TnT/Multi_CodeNet4Repair huggingface dataset:
Problem: There are 3N participants in _AtCoder Group Contest_. The _strength_ of the i-th participant is represented by an integer a_i. They will form N teams, each consisting of three participants. No participant may belong to multiple teams.  The strength of a team is defined as the second largest strength among its members. For example, a team of participants of strength 1, 5, 2 has a strength 2, and a team of three participants of strength 3, 2, 3 has a strength 3.  Find the maximum possible sum of the strengths of N teams.

Submission attempts (in order):
n = int(input())
a = sorted(list(map(int, input().split())))
print(a)

print(sum(a[n:2*n]))

n = int(input())
a = sorted(list(map(int, input().split())))
print(sum(a[n::2]))

Verdicts: Wrong Answer

Accepted
Problem: There are 3N participants in _AtCoder Group Contest_. The _strength_ of the i-th participant is represented by an integer a_i. They will form N teams, each consisting of three participants. No participant may belong to multiple teams.  The strength of a team is defined as the second largest strength among its members. For example, a team of participants of strength 1, 5, 2 has a strength 2, and a team of three participants of strength 3, 2, 3 has a strength 3.  Find the maximum possible sum of the strengths of N teams.

Submission attempts (in order):
a =int(input())
x =list(map(int, input().split()))
x.sort()
s=0
for i in range(0,(a+1)//3,1):
  s+=x[i+1]
print(s)


a =int(input())
x =list(map(int, input().split()))
x.sort()
s=0
for i in reversed(range(0,a//3,1):
  s+=x[i+1]
print(s)


a =int(input())
x =list(map(int, input().split()))
x.sort()
s=0
for i in reversed(range(1,(a+1)//3,1):
  s+=x[i]
print(s)


a =int(input())
x =list(map(int, input().split()))
x.sort()
s=0
for I in reversed(range(1,a//3,1):
  s+=x[i]
print(s)

a =int(input())
x =list(map(int, input().split()))
x.sort()
s=0
for i in range(0,a//3,1):
  s+=x[i+1]
print(s)



a =int(input())
x =list(map(int, input().split()))
x.sort()
x =x[::-1]
s=0
for i in range(1,a//3+1,1):
  s+=x[i]
print(s)


n = int(input())
A = list(map(int,input().split()))
A.sort()

print(sum(A[n:3*n:2]))

Verdicts: Wrong Answer

Runtime Error

Runtime Error

Runtime Error

Wrong Answer

Wrong Answer

Accepted
Problem: There are 3N participants in _AtCoder Group Contest_. The _strength_ of the i-th participant is represented by an integer a_i. They will form N teams, each consisting of three participants. No participant may belong to multiple teams.  The strength of a team is defined as the second largest strength among its members. For example, a team of participants of strength 1, 5, 2 has a strength 2, and a team of three participants of strength 3, 2, 3 has a strength 3.  Find the maximum possible sum of the strengths of N teams.

Submission attempts (in order):
N = int(input())
li = list(map(int,input().split()))
li.sort(reverse = True)
total = []
for i in li[1:3 * N:3]:
  total.append(i)
print(sum(total))

N = int(input())
li = list(map(int,input().split()))
li.sort(reverse = True)
total = []
for i in li[1:2 * N:2]:
  total.append(i)
print(sum(total))

Verdicts: Wrong Answer

Accepted
Problem: There are 3N participants in _AtCoder Group Contest_. The _strength_ of the i-th participant is represented by an integer a_i. They will form N teams, each consisting of three participants. No participant may belong to multiple teams.  The strength of a team is defined as the second largest strength among its members. For example, a team of participants of strength 1, 5, 2 has a strength 2, and a team of three participants of strength 3, 2, 3 has a strength 3.  Find the maximum possible sum of the strengths of N teams.

Submission attempts (in order):
N = int(input)

a_s = input().split()

for i in ragne(3*N):
  a_s[i] = int(a_s[i])
  
a_s = sorted(a_s)

a_1 = a_s[:N]
a_2 = a_s[N:]

ans = sum(a_2[0::2])

print(ans)

N = int(input())

a_s = input().split()

for i in ragne(3*N):
  a_s[i] = int(a_s[i])
  
a_s = sorted(a_s)

a_1 = a_s[:N]
a_2 = a_s[N:]

ans = sum(a_2[0::2])

print(ans)


N = int(input())

a_s = input().split()

for i in range(3*N):
  a_s[i] = int(a_s[i])
  
a_s = sorted(a_s)

a_1 = a_s[:N]
a_2 = a_s[N:]

ans = sum(a_2[0::2])

print(ans)


Verdicts: Runtime Error

Runtime Error

Accepted
Problem: There are 3N participants in _AtCoder Group Contest_. The _strength_ of the i-th participant is represented by an integer a_i. They will form N teams, each consisting of three participants. No participant may belong to multiple teams.  The strength of a team is defined as the second largest strength among its members. For example, a team of participants of strength 1, 5, 2 has a strength 2, and a team of three participants of strength 3, 2, 3 has a strength 3.  Find the maximum possible sum of the strengths of N teams.

Submission attempts (in order):
N = int(input())
A = sorted(list(map(int, input().split())))
print(A)
ans = 0
for i in range(2 * N):
    if i % 2 == 1:
        print(3 * N - i - 1)
        ans += A[3 * N - i - 1]

print(ans)

N = int(input())
A = sorted(list(map(int, input().split())))[N:]
print(A)
ans = 0
for i in range(2 * N):
    if i % N == 0:
        ans += A[i]

print(ans)

N = int(input())
A = sorted(list(map(int, input().split())))

ans = 0
for i in range(2 * N):
    if i % 2 == 1:
        ans += A[3 * N - i - 1]

print(ans)

Verdicts: Wrong Answer

Wrong Answer

Accepted
Problem: There are 3N participants in _AtCoder Group Contest_. The _strength_ of the i-th participant is represented by an integer a_i. They will form N teams, each consisting of three participants. No participant may belong to multiple teams.  The strength of a team is defined as the second largest strength among its members. For example, a team of participants of strength 1, 5, 2 has a strength 2, and a team of three participants of strength 3, 2, 3 has a strength 3.  Find the maximum possible sum of the strengths of N teams.

Submission attempts (in order):
n = int(input())
xs = [int(c) for c in input().strip().split()]
xs.sort(reverse=True)
ans = 0
for i in range(n):
  ans += xs[2 * i - 1]
print(ans)

n = int(input())
xs = [int(c) for c in input().strip().split()]

ans = 0
for i in range(n):
  ans += xs[2 * i - 1]
print(ans)

n = int(input())
xs = [int(c) for c in input().strip().split()]
xs.sort(reverse=True)
ans = 0
for i in range(n):
    ans += xs[2 * i + 1]
print(ans)


Verdicts: Wrong Answer

Wrong Answer

Accepted
Problem: There are 3N participants in _AtCoder Group Contest_. The _strength_ of the i-th participant is represented by an integer a_i. They will form N teams, each consisting of three participants. No participant may belong to multiple teams.  The strength of a team is defined as the second largest strength among its members. For example, a team of participants of strength 1, 5, 2 has a strength 2, and a team of three participants of strength 3, 2, 3 has a strength 3.  Find the maximum possible sum of the strengths of N teams.

Submission attempts (in order):
N = int(input())
a = list(map(int, input().split()))

print(sum(a[1:N*2:3]))


N = int(input())
a = list(map(int, input().split()))

print(a[1:N*2:3])


N = int(input())
a = list(map(int, input().split()))

a.sort(reverse=True)
print(sum(a[1:N*2+1:2]))


Verdicts: Wrong Answer

Wrong Answer

Accepted
Problem: There are 3N participants in _AtCoder Group Contest_. The _strength_ of the i-th participant is represented by an integer a_i. They will form N teams, each consisting of three participants. No participant may belong to multiple teams.  The strength of a team is defined as the second largest strength among its members. For example, a team of participants of strength 1, 5, 2 has a strength 2, and a team of three participants of strength 3, 2, 3 has a strength 3.  Find the maximum possible sum of the strengths of N teams.

Submission attempts (in order):
if __name__ == '__main__':
    n = int(input())
    alist = [int(i) for i in input().split()]
    alist.sort(reverse=True)
    print(alist)
    st = 0
    for i in range(n):
        st += alist[2*i+1]
    print(st)


if __name__ == '__main__':
    n = int(input())
    alist = [int(i) for i in input().split()]
    alist.sort(reverse=True)
    st = 0
    for i in range(n):
        st += alist[2*i+1]
    print(st)


Verdicts: Wrong Answer

Accepted
Problem: There are 3N participants in _AtCoder Group Contest_. The _strength_ of the i-th participant is represented by an integer a_i. They will form N teams, each consisting of three participants. No participant may belong to multiple teams.  The strength of a team is defined as the second largest strength among its members. For example, a team of participants of strength 1, 5, 2 has a strength 2, and a team of three participants of strength 3, 2, 3 has a strength 3.  Find the maximum possible sum of the strengths of N teams.

Submission attempts (in order):
N=int(input())
a=list(map(int,input().split()))
b=sorted(a)
ans1=sum(b[N:2*N])

ans2=0
for i in range(N):
    ans2 += b[i*3+1]

ans3=0
for i in range(N):
    ans3 += b[i*2+4]

print(max(ans1,ans2,ans3))

N=int(input())
a=list(map(int,input().split()))
b=sorted(a)

ans3=0
for i in range(N):
    ans3 += b[i*2+N]

print(ans3)


Verdicts: Runtime Error

Accepted
Problem: There are 3N participants in _AtCoder Group Contest_. The _strength_ of the i-th participant is represented by an integer a_i. They will form N teams, each consisting of three participants. No participant may belong to multiple teams.  The strength of a team is defined as the second largest strength among its members. For example, a team of participants of strength 1, 5, 2 has a strength 2, and a team of three participants of strength 3, 2, 3 has a strength 3.  Find the maximum possible sum of the strengths of N teams.

Submission attempts (in order):
N = int(input())
a = [int(i) for i in input().split()]
s_sort = sorted(a, reverse=True)
score = 0
for i in range(1, 3N, 2):
    score += s_sort[i]
print(score)

N = int(input())
a = [int(i) for i in input().split()]
s_sort = sorted(a, reverse=True)
score = 0
for i in range(1, 3*N, 2):
    score += s_sort[i]
print(score)

N = int(input())
a = sorted([int(i) for i in input().split()], reverse=True)

ans = []
for i in range(1, len(a), 2):
    if len(ans) < N:
        ans.append(a[i])

print(sum(ans))

Verdicts: Runtime Error

Wrong Answer

Accepted
Problem: There are 3N participants in _AtCoder Group Contest_. The _strength_ of the i-th participant is represented by an integer a_i. They will form N teams, each consisting of three participants. No participant may belong to multiple teams.  The strength of a team is defined as the second largest strength among its members. For example, a team of participants of strength 1, 5, 2 has a strength 2, and a team of three participants of strength 3, 2, 3 has a strength 3.  Find the maximum possible sum of the strengths of N teams.

Submission attempts (in order):
n = int(input())
a = list(map(int,input().split()))

a = sorted(a)

ans = 0

for i in range(3,n*3):
	if i%2 == 1:
		ans = ans + a[i]
	
print(ans)


n = int(input())
a = list(map(int,input().split()))

a = sorted(a)

ans = 0

for i in range(n,n*3,2):
		ans = ans + a[i]
	
print(ans)


Verdicts: Wrong Answer

Accepted
Problem: There are 3N participants in _AtCoder Group Contest_. The _strength_ of the i-th participant is represented by an integer a_i. They will form N teams, each consisting of three participants. No participant may belong to multiple teams.  The strength of a team is defined as the second largest strength among its members. For example, a team of participants of strength 1, 5, 2 has a strength 2, and a team of three participants of strength 3, 2, 3 has a strength 3.  Find the maximum possible sum of the strengths of N teams.

Submission attempts (in order):
n = int(input())
a = map(int, input().split())

a = reversed(sorted(a))
m = 0

for i in range(n, 2 * n):
    m += a[i]

print(m)

n = int(input())
a = map(int, input().split())

a = sorted(a, reverse=True)
m = 0

for i in range(1, 2 * n, 2):
    m += a[i]

print(m)

Verdicts: Runtime Error

Accepted
Problem: There are 3N participants in _AtCoder Group Contest_. The _strength_ of the i-th participant is represented by an integer a_i. They will form N teams, each consisting of three participants. No participant may belong to multiple teams.  The strength of a team is defined as the second largest strength among its members. For example, a team of participants of strength 1, 5, 2 has a strength 2, and a team of three participants of strength 3, 2, 3 has a strength 3.  Find the maximum possible sum of the strengths of N teams.

Submission attempts (in order):
#bin/usr/env python3

N = int(input())
a = input()
x = 0
b = []

a = a.split()
for i in a:
	b.append(int(i))
b.sort()
for i in range(N):
	x += int(b[2*(i+1)+1])
print(x)

#bin/usr/env python3

N = int(input())
a = input()
x = 0
b = []

a = a.split()
for i in a:
	b.append(int(i))
b.sort()
for i in range(N):
	c = b[-(2*(i+1))]
	x += int(c)
print(x)

Verdicts: Wrong Answer

Accepted
Problem: There are 3N participants in _AtCoder Group Contest_. The _strength_ of the i-th participant is represented by an integer a_i. They will form N teams, each consisting of three participants. No participant may belong to multiple teams.  The strength of a team is defined as the second largest strength among its members. For example, a team of participants of strength 1, 5, 2 has a strength 2, and a team of three participants of strength 3, 2, 3 has a strength 3.  Find the maximum possible sum of the strengths of N teams.

Submission attempts (in order):
N = int(input())
A = list(map(int, input().split()))
A.sort(reverse=True)
k = 0
for i in range(1, len(A)-N, 2):
  k += A[i]
  print(k)


N = int(input())
A = list(map(int, input().split()))
A.sort(reverse=True)
k = 0
for i in range(1, len(A)-N, 2):
  k += A[i]
print(k)

Verdicts: Wrong Answer

Accepted
Problem: There are 3N participants in _AtCoder Group Contest_. The _strength_ of the i-th participant is represented by an integer a_i. They will form N teams, each consisting of three participants. No participant may belong to multiple teams.  The strength of a team is defined as the second largest strength among its members. For example, a team of participants of strength 1, 5, 2 has a strength 2, and a team of three participants of strength 3, 2, 3 has a strength 3.  Find the maximum possible sum of the strengths of N teams.

Submission attempts (in order):
n = int(input())
a = list(map(int,input().split()))
a.sort(reverse=True)
x,y=[],[]
for i in range(n*3):
    if i%2==0:
        x.append(a[i])
    else:
        y.append(a[i])
print(x[1],y[1])

n = int(input())
a = list(map(int,input().split()))
a.sort(reverse=True)
print(sum(a[1:n*2:2]))

Verdicts: Wrong Answer

Accepted
Problem: There are 3N participants in _AtCoder Group Contest_. The _strength_ of the i-th participant is represented by an integer a_i. They will form N teams, each consisting of three participants. No participant may belong to multiple teams.  The strength of a team is defined as the second largest strength among its members. For example, a team of participants of strength 1, 5, 2 has a strength 2, and a team of three participants of strength 3, 2, 3 has a strength 3.  Find the maximum possible sum of the strengths of N teams.

Submission attempts (in order):
n = int(input())
a = list(map(int,input().split()))
print(sum(sorted(a,reverse=True)[:n*2:2]))

n = int(input())
a = list(map(int,input().split()))
print(sum(sorted(a,reverse=True)[1:n*2:2]))

Verdicts: Wrong Answer

Accepted
Problem: There are 3N participants in _AtCoder Group Contest_. The _strength_ of the i-th participant is represented by an integer a_i. They will form N teams, each consisting of three participants. No participant may belong to multiple teams.  The strength of a team is defined as the second largest strength among its members. For example, a team of participants of strength 1, 5, 2 has a strength 2, and a team of three participants of strength 3, 2, 3 has a strength 3.  Find the maximum possible sum of the strengths of N teams.

Submission attempts (in order):
# -*- coding: UTF-8 -*-
N = int(input())
d = list(map(int,input().split(" ")))
d.sort(reverse = True)
_ans = []
for i in range(N-1):
    _ans.append(d[2*i+1])
print(sum(_ans))


# -*- coding: UTF-8 -*-
N = int(input())
d = list(map(int,input().split(" ")))
d.sort(reverse = True)
_ans = []
for i in range(N):
    _ans.append(d[2*i+1])
print(sum(_ans))

Verdicts: Wrong Answer

Accepted
Problem: There are 3N participants in _AtCoder Group Contest_. The _strength_ of the i-th participant is represented by an integer a_i. They will form N teams, each consisting of three participants. No participant may belong to multiple teams.  The strength of a team is defined as the second largest strength among its members. For example, a team of participants of strength 1, 5, 2 has a strength 2, and a team of three participants of strength 3, 2, 3 has a strength 3.  Find the maximum possible sum of the strengths of N teams.

Submission attempts (in order):
n=int(input())
a=list(map(int,input().split()))
a.sort(reverse=True)
print(sum(a[1:2*n+2:2]))

n=int(input())
a=list(map(int,input().split()))
a.sort(reverse=True)
print(sum[1:2*n+2:2])

n=int(input())
a=list(map(int,input().split()))
a.sort()
ans=0
for i in range(3*n-2,2*n,-2):
  ans+=a[i]
print(ans)

n=int(input())
a=list(map(int,input().split()))
ans=0
for i in range(3*n-2,2*n,-2):
  ans+=a[i]
print(ans)

n=int(input())
a=list(map(int,input().split()))
a.sort(reverse=True)
print(sum(a[1:2*n:2]))

Verdicts: Wrong Answer

Runtime Error

Wrong Answer

Wrong Answer

Accepted
Problem: There are 3N participants in _AtCoder Group Contest_. The _strength_ of the i-th participant is represented by an integer a_i. They will form N teams, each consisting of three participants. No participant may belong to multiple teams.  The strength of a team is defined as the second largest strength among its members. For example, a team of participants of strength 1, 5, 2 has a strength 2, and a team of three participants of strength 3, 2, 3 has a strength 3.  Find the maximum possible sum of the strengths of N teams.

Submission attempts (in order):
N = int(input())
a = list(map(int,input().split(" ")))
a.sort()
print(a)
total=0
for i in range(N):
    total+=a[len(a)-i-2]
print(total)


N = int(input())
a = list(map(int,input().split(" ")))
a.sort(reverse=True)
#print(a)
total=0
for i in range(1,N+1):
    total+=a[2*i-1]
print(total)



Verdicts: Wrong Answer

Accepted
Problem: There are 3N participants in _AtCoder Group Contest_. The _strength_ of the i-th participant is represented by an integer a_i. They will form N teams, each consisting of three participants. No participant may belong to multiple teams.  The strength of a team is defined as the second largest strength among its members. For example, a team of participants of strength 1, 5, 2 has a strength 2, and a team of three participants of strength 3, 2, 3 has a strength 3.  Find the maximum possible sum of the strengths of N teams.

Submission attempts (in order):
from typing import Tuple, List

def main()->None:
    
    N, x = map(int, input().split())
    a:List[int] = list(map(int, input().split()))
    sorted_a:List[int] = sorted(a) 
    count:int = 0 
    for d in sorted_a:
        x -= d
        if x < 0: 
            break
        count += 1
    
    
    if x > 0:
        print(count-1)
    else: 
        print(count)

if __name__ == "__main__":
    main()



def main()->None:
    
    
    
    N = int(input()) 
    a = list(map(int, input().split()))
    sorted_a = sorted(a)
    
    sum_a = 0 
    range_N = range(N)
    for i in range(N):
        sum_a += sorted_a[3*N - 2*i - 2]
    print(sum_a)


if __name__ == "__main__":
    main()

Verdicts: Runtime Error

Accepted
Problem: There are 3N participants in _AtCoder Group Contest_. The _strength_ of the i-th participant is represented by an integer a_i. They will form N teams, each consisting of three participants. No participant may belong to multiple teams.  The strength of a team is defined as the second largest strength among its members. For example, a team of participants of strength 1, 5, 2 has a strength 2, and a team of three participants of strength 3, 2, 3 has a strength 3.  Find the maximum possible sum of the strengths of N teams.

Submission attempts (in order):
N=int(input())
A=list(map(int,input().split()))
A.sort()
Ans=0
for i in range(N):
    Ans+=A[3*N-1-2*i]
print(Ans)

N=int(input())
A=list(map(int,input().split()))
A.sort()
Ans=0
for i in range(N):
    Ans+=A[3*N-2-2*i]
print(Ans)

Verdicts: Wrong Answer

Accepted
Problem: There are 3N participants in _AtCoder Group Contest_. The _strength_ of the i-th participant is represented by an integer a_i. They will form N teams, each consisting of three participants. No participant may belong to multiple teams.  The strength of a team is defined as the second largest strength among its members. For example, a team of participants of strength 1, 5, 2 has a strength 2, and a team of three participants of strength 3, 2, 3 has a strength 3.  Find the maximum possible sum of the strengths of N teams.

Submission attempts (in order):
n=int(input())
l=list(map(int,input().split()))
if ' '.join([str(x) for x in l])=='5 2 8 5 1 5':
  exit()
l=sorted(l,key=lambda x:x,reverse=True)
s=0
ind=1
while True:
  s+=l[ind]
  ind+=2
  if ind>2*n:
    break
print(s)

n=int(input())
l=list(map(int,input().split()))
l=sorted(l,key=lambda x:x,reverse=True)
s=0
ind=1
while True:
  s+=l[ind]
  ind+=2
  if ind>2*n:
    break
print(s)

Verdicts: Wrong Answer

Accepted
Problem: There are 3N participants in _AtCoder Group Contest_. The _strength_ of the i-th participant is represented by an integer a_i. They will form N teams, each consisting of three participants. No participant may belong to multiple teams.  The strength of a team is defined as the second largest strength among its members. For example, a team of participants of strength 1, 5, 2 has a strength 2, and a team of three participants of strength 3, 2, 3 has a strength 3.  Find the maximum possible sum of the strengths of N teams.

Submission attempts (in order):
n = int(input())
a = sorted(list(map(int, input().split())), reverse=True)
x = len(a) // n
l = []
for i in range(n):
    for j in range(x):
        ll.append(a[i+j::x-1])
      
ans = 0
for i in range(n):
    ans += l[i][1]
print(ans)

n = int(input())
a = sorted(list(map(int, input().split())), reverse=True)
x = len(a) // n
l = []
for i in range(n):
    l.append(a[i*x::x-1])
      
ans = 0
for i in range(n):
    ans += l[i][1]
print(ans)

n = int(input())
a = sorted(list(map(int, input().split())), reverse=True)
x = len(a) // n
l = []
for i in range(n):
    tmp = a[i::x-1]
    l.append(tmp[:x])
      
ans = 0
for i in range(n):
    ans += l[i][1]
print(ans)
print(l)

n = int(input())
a = sorted(list(map(int, input().split())), reverse=True)
x = len(a) // n
l = []
for i in range(n):
    tmp = a[i::x-1]
    l.append(tmp[:x])
      
ans = 0
for i in range(n):
    ans += l[i][1]
print(ans)
print(l)

n = int(input())
a = sorted(map(int, input().split()))
print(sum(a[n::2]))

Verdicts: Runtime Error

Wrong Answer

Wrong Answer

Wrong Answer

Accepted
Problem: There are 3N participants in _AtCoder Group Contest_. The _strength_ of the i-th participant is represented by an integer a_i. They will form N teams, each consisting of three participants. No participant may belong to multiple teams.  The strength of a team is defined as the second largest strength among its members. For example, a team of participants of strength 1, 5, 2 has a strength 2, and a team of three participants of strength 3, 2, 3 has a strength 3.  Find the maximum possible sum of the strengths of N teams.

Submission attempts (in order):

N = int(input())


A_list = list(map(int, input().split()))

A_list = list(reversed(sorted(A_list)))
print(A_list)



ans  = 0
for i in range(1,2*N,2):
  #print(i)
  #print(A_list[i])
  ans = ans + A_list[i]
print(ans)


N = int(input())


A_list = list(map(int, input().split()))

A_list = list(reversed(sorted(A_list)))
#print(A_list)



ans  = 0
for i in range(1,2*N,2):
  ans = ans + A_list[i]
print(ans)

Verdicts: Wrong Answer

Accepted
Problem: There are 3N participants in _AtCoder Group Contest_. The _strength_ of the i-th participant is represented by an integer a_i. They will form N teams, each consisting of three participants. No participant may belong to multiple teams.  The strength of a team is defined as the second largest strength among its members. For example, a team of participants of strength 1, 5, 2 has a strength 2, and a team of three participants of strength 3, 2, 3 has a strength 3.  Find the maximum possible sum of the strengths of N teams.

Submission attempts (in order):
print(sum(sorted(map(int,input().split()))[::-1][1:2*int(input()):2]))

N=int(input());print(sum(sorted(map(int,input().split()))[-2:N-1:-2]))

Verdicts: Runtime Error

Accepted
Problem: There are 3N participants in _AtCoder Group Contest_. The _strength_ of the i-th participant is represented by an integer a_i. They will form N teams, each consisting of three participants. No participant may belong to multiple teams.  The strength of a team is defined as the second largest strength among its members. For example, a team of participants of strength 1, 5, 2 has a strength 2, and a team of three participants of strength 3, 2, 3 has a strength 3.  Find the maximum possible sum of the strengths of N teams.

Submission attempts (in order):
N = int(input())
A = list(map(int, input().split()))

A.sort()
total = 0
for i in range(N):
	total = A[N + 1 + 2*i]

print(total)

N = int(input())
A = list(map(int, input().split()))

A.sort()
total = 0
for i in range(N):
  #print(N + 1 + 2*i)
  total += A[N + 2*i]

print(total)

Verdicts: Wrong Answer

Accepted
Problem: There are 3N participants in _AtCoder Group Contest_. The _strength_ of the i-th participant is represented by an integer a_i. They will form N teams, each consisting of three participants. No participant may belong to multiple teams.  The strength of a team is defined as the second largest strength among its members. For example, a team of participants of strength 1, 5, 2 has a strength 2, and a team of three participants of strength 3, 2, 3 has a strength 3.  Find the maximum possible sum of the strengths of N teams.

Submission attempts (in order):
n = int(input())




# m = [int(input()) for _ in range(N)]
a = list(map(int, input().split()))

a.sort()


print(sum(a[::-1][n::2]))

n = int(input())




# m = [int(input()) for _ in range(N)]
a = list(map(int, input().split()))

a.sort()


print(sum(a[::-1][n-1::2]))

n = int(input())




# m = [int(input()) for _ in range(N)]
a = list(map(int, input().split()))

a.sort()


print(sum(a[::-1][1:2*n:2]))

Verdicts: Wrong Answer

Wrong Answer

Accepted
Problem: There are 3N participants in _AtCoder Group Contest_. The _strength_ of the i-th participant is represented by an integer a_i. They will form N teams, each consisting of three participants. No participant may belong to multiple teams.  The strength of a team is defined as the second largest strength among its members. For example, a team of participants of strength 1, 5, 2 has a strength 2, and a team of three participants of strength 3, 2, 3 has a strength 3.  Find the maximum possible sum of the strengths of N teams.

Submission attempts (in order):

def f(n):
    if n == 1:
        return []
    if n % 2 == 0:
        r = f(n//2)
        k = len(r)
        return r + [k+1]
    if n % 2 == 1:
        r = f(n - 1)
        k = len(r)
        return [k+1] + r

n = int(input())
r = f(n+1)
k = len(r)
res = r + list(range(1, k+1))
print(len(res))
for x in res:
    print(x, end=' ')
print()



n = int (input())
a = list(map(int, input().split()))
a.sort()
res = sum(a[n::2])
print(res)

Verdicts: Wrong Answer

Accepted
Problem: There are 3N participants in _AtCoder Group Contest_. The _strength_ of the i-th participant is represented by an integer a_i. They will form N teams, each consisting of three participants. No participant may belong to multiple teams.  The strength of a team is defined as the second largest strength among its members. For example, a team of participants of strength 1, 5, 2 has a strength 2, and a team of three participants of strength 3, 2, 3 has a strength 3.  Find the maximum possible sum of the strengths of N teams.

Submission attempts (in order):
N = int(input())
a = list(map(int, input().split()))
a.sort()
score = 0

for i in range(1, N*3, 3):
    score += a[i]

print(score)

N = int(input())
a = list(map(int, input().split()))
a.sort()
score = 0

for i in range(N, N*3, 2):
    score += a[i]

print(score)

Verdicts: Wrong Answer

Accepted
Problem: Squid loves painting vertices in graphs.  There is a simple undirected graph consisting of N vertices numbered 1 through N, and M edges. Initially, all the vertices are painted in color 0. The i-th edge bidirectionally connects two vertices a_i and b_i. The length of every edge is 1.  Squid performed Q operations on this graph. In the i-th operation, he repaints all the vertices within a distance of d_i from vertex v_i, in color c_i.  Find the color of each vertex after the Q operations.

Submission attempts (in order):
import numpy as np

def dp(v, d, c):
    if res[v-1] == 0:
        res[v-1] = c

    if d == 0:
        return None

    if np.all(res > 0):
        return None



    for e in dic.get(v, []):
        dp(e, d-1, c)


N, M = tuple(map(int, input().split(' ')))

dic = {}

for m in range(M):
    a, b = tuple(map(int, input().split(' ')))
    dic[a] = dic.get(a, []) + [b]
    dic[b] = dic.get(b, []) + [a]

Q = int(input())

vdc = []

for q in range(Q):
    vdc.append(tuple(map(int, input().split(' '))))

vdc.reverse()

res = np.zeros(N)

for q in range(Q):
    v, d, c = vdc[q]
    dp(v, d, c)

print(res)


import numpy as np

def dp(v, d, c):
    if memo[v-1][d]:
        return

    memo[v-1][d] = True
    if d == 0:
        res[v-1] = str(c)
        return

    dp(v, d-1, c)
    for e in dic[v-1]:
        dp(e, d-1, c)


N, M = [int(e) for e in input().split(' ')]

dic = [[] for _ in range(N)]
memo = [[False]*11 for _ in range(N)]

for m in range(M):
    a, b = [int(e) for e in input().split(' ')]
    dic[a-1].append(b)
    dic[b-1].append(a)

Q = int(input())


vdc = [[int(e) for e in input().split(' ')] for _ in range(Q)]

res = ['0']*N

[dp(v, d, c) for v, d, c in vdc[::-1]]


print('\n'.join(res))


Verdicts: Wrong Answer

Accepted
Problem: Squid loves painting vertices in graphs.  There is a simple undirected graph consisting of N vertices numbered 1 through N, and M edges. Initially, all the vertices are painted in color 0. The i-th edge bidirectionally connects two vertices a_i and b_i. The length of every edge is 1.  Squid performed Q operations on this graph. In the i-th operation, he repaints all the vertices within a distance of d_i from vertex v_i, in color c_i.  Find the color of each vertex after the Q operations.

Submission attempts (in order):
from collections import deque

G = dict()
N, M = map(int, input().split())
color = [0]*(N+1)
for n in range(1, N+1):
    G[n] = []
for _ in range(M):
    a, b = map(int, input().split())
    G[a].append(b)
    G[b].append(a)
Q = int(input())




data = [list(map(int, input().split())) for _ in range(Q)]
data.reverse()
color_d = [-1]*(N+1)

for v, d, c in data:
    if color_d[v] >= d:
        pass
    # BFS
    deq = deque([v])
    if color[v] != 0:
        color[v] = c
    color_d[v] = d
    while len(deq) > 0:
        s = deq.popleft()
        for t in G[s]:
            if color_d[t] >= color_d[s] - 1:
                pass
            else:
                color_d[t] = color_d[s] - 1
                if color[t] == 0:
                    color[t] = c
                if color_d[t] > 0:
                    deq.append(t)
                
for n in range(1, N+1):
    print(color[n])
                



from collections import deque

G = dict()
N, M = map(int, input().split())
color = [0]*(N+1)
for n in range(1, N+1):
    G[n] = []
for _ in range(M):
    a, b = map(int, input().split())
    G[a].append(b)
    G[b].append(a)
Q = int(input())

for _ in range(Q):
    v, d, c = map(int, input().split())
    # BFS
    dist = dict()
    deq = deque([v])
    color[v] = c
    dist[v] = 0
    while len(deq) > 0:
        s = deq.popleft()
        if dist[s] >= d:
            break
        for t in G[s]:
            if t in dist.keys()
                pass
            else:
                dist[t] = dist[s] + 1
                color[t] = c
                deq.append(t)
                
for n in range(1, N+1):
    print(color[n])
                


from collections import deque

G = dict()
N, M = map(int, input().split())
color = [0]*(N+1)
for n in range(1, N+1):
    G[n] = []
for _ in range(M):
    a, b = map(int, input().split())
    G[a].append(b)
    G[b].append(a)
Q = int(input())




data = [list(map(int, input().split())) for _ in range(Q)]
data.reverse()
color_d = [-1]*(N+1)

for v, d, c in data:
    if color_d[v] >= d:
        pass
    # BFS
    deq = deque([v])
    if color[v] == 0:
        color[v] = c
    color_d[v] = d
    while len(deq) > 0:
        s = deq.popleft()
        for t in G[s]:
            if color_d[t] >= color_d[s] - 1:
                pass
            else:
                color_d[t] = color_d[s] - 1
                if color[t] == 0:
                    color[t] = c
                if color_d[t] > 0:
                    deq.append(t)
                
for n in range(1, N+1):
    print(color[n])
                



Verdicts: Wrong Answer

Runtime Error

Accepted
Problem: Squid loves painting vertices in graphs.  There is a simple undirected graph consisting of N vertices numbered 1 through N, and M edges. Initially, all the vertices are painted in color 0. The i-th edge bidirectionally connects two vertices a_i and b_i. The length of every edge is 1.  Squid performed Q operations on this graph. In the i-th operation, he repaints all the vertices within a distance of d_i from vertex v_i, in color c_i.  Find the color of each vertex after the Q operations.

Submission attempts (in order):
N, M = map(int, input().split())
G = [[] for i in range(N)]
for i in range(M):
    a, b = map(int, input().split())
    G[a-1].append(b-1)
    G[b-1].append(a-1)

dp = [{-1: -1} for i in range(N)]
mark = {}
Q = int(input())
col = [0]*(Q+1)
col[-1] = 0
C = []
for i in range(Q):
    v, d, c = map(int, input().split())
    col[i] = c
    C.append((v-1, d))

def dfs(v, d, idx):
    if d <= max(dp[v]):
        return
    print(v, d, idx)
    dp[v][d] = idx
    if d:
        for w in G[v]:
            dfs(w, d-1, idx)

for i in range(Q-1, -1, -1):
    v, d = C[i]
    dfs(v, d, i)
print(*(col[max(e.values())] for e in dp), sep='\n')


N, M = map(int, input().split())
G = [[] for i in range(N)]
for i in range(M):
    a, b = map(int, input().split())
    G[a-1].append(b-1)
    G[b-1].append(a-1)

dp = [{-1: -1} for i in range(N)]
mark = {}
Q = int(input())
col = [0]*(Q+1)
col[-1] = 0
C = []
for i in range(Q):
    v, d, c = map(int, input().split())
    col[i] = c
    C.append((v-1, d))

def dfs(v, d, idx):
    if d <= max(dp[v]):
        return
    dp[v][d] = idx
    if d:
        for w in G[v]:
            dfs(w, d-1, idx)

for i in range(Q-1, -1, -1):
    v, d = C[i]
    dfs(v, d, i)
print(*(col[max(e.values())] for e in dp), sep='\n')

Verdicts: Wrong Answer

Accepted
Problem: Squid loves painting vertices in graphs.  There is a simple undirected graph consisting of N vertices numbered 1 through N, and M edges. Initially, all the vertices are painted in color 0. The i-th edge bidirectionally connects two vertices a_i and b_i. The length of every edge is 1.  Squid performed Q operations on this graph. In the i-th operation, he repaints all the vertices within a distance of d_i from vertex v_i, in color c_i.  Find the color of each vertex after the Q operations.

Submission attempts (in order):
from queue import deque


def paint(s, d, c):
    queue = deque([(d, s)])
    while queue:
        remains, v = queue.popleft()
        if painted[v] >= remains:
            continue
        if colors[v] == 0:
            colors[v] = c
        painted[v] = remains
        if remains:
            queue.extend((remains - 1, u) for u in links[v])


n, m = map(int, input().split())
links = [None] + [set() for _ in range(n)]
for _ in range(m):
    a, b = map(int, input().split())
    links[a].add(b)
    links[b].add(a)

q = int(input())
queries = [map(int, input().split()) for _ in range(q)]
colors = [0] * (n + 1)
painted = [0] * (n + 1)
while queries:
    paint(*queries.pop())

for c in colors[1:]:
    print(c)


from queue import deque


def paint(s, d, c):
    queue = deque([(d, s)])
    while queue:
        remains, v = queue.popleft()
        if painted[v] >= remains:
            continue
        if colors[v] == 0:
            colors[v] = c
        painted[v] = remains
        if remains:
            queue.extend((remains - 1, u) for u in links[v])


n, m = map(int, input().split())
links = [None] + [set() for _ in range(n)]
for _ in range(m):
    a, b = map(int, input().split())
    links[a].add(b)
    links[b].add(a)

q = int(input())
queries = [map(int, input().split()) for _ in range(q)]
colors = [0] * (n + 1)
painted = [-1] * (n + 1)
while queries:
    paint(*queries.pop())

for c in colors[1:]:
    print(c)


Verdicts: Wrong Answer

Accepted
Problem: Squid loves painting vertices in graphs.  There is a simple undirected graph consisting of N vertices numbered 1 through N, and M edges. Initially, all the vertices are painted in color 0. The i-th edge bidirectionally connects two vertices a_i and b_i. The length of every edge is 1.  Squid performed Q operations on this graph. In the i-th operation, he repaints all the vertices within a distance of d_i from vertex v_i, in color c_i.  Find the color of each vertex after the Q operations.

Submission attempts (in order):
q = int(input())
node_color = [0 for _ in range(n + 1)]
vdc = [list(map(int, input().split())) for _ in range(q)]

vdc = vdc[::-1]
painted = set()
visited = [[0] * 11 for _ in range(n + 1)]


def dfs(now, dist, color):
    if dist < 0:
        return
    if visited[now][dist]:
        return

    visited[now][dist] += 1
    if not node_color[now]:
        node_color[now] = color

    for x in node[now]:
        dfs(x, dist - 1, color)


for i in vdc:
    v, d, c = i
    dfs(v, d, c)

for m in range(n):
    print(node_color[m + 1])


n, m = map(int, input().split())
node = [[] for _ in range(n + 1)]
for i in range(m):
    a, b = map(int, input().split())
    node[a].append(b)
    node[b].append(a)

q = int(input())
node_color = [0 for _ in range(n + 1)]
vdc = [list(map(int, input().split())) for _ in range(q)]

vdc = vdc[::-1]
painted = set()
visited = [[0] * 11 for _ in range(n + 1)]


def dfs(now, dist, color):
    if dist < 0:
        return
    if visited[now][dist]:
        return

    visited[now][dist] += 1
    if not node_color[now]:
        node_color[now] = color

    for x in node[now]:
        dfs(x, dist - 1, color)


for i in vdc:
    v, d, c = i
    dfs(v, d, c)

for m in range(n):
    print(node_color[m + 1])


Verdicts: Runtime Error

Accepted
Problem: Squid loves painting vertices in graphs.  There is a simple undirected graph consisting of N vertices numbered 1 through N, and M edges. Initially, all the vertices are painted in color 0. The i-th edge bidirectionally connects two vertices a_i and b_i. The length of every edge is 1.  Squid performed Q operations on this graph. In the i-th operation, he repaints all the vertices within a distance of d_i from vertex v_i, in color c_i.  Find the color of each vertex after the Q operations.

Submission attempts (in order):
from collections import deque

n, m = map(int, input().split())
G = [list() for _ in range(n)]
for _ in range(m):
  a, b = map(int, input().split())
  a -= 1; b -= 1
  G[a].append(b)
  G[b].append(a)
Q = int(input())
queries = [tuple(map(int, input().split())) for _ in range(Q)][::-1]
dp = [[Q]*n for _ in range(11)]
for i, (v, d, _) in enumerate(queries):
  dp[v-1][d] = i
for d in range(10, 0, -1):
  for v in range(n):
    if dp[v][d] < Q:
      dp[v][d-1] = min(dp[v][d-1], dp[v][d])
      for x in G[v]:
        dp[x][d-1] = min(dp[x][d-1], dp[v][d])
queries.append((-1, -1, 0))
color = [queries[dp[v][0]][2] for v in range(n)]
print(*color, sep='\n')


from collections import deque

n, m = map(int, input().split())
G = [list() for _ in range(n)]
for _ in range(m):
  a, b = map(int, input().split())
  a -= 1; b -= 1
  G[a].append(b)
  G[b].append(a)
Q = int(input())
queries = [tuple(map(int, input().split())) for _ in range(Q)][::-1]
dp = [[Q]*n for _ in range(n)]
for i, (v, d, _) in enumerate(queries):
  dp[v-1][d] = i
for d in range(10, 0, -1):
  for v in range(n):
    if dp[v][d] < Q:
      dp[v][d-1] = min(dp[v][d-1], dp[v][d])
      for x in G[v]:
        dp[x][d-1] = min(dp[x][d-1], dp[v][d])
queries.append((-1, -1, 0))
color = [queries[dp[v][0]][2] for v in range(n)]
print(*color, sep='\n')

import sys
from collections import deque

readline = sys.stdin.readline

ns = lambda: readline().rstrip()
ni = lambda: int(readline().rstrip())
nm = lambda: map(int, readline().split())
nl = lambda: list(map(int, readline().split()))

def solve():
  n, m = nm()
  G = [list() for _ in range(n)]
  for _ in range(m):
    a, b = nm()
    a -= 1; b -= 1
    G[a].append(b)
    G[b].append(a)
  Q = int(input())
  color = [0]*n
  dist = [-1]*n
  queries = [tuple(nm()) for _ in range(Q)][::-1]
  for v, d, c in queries:
    v -= 1
    q = deque()
    q.append((v, d))
    while q:
      v, d = q.popleft()
      if dist[v] >= d:
        continue
      dist[v] = d
      if not color[v]: color[v] = c
      for x in G[v]:
        if d - 1 > dist[x]:
          q.append((x, d-1))
  print(*color, sep='\n')
  return

solve()


Verdicts: Runtime Error

Runtime Error

Accepted
Problem: Dolphin loves programming contests. Today, he will take part in a contest in AtCoder.   In this country, 24-hour clock is used. For example, 9:00 p.m. is referred to as "21 o'clock".   The current time is A o'clock, and a contest will begin in exactly B hours. When will the contest begin? Answer in 24-hour time.

Submission attempts (in order):
a = int(input())
b = int(input())

if (a+b>=24):
    print((a+b)-24)
else:
    print(a+b)

a, b = map(int, input().split())

A = a+b
B = A%24
print(B)

Verdicts: Runtime Error

Accepted
Problem: Dolphin loves programming contests. Today, he will take part in a contest in AtCoder.   In this country, 24-hour clock is used. For example, 9:00 p.m. is referred to as "21 o'clock".   The current time is A o'clock, and a contest will begin in exactly B hours. When will the contest begin? Answer in 24-hour time.

Submission attempts (in order):
a,b = map(int,input().split())
if a + b < 24:
    print(a + b)
    else:
    print(a + b -24)

a,b = map(int,input().split())
if a + b < 24:
    print(a + b)
else:
    print(a + b -24)

Verdicts: Runtime Error

Accepted
Problem: Dolphin loves programming contests. Today, he will take part in a contest in AtCoder.   In this country, 24-hour clock is used. For example, 9:00 p.m. is referred to as "21 o'clock".   The current time is A o'clock, and a contest will begin in exactly B hours. When will the contest begin? Answer in 24-hour time.

Submission attempts (in order):
A,B = map(int,input().split())
print((A+B)//24)

print(sum(map(int,input().split()))%24)

Verdicts: Wrong Answer

Accepted
Problem: Dolphin loves programming contests. Today, he will take part in a contest in AtCoder.   In this country, 24-hour clock is used. For example, 9:00 p.m. is referred to as "21 o'clock".   The current time is A o'clock, and a contest will begin in exactly B hours. When will the contest begin? Answer in 24-hour time.

Submission attempts (in order):
a, b=map(int, input().split())
2.c=a+b
3.if c<24:
4.  print (c)
5.elif c>=24:
6.  print (c-24)


a, b=map(int, input().split())
c=a+b
if c<24:
  print (c)
elif c>=24:
  print (c-24)


Verdicts: Runtime Error

Accepted
Problem: Dolphin loves programming contests. Today, he will take part in a contest in AtCoder.   In this country, 24-hour clock is used. For example, 9:00 p.m. is referred to as "21 o'clock".   The current time is A o'clock, and a contest will begin in exactly B hours. When will the contest begin? Answer in 24-hour time.

Submission attempts (in order):
n=sum(list(map(int,input().split())));print(n-24 if n>=4 else n)

a,b=map(int,input().split());print((a+b)%24)

Verdicts: Wrong Answer

Accepted
Problem: Dolphin loves programming contests. Today, he will take part in a contest in AtCoder.   In this country, 24-hour clock is used. For example, 9:00 p.m. is referred to as "21 o'clock".   The current time is A o'clock, and a contest will begin in exactly B hours. When will the contest begin? Answer in 24-hour time.

Submission attempts (in order):
#ABC057.B
N,M = map(int,input().split())
L1 = []
L2 = []

for i in range(N):
    a,b = map(int,input().split())
    L1.append([a,b])
for m in range(M):
    c,d = map(int,input().split())
    L2.append([c,d])

for j in range(N):
    L3 = []
    for l in range(M):
        L3.append(abs(L1[j][0]-L2[l][0]) + abs(L1[j][1]-L2[l][1]))
    print(L3.index(min(L3))+1)

#ABC057.B
N,M = map(int(input().split()))
L1 = []
L2 = []

for i in range(N):
    a,b = map(int,input().split())
    L1.append([a,b])
for m in range(M):
    c,d = map(int,input().split())
    L2.append([c,d])

for j in range(N):
    L3 = []
    for l in range(M):
        L3.append(abs(L1[j][0]-L2[l][0]) + abs(L1[j][1]-L2[l][1]))
    print(L3.index(min(L3)+1)

#ABC057.A
A,B = map(int,input().split())
if A+B <24:
    print(A+B)
else:
    print(A+B-24)


Verdicts: Runtime Error

Runtime Error

Accepted
Problem: Dolphin loves programming contests. Today, he will take part in a contest in AtCoder.   In this country, 24-hour clock is used. For example, 9:00 p.m. is referred to as "21 o'clock".   The current time is A o'clock, and a contest will begin in exactly B hours. When will the contest begin? Answer in 24-hour time.

Submission attempts (in order):
a,b=map(int,input().split())

if a+b>=23:
  print(a+b)
else:
  print(a+b-24)

a,b=map(int,input().split())

if a+b<=23:
  print(a+b)
else:
  print(a+b-24)

Verdicts: Wrong Answer

Accepted
Problem: Dolphin loves programming contests. Today, he will take part in a contest in AtCoder.   In this country, 24-hour clock is used. For example, 9:00 p.m. is referred to as "21 o'clock".   The current time is A o'clock, and a contest will begin in exactly B hours. When will the contest begin? Answer in 24-hour time.

Submission attempts (in order):
a,b=map(int,input().split())
if a+b<24:
  print(a+b)
else:
  print(a+b-24

a,b=map(int,input().split())
if a+b<24:
  print(a+b)
else:
  print(a+b-24)


Verdicts: Runtime Error

Accepted
Problem: Dolphin loves programming contests. Today, he will take part in a contest in AtCoder.   In this country, 24-hour clock is used. For example, 9:00 p.m. is referred to as "21 o'clock".   The current time is A o'clock, and a contest will begin in exactly B hours. When will the contest begin? Answer in 24-hour time.

Submission attempts (in order):
import math
N,A,B=map(int,input().split())
v=list(map(int,input().split()))
a=v.sort()[-A:]
print(sum(a)/A)
b=v.count(a[0])
c=a.count(a[0])
if c!=A:
    print(math.factorial(b)//(math.factorial(c)*math.factorial(b-c)))
else:
    l=[]
    i=min([b-c,B-A])
    for k in range(0,i+1):
        l.append(math.factorial(b)//(math.factorial(c+k)*math.factorial(b-c-k)))
    print(sum(l))

a,b=map(int,input().split())
print((a+b)%24)

Verdicts: Runtime Error

Accepted