()
endtime = time.time()
# print(f"Runtime of the program is {endtime - starttime}")
``` | output | 1 | 89,622 | 17 | 179,245 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Two semifinals have just been in the running tournament. Each semifinal had n participants. There are n participants advancing to the finals, they are chosen as follows: from each semifinal, we choose k people (0 β€ 2k β€ n) who showed the best result in their semifinals and all other places in the finals go to the people who haven't ranked in the top k in their semifinal but got to the n - 2k of the best among the others.
The tournament organizers hasn't yet determined the k value, so the participants want to know who else has any chance to get to the finals and who can go home.
Input
The first line contains a single integer n (1 β€ n β€ 105) β the number of participants in each semifinal.
Each of the next n lines contains two integers ai and bi (1 β€ ai, bi β€ 109) β the results of the i-th participant (the number of milliseconds he needs to cover the semifinals distance) of the first and second semifinals, correspondingly. All results are distinct. Sequences a1, a2, ..., an and b1, b2, ..., bn are sorted in ascending order, i.e. in the order the participants finished in the corresponding semifinal.
Output
Print two strings consisting of n characters, each equals either "0" or "1". The first line should correspond to the participants of the first semifinal, the second line should correspond to the participants of the second semifinal. The i-th character in the j-th line should equal "1" if the i-th participant of the j-th semifinal has any chances to advance to the finals, otherwise it should equal a "0".
Examples
Input
4
9840 9920
9860 9980
9930 10020
10040 10090
Output
1110
1100
Input
4
9900 9850
9940 9930
10000 10020
10060 10110
Output
1100
1100
Note
Consider the first sample. Each semifinal has 4 participants. The results of the first semifinal are 9840, 9860, 9930, 10040. The results of the second semifinal are 9920, 9980, 10020, 10090.
* If k = 0, the finalists are determined by the time only, so players 9840, 9860, 9920 and 9930 advance to the finals.
* If k = 1, the winners from both semifinals move to the finals (with results 9840 and 9920), and the other places are determined by the time (these places go to the sportsmen who run the distance in 9860 and 9930 milliseconds).
* If k = 2, then first and second places advance from each seminfial, these are participants with results 9840, 9860, 9920 and 9980 milliseconds. | instruction | 0 | 89,623 | 17 | 179,246 |
Tags: implementation, sortings
Correct Solution:
```
n=int(input())
k=[]
a=[]
b=[]
for i in range(n):
A,B=map(int, input().split())
a.append(A)
b.append(B)
la=0
lb=0
aind=[0]*n
bind=[0]*n
for i in range(n):
if a[la]<b[lb]:
aind[la]=1
la+=1
else:
bind[lb]=1
lb+=1
for i in range(n//2):
aind[i]=1
bind[i]=1
for i in aind:
print(i,end='')
print('')
for i in bind:
print(i,end='')
``` | output | 1 | 89,623 | 17 | 179,247 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Two semifinals have just been in the running tournament. Each semifinal had n participants. There are n participants advancing to the finals, they are chosen as follows: from each semifinal, we choose k people (0 β€ 2k β€ n) who showed the best result in their semifinals and all other places in the finals go to the people who haven't ranked in the top k in their semifinal but got to the n - 2k of the best among the others.
The tournament organizers hasn't yet determined the k value, so the participants want to know who else has any chance to get to the finals and who can go home.
Input
The first line contains a single integer n (1 β€ n β€ 105) β the number of participants in each semifinal.
Each of the next n lines contains two integers ai and bi (1 β€ ai, bi β€ 109) β the results of the i-th participant (the number of milliseconds he needs to cover the semifinals distance) of the first and second semifinals, correspondingly. All results are distinct. Sequences a1, a2, ..., an and b1, b2, ..., bn are sorted in ascending order, i.e. in the order the participants finished in the corresponding semifinal.
Output
Print two strings consisting of n characters, each equals either "0" or "1". The first line should correspond to the participants of the first semifinal, the second line should correspond to the participants of the second semifinal. The i-th character in the j-th line should equal "1" if the i-th participant of the j-th semifinal has any chances to advance to the finals, otherwise it should equal a "0".
Examples
Input
4
9840 9920
9860 9980
9930 10020
10040 10090
Output
1110
1100
Input
4
9900 9850
9940 9930
10000 10020
10060 10110
Output
1100
1100
Note
Consider the first sample. Each semifinal has 4 participants. The results of the first semifinal are 9840, 9860, 9930, 10040. The results of the second semifinal are 9920, 9980, 10020, 10090.
* If k = 0, the finalists are determined by the time only, so players 9840, 9860, 9920 and 9930 advance to the finals.
* If k = 1, the winners from both semifinals move to the finals (with results 9840 and 9920), and the other places are determined by the time (these places go to the sportsmen who run the distance in 9860 and 9930 milliseconds).
* If k = 2, then first and second places advance from each seminfial, these are participants with results 9840, 9860, 9920 and 9980 milliseconds. | instruction | 0 | 89,624 | 17 | 179,248 |
Tags: implementation, sortings
Correct Solution:
```
n = int(input())
a = [None] * n
b = [None] * n
for i in range(n):
a[i], b[i] = map(int, input().split())
for i in range(n):
if i < n // 2 or a[i] < b[n - i - 1]:
print(1, end = "")
else:
print(0, end = "")
print()
for i in range(n):
if i < n // 2 or b[i] < a[n - i - 1]:
print(1, end = "")
else:
print(0, end = "")
``` | output | 1 | 89,624 | 17 | 179,249 |
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