TnT/Multi_CodeNet4Repair huggingface dataset:
Problem: We have two bottles for holding water.  Bottle 1 can hold up to A milliliters of water, and now it contains B milliliters of water.  Bottle 2 contains C milliliters of water.  We will transfer water from Bottle 2 to Bottle 1 as much as possible.  How much amount of water will remain in Bottle 2?

Submission attempts (in order):
nokori = a - b
ans = c - nokori
if ans < 0:
    ans = 0
print(ans)


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

nokori = a - b
ans = c - nokori
if ans < 0:
    ans = 0
print(ans)


Verdicts: Runtime Error

Accepted
Problem: We have two bottles for holding water.  Bottle 1 can hold up to A milliliters of water, and now it contains B milliliters of water.  Bottle 2 contains C milliliters of water.  We will transfer water from Bottle 2 to Bottle 1 as much as possible.  How much amount of water will remain in Bottle 2?

Submission attempts (in order):
N=int(input())
if 1<=N and N<10:
	print(N)
elif 10<=N and N<100:
	print("9")
elif 100<=N and N<1000:
	print(9+N-99)
elif 1000<=N and N<10000:
	print("909")
elif 10000<=N and N<100000:
	print(909+N-999)
else:
	print("90909")

A,B,C=map(int,input().split())
if B+C>A:
	print(B+C-A)
else:
	print("0")

Verdicts: Runtime Error

Accepted
Problem: We have two bottles for holding water.  Bottle 1 can hold up to A milliliters of water, and now it contains B milliliters of water.  Bottle 2 contains C milliliters of water.  We will transfer water from Bottle 2 to Bottle 1 as much as possible.  How much amount of water will remain in Bottle 2?

Submission attempts (in order):


a, b, c = map(int, input().split())
print(a)
cap = a-b
use = min(a-b, c)
ans = c - use
print(ans)




a, b, c = map(int, input().split())
cap = a-b
use = min(a-b, c)
ans = c - use
print(ans)


Verdicts: Wrong Answer

Accepted
Problem: We have two bottles for holding water.  Bottle 1 can hold up to A milliliters of water, and now it contains B milliliters of water.  Bottle 2 contains C milliliters of water.  We will transfer water from Bottle 2 to Bottle 1 as much as possible.  How much amount of water will remain in Bottle 2?

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

if a > b + c:
  print(0)
else:
  print(c-b)

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

if a > b + c:
  print(0)
else:
  print(c+b-a)

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

if a > b + c:
  print(0)
else:
  print(c-b)

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

if a > b + c:
  print(0)
else:
  print(c+b-a)

Verdicts: Wrong Answer

Runtime Error

Runtime Error

Accepted
Problem: We have two bottles for holding water.  Bottle 1 can hold up to A milliliters of water, and now it contains B milliliters of water.  Bottle 2 contains C milliliters of water.  We will transfer water from Bottle 2 to Bottle 1 as much as possible.  How much amount of water will remain in Bottle 2?

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

cnt=0
for i in range(1,n+1):
  if len(str(i)) % 2 == 1:
    cnt += 1
print(cnt)

A,B,C = map(int, input().split())
print(max(C-(A-B), 0))

Verdicts: Runtime Error

Accepted
Problem: We have two bottles for holding water.  Bottle 1 can hold up to A milliliters of water, and now it contains B milliliters of water.  Bottle 2 contains C milliliters of water.  We will transfer water from Bottle 2 to Bottle 1 as much as possible.  How much amount of water will remain in Bottle 2?

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

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

flag = "Yes"
cnt = 0
if (N > 1):
  for i in range(len(lst) - 1):
    if (lst[i+1] - lst[i] < 0):
      cnt += 1
    else:
      cnt = 0
      
    if (cnt > 1):
      flag = "No"

print(flag)

A, B, C = map(int, input().split())

if (A - B >= C):
  print(0)
else:
  print(C - (A - B))

Verdicts: Runtime Error

Accepted
Problem: We have two bottles for holding water.  Bottle 1 can hold up to A milliliters of water, and now it contains B milliliters of water.  Bottle 2 contains C milliliters of water.  We will transfer water from Bottle 2 to Bottle 1 as much as possible.  How much amount of water will remain in Bottle 2?

Submission attempts (in order):
a,b,c=map(int,input().split())
if a-b==0:
    print(0)
else:
    one=a-b
    two=c-one
    if one>two:
        print(0)
    else:print(two)

a,b,c=map(int,input().split())
if a-b==0:
    print(c)
else:
    one=a-b
    two=c-one
    if one>=c:
        print(0)
    else:print(two)

Verdicts: Wrong Answer

Accepted
Problem: We have two bottles for holding water.  Bottle 1 can hold up to A milliliters of water, and now it contains B milliliters of water.  Bottle 2 contains C milliliters of water.  We will transfer water from Bottle 2 to Bottle 1 as much as possible.  How much amount of water will remain in Bottle 2?

Submission attempts (in order):
A, B, C = map(int, input().split())
print(max(C - (A - B), 0)

A, B, C = map(int, input().split())

print(max(C - (A - B), 0))

Verdicts: Runtime Error

Accepted
Problem: We have two bottles for holding water.  Bottle 1 can hold up to A milliliters of water, and now it contains B milliliters of water.  Bottle 2 contains C milliliters of water.  We will transfer water from Bottle 2 to Bottle 1 as much as possible.  How much amount of water will remain in Bottle 2?

Submission attempts (in order):
N = int(input())
if N < 10 ** 1:
    print(N)
elif N < 10 ** 2:
    print(9)
elif N < 10 ** 3:
    print(N - 90)
elif N < 10 ** 4:
    print(909)
elif N < 10 ** 5:
    print(N - 9090)
else:
     print(90909)

A, B, C = map(int, input().split())
if A >= 1 and A <=20 and B >= 1 and B <=20 and A >= B and C >= 1 and C <=20:
    if C - (A - B) >= 0:
        print(C - (A - B))
    else:
        print(0)

Verdicts: Runtime Error

Accepted
Problem: We have two bottles for holding water.  Bottle 1 can hold up to A milliliters of water, and now it contains B milliliters of water.  Bottle 2 contains C milliliters of water.  We will transfer water from Bottle 2 to Bottle 1 as much as possible.  How much amount of water will remain in Bottle 2?

Submission attempts (in order):
A, B, C = map(int, input().split())
print(C - A - B)

A, B, C = map(int, input().split())
print(max(0, C - (A - B)))


Verdicts: Wrong Answer

Accepted
Problem: We have two bottles for holding water.  Bottle 1 can hold up to A milliliters of water, and now it contains B milliliters of water.  Bottle 2 contains C milliliters of water.  We will transfer water from Bottle 2 to Bottle 1 as much as possible.  How much amount of water will remain in Bottle 2?

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

count = 0
for i in range(1, n + 1):
    s = len(str(i))
    if s%2 == 1:
        count += 1

print(count)


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

print(0 if b+c <= a else c-(a-b))


Verdicts: Runtime Error

Accepted
Problem: We have two bottles for holding water.  Bottle 1 can hold up to A milliliters of water, and now it contains B milliliters of water.  Bottle 2 contains C milliliters of water.  We will transfer water from Bottle 2 to Bottle 1 as much as possible.  How much amount of water will remain in Bottle 2?

Submission attempts (in order):
A,B,C = map(int,input().split())
x = C - A + B
if X < 0:X = 0
print(X)

A,B,C = map(int(),input().split())
print(C-A+B)

A,B,C=input()
print(C-A+B)

A,B,C=(int,input().split()))
print(C-A+B)

A,B,C = map(int,input().split())
x = C - A + B
if X < 0:
    X = 0
print(X)

A,B,C=map(int(),input().split())
print(C-A+B)

A,B,C = map(int,input.split)
print(C-A+B)

A,B,C = map(int,input().split())
x = C-A+B
if X < 0:
X = 0
print(X)

A,B,C=map(int,input(),split())
print(C-A+B)

A,B,C = map(int,input().split())
x = C-A+B
print(X)
if X < 0
print(X)

A,B,C = map(int,input().split())
x = C-A+B
print(X)
if X < 0
print(0)

A,B,C=map(int(),input(),split())
print(C-A+B)

A,B,C = map(int,input().split())
X = C - A + B
if X < 0:
    X = 0
print(X)

Verdicts: Runtime Error

Runtime Error

Runtime Error

Runtime Error

Runtime Error

Runtime Error

Runtime Error

Runtime Error

Runtime Error

Runtime Error

Runtime Error

Runtime Error

Accepted
Problem: We have two bottles for holding water.  Bottle 1 can hold up to A milliliters of water, and now it contains B milliliters of water.  Bottle 2 contains C milliliters of water.  We will transfer water from Bottle 2 to Bottle 1 as much as possible.  How much amount of water will remain in Bottle 2?

Submission attempts (in order):
s=list(map(int,input().split()))
a=s[0]-[1]
if a>s[2]:
  print(0)
else:
  print(s[2]-a)

s=list(map(int,input().split()))
s[0]=int(s[0])
s[1]=int(s[1])
s[2]=int(s[2])
a=0
if s[0]-s[1]>s[2]:
  print(0)
else:
  print(s[2]-s[0]+s[1])

Verdicts: Runtime Error

Accepted
Problem: We have two bottles for holding water.  Bottle 1 can hold up to A milliliters of water, and now it contains B milliliters of water.  Bottle 2 contains C milliliters of water.  We will transfer water from Bottle 2 to Bottle 1 as much as possible.  How much amount of water will remain in Bottle 2?

Submission attempts (in order):
a, b, c = input().spilt()

if a>=b+c:
  print(0)

else:
  print(b+c-a)

a, b, c = input().split()
a = int(a)
b = int(b)
c = int(c)
if a>=b+c:
    print(0)
     
else:
    print(b+c-a)

Verdicts: Runtime Error

Accepted
Problem: We have two bottles for holding water.  Bottle 1 can hold up to A milliliters of water, and now it contains B milliliters of water.  Bottle 2 contains C milliliters of water.  We will transfer water from Bottle 2 to Bottle 1 as much as possible.  How much amount of water will remain in Bottle 2?

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

print(a-(b+c) if (b + c) < a else 0)

a,b,c = list(map(int, input().split(" ")))

print((b+c)-a if (b + c) > a else 0)

Verdicts: Wrong Answer

Accepted
Problem: We have two bottles for holding water.  Bottle 1 can hold up to A milliliters of water, and now it contains B milliliters of water.  Bottle 2 contains C milliliters of water.  We will transfer water from Bottle 2 to Bottle 1 as much as possible.  How much amount of water will remain in Bottle 2?

Submission attempts (in order):
a,b,c= map(int,input().split())
t = max(a-b,c)
print(c-t)

a,b,c= map(int,input().split())
t = min(a-b,c)
print(c-t)

Verdicts: Wrong Answer

Accepted
Problem: We have two bottles for holding water.  Bottle 1 can hold up to A milliliters of water, and now it contains B milliliters of water.  Bottle 2 contains C milliliters of water.  We will transfer water from Bottle 2 to Bottle 1 as much as possible.  How much amount of water will remain in Bottle 2?

Submission attempts (in order):
# -*- coding: utf-8 -*-
N = int(input())
H = list(map(int,input().split()))

ans = 'Yes'
diff = 0

for i in range(N - 1):
	diff = H[i + 1] - H[i]

	if diff > 1:
		H[i + 1] -= 1
	elif diff < 0:
		ans = 'No'
print(ans)

# -*- coding: utf-8 -*-

A = list(map(int,input().split()))
remain = 0

if A[2] - (A[0] - A[1]) <= 0:
	print(remain)
else:
	remain = A[2] - (A[0] - A[1])
	print(remain)

Verdicts: Runtime Error

Accepted
Problem: We have two bottles for holding water.  Bottle 1 can hold up to A milliliters of water, and now it contains B milliliters of water.  Bottle 2 contains C milliliters of water.  We will transfer water from Bottle 2 to Bottle 1 as much as possible.  How much amount of water will remain in Bottle 2?

Submission attempts (in order):
import sys

N = int(input())
H = [int(x) for x in input().split()]

flag = 0

for i in range(len(H) - 1):
    if H[i] > H[i+1] + 1:
        print('No')
        sys.exit()
	
    elif H[i] == H[i+1] + 1:
	flag = 1
	
    if flag == 1:
        if H[i]	> H[i+1]:
            print('No')
            sys.exit()

print('Yes')

a, b, c = (int(x) for x in input().split())
if c-a+b>0:
	print(c-a+b)
else:
 	print(0)

Verdicts: Runtime Error

Accepted
Problem: We have two bottles for holding water.  Bottle 1 can hold up to A milliliters of water, and now it contains B milliliters of water.  Bottle 2 contains C milliliters of water.  We will transfer water from Bottle 2 to Bottle 1 as much as possible.  How much amount of water will remain in Bottle 2?

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

for x in range(n-1):
    if h[x] <= h[x+1]:
        pass
    else:
        h[x] = h[x]-1

if n == 1:
    ans ='Yes'
for x in range(n-1):
    if h[x] <= h[x+1]:
        ans = 'Yes'
    else:
        ans = 'No'
        break
print(ans)

N = int (input())
H = list(map(int,input().split()))
ans = 'Yes'
for  i in range(N-1):
    diff = H[i+1] - H[i]
    if diff >= 1:
        H[i+1] -= 1
    elif diff < 0:
        ans ='No'
print(ans)


a,b,c = map(int,(input().split()))
ans = c-(a-b)
if ans <= 0:
    ans = 0
else:
    pass
print(ans)

Verdicts: Runtime Error

Runtime Error

Accepted
Problem: We have two bottles for holding water.  Bottle 1 can hold up to A milliliters of water, and now it contains B milliliters of water.  Bottle 2 contains C milliliters of water.  We will transfer water from Bottle 2 to Bottle 1 as much as possible.  How much amount of water will remain in Bottle 2?

Submission attempts (in order):
a,b,c=int(input())
if a-b>=c:
  print(0)
else:
  print(c-(a-b))

a,b,c=map(int,input().split())
if a-b>=c:
  print(0)
else:
  print(c-(a-b))

Verdicts: Runtime Error

Accepted
Problem: We have two bottles for holding water.  Bottle 1 can hold up to A milliliters of water, and now it contains B milliliters of water.  Bottle 2 contains C milliliters of water.  We will transfer water from Bottle 2 to Bottle 1 as much as possible.  How much amount of water will remain in Bottle 2?

Submission attempts (in order):
if (A-B) > C:
  print (0)
else:
  print (A-B)

def main():
    A, B, C = [int(_) for _ in input().split()]
    if (A-B) < C:
      print (C-A+B)
    else:
      print(0)
main()

Verdicts: Runtime Error

Accepted
Problem: We have two bottles for holding water.  Bottle 1 can hold up to A milliliters of water, and now it contains B milliliters of water.  Bottle 2 contains C milliliters of water.  We will transfer water from Bottle 2 to Bottle 1 as much as possible.  How much amount of water will remain in Bottle 2?

Submission attempts (in order):
N = int(input())
H = list(map(int, input().split()))
count = 0
if(N==1):
    print("Yes")
else:
    for i in range(N-1):
        if(H[i+1]-H[i]==-1):
            count +=1
            if(count==1):
                H[i]=H[i]-1
        elif(H[i]>H[i+1]):
            count +=1
    print(count)
    if(count<=1):
        print("Yes")
    else:
        print("No")


a, b, c= (int(i) for i in input().split())  

cap=a-b
c=c-cap
if(c>0):
    print(c)
else:
    print("0")







Verdicts: Runtime Error

Accepted
Problem: We have two bottles for holding water.  Bottle 1 can hold up to A milliliters of water, and now it contains B milliliters of water.  Bottle 2 contains C milliliters of water.  We will transfer water from Bottle 2 to Bottle 1 as much as possible.  How much amount of water will remain in Bottle 2?

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

a,b,c=map(int,input().split())
res=c-(a-b)
if res<0:
    print(0)
else:
    print(res)

Verdicts: Wrong Answer

Accepted
Problem: We have two bottles for holding water.  Bottle 1 can hold up to A milliliters of water, and now it contains B milliliters of water.  Bottle 2 contains C milliliters of water.  We will transfer water from Bottle 2 to Bottle 1 as much as possible.  How much amount of water will remain in Bottle 2?

Submission attempts (in order):
a, b, c = map(int, input().split())
if b + c > a:
  print(c-(a-b))
elif b + c =< a:
  print(0)

a, b, c = map(int, input().split())
if b + c > a:
  print(c-(a-b))
elif b + c <= a:
  print(0)


Verdicts: Runtime Error

Accepted
Problem: We have two bottles for holding water.  Bottle 1 can hold up to A milliliters of water, and now it contains B milliliters of water.  Bottle 2 contains C milliliters of water.  We will transfer water from Bottle 2 to Bottle 1 as much as possible.  How much amount of water will remain in Bottle 2?

Submission attempts (in order):
 from sys import stdin

a,b,c = [int(x) for x in stdin.readline().rstrip().split()]

print(b+c-a)

from sys import stdin

a,b,c = [int(x) for x in stdin.readline().rstrip().split()]

if b+c-a>0:
    print(b+c-a)
else:
    print(0)

Verdicts: Runtime Error

Accepted
Problem: We have two bottles for holding water.  Bottle 1 can hold up to A milliliters of water, and now it contains B milliliters of water.  Bottle 2 contains C milliliters of water.  We will transfer water from Bottle 2 to Bottle 1 as much as possible.  How much amount of water will remain in Bottle 2?

Submission attempts (in order):
#Time up and try
A,B,C = map(int,input().split())
answer = 0

for i in range(1,B+1):
    if  A != B:
        B += 1
    elif A == B:
        print(i)
        answer = C - i+1
        print(answer)
        exit()

A,B,C = map(int,input().split())
answer = 0

for i in range(1,B+C):
    if A >= B + C:
        print(answer)
    elif  A != B:
        B += 1
    elif A == B:
        answer = C - i+1
        print(answer)

A,B,C = map(int,input().split())
answer = 0

for i in range(1,B+C):
    if A >= B + C:
        print(answer)
    elif  A != B:
        B += 1
    elif A == B:
        answer = C - i+1
        print(answer)

A,B,C = map(int,input().split())
answer = 0

for i in range(1,B+C):
    if A >= B + C:
        print(answer)
    elif  A != B:
        B += 1
    elif A == B:
        answer = C - i+1
        print(answer)

A,B,C = map(int,input().split())
answer = 0

for i in range(1,B+C):
    if A >= B + C:
        print(answer)
        exit()
    elif  A != B:
        B += 1
    elif A == B:
        answer = C - i+1
        print(answer)
        exit()

Verdicts: Wrong Answer

Wrong Answer

Wrong Answer

Wrong Answer

Accepted
Problem: We have two bottles for holding water.  Bottle 1 can hold up to A milliliters of water, and now it contains B milliliters of water.  Bottle 2 contains C milliliters of water.  We will transfer water from Bottle 2 to Bottle 1 as much as possible.  How much amount of water will remain in Bottle 2?

Submission attempts (in order):
import sys
A = int(sys.argv[1])
B = int(sys.argv[2])
C = int(sys.argv[3])
if C - (A - B) >= 0:
    print(C - (A - B))
else:
    print(0)

A, B, C = map(int, input().split())
if C - (A - B) >= 0:
    print(C - (A - B))
else:
    print(0)

Verdicts: Runtime Error

Accepted
Problem: We have two bottles for holding water.  Bottle 1 can hold up to A milliliters of water, and now it contains B milliliters of water.  Bottle 2 contains C milliliters of water.  We will transfer water from Bottle 2 to Bottle 1 as much as possible.  How much amount of water will remain in Bottle 2?

Submission attempts (in order):
import sys

def main():
    a = int(sys.argv[1])
    b = int(sys.argv[2])
    c = int(sys.argv[3])
    result = c-(a-b)
    if result < 0:
        result = 0

    print(result)
    sys.exit(2)

if __name__ == "__main__":
    main()

import sys

def main():
    a = int(sys.argv[1])
    b = int(sys.argv[2])
    c = int(sys.argv[3])
    result = c-(a-b)
    if result < 0:
        result = 0

    print(result)
    sys.exit(2)

if __name__ == "__main__":
    main()

def main():
    a, b, c = map(int, input().split())
    result = c-(a-b)
    if result < 0:
        result = 0

    print(result)

if __name__ == "__main__":
    main()

Verdicts: Runtime Error

Runtime Error

Accepted
Problem: We have two bottles for holding water.  Bottle 1 can hold up to A milliliters of water, and now it contains B milliliters of water.  Bottle 2 contains C milliliters of water.  We will transfer water from Bottle 2 to Bottle 1 as much as possible.  How much amount of water will remain in Bottle 2?

Submission attempts (in order):
a=list(map(int,split().input()))
print(max(0,a[2]-(a[0]-a[1])))

a = list(map(int, input().split()))
print(max(0, a[2]-(a[0]-a[1])))


Verdicts: Runtime Error

Accepted
Problem: We have two bottles for holding water.  Bottle 1 can hold up to A milliliters of water, and now it contains B milliliters of water.  Bottle 2 contains C milliliters of water.  We will transfer water from Bottle 2 to Bottle 1 as much as possible.  How much amount of water will remain in Bottle 2?

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

for i in range(N-1):
    minh = min(H[i+1:]) 
    if H[i] - minh >= 2:
        print('No')
        break
    else:
        continue
else:
    print('Yes')



A, B, C = map(int, input().split())

ans = max(0, C-(A-B))

print(ans)

Verdicts: Runtime Error

Accepted
Problem: We have two bottles for holding water.  Bottle 1 can hold up to A milliliters of water, and now it contains B milliliters of water.  Bottle 2 contains C milliliters of water.  We will transfer water from Bottle 2 to Bottle 1 as much as possible.  How much amount of water will remain in Bottle 2?

Submission attempts (in order):
a,b,c=map(int,input().split())
if c>b-a:
    print(c-b-a)
else:
    print(0)

a,b,c=map(int,input().split())
if c>a-b:
    print(c+b-a)
else:
    print(0)

Verdicts: Wrong Answer

Accepted
Problem: We have two bottles for holding water.  Bottle 1 can hold up to A milliliters of water, and now it contains B milliliters of water.  Bottle 2 contains C milliliters of water.  We will transfer water from Bottle 2 to Bottle 1 as much as possible.  How much amount of water will remain in Bottle 2?

Submission attempts (in order):
a,b,c=map(int,input().split())
s=c-b+a
print(int(s))

N = int(input())
ans = 0
for i in range(1, N + 1):
	if len(i) % 2 == 1:
		ans += 1
print(ans)

N = int(input())
ans = 0
for i in range(1, N + 1):
	if len(str(i)) % 2 == 1:
		ans += 1
print(ans)

N = int(input())
ans = 0
for i in range(1, N + 1):
	if len(int(i)) % 2 == 1:
		ans += 1
print(ans)

a,b,c=map(int,input().split())
s=c-b+a
print(s)

a,b,c=map(int,input().split())
s=c-min(c,a-b)
print(s)

Verdicts: Wrong Answer

Runtime Error

Runtime Error

Runtime Error

Wrong Answer

Accepted
Problem: We have two bottles for holding water.  Bottle 1 can hold up to A milliliters of water, and now it contains B milliliters of water.  Bottle 2 contains C milliliters of water.  We will transfer water from Bottle 2 to Bottle 1 as much as possible.  How much amount of water will remain in Bottle 2?

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

print(a + b < c)

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

i = b + c - a
if b + c > a:
    print(i)
else:
    print(0)

Verdicts: Wrong Answer

Accepted
Problem: We have two bottles for holding water.  Bottle 1 can hold up to A milliliters of water, and now it contains B milliliters of water.  Bottle 2 contains C milliliters of water.  We will transfer water from Bottle 2 to Bottle 1 as much as possible.  How much amount of water will remain in Bottle 2?

Submission attempts (in order):
n = int(input())
h = [int(x) for x in input().split(" ")]
 
for i in range(1, n):
  if h[-i]-h[-i-1] >= 0:
    continue  
  elif h[-i]-h[-i-1] == -1:
    h[-i-1] = h[i-1] - 1
    continue
  elif h[i]-h[-i-1] < -1:
    ans = "No"
    break
else:
  ans = "Yes"
 
print(ans)

a, b, c = [int(x) for x in input().split(" ")]
 
rest = a - b
ans = c - rest
if ans < 0:
  ans = 0
 
print(ans)

Verdicts: Runtime Error

Accepted
Problem: We have two bottles for holding water.  Bottle 1 can hold up to A milliliters of water, and now it contains B milliliters of water.  Bottle 2 contains C milliliters of water.  We will transfer water from Bottle 2 to Bottle 1 as much as possible.  How much amount of water will remain in Bottle 2?

Submission attempts (in order):
a,b,c = map(int, input().split())
print(c-(a-b) if c<a-b else 0)

a,b,c = map(int, input().split())
print(c-(a-b) if c>a-b else 0)

Verdicts: Wrong Answer

Accepted
Problem: We have two bottles for holding water.  Bottle 1 can hold up to A milliliters of water, and now it contains B milliliters of water.  Bottle 2 contains C milliliters of water.  We will transfer water from Bottle 2 to Bottle 1 as much as possible.  How much amount of water will remain in Bottle 2?

Submission attempts (in order):
A, B, C = map(int,input().split())
if B+C <=A{
  C=0
}else{
C=B+C-A
}

A, B, C = map(int,input().split())
if B+C <=A {
  print(0)
}else{
C=B+C-A
  print(C)
}

A, B, C = map(int,input().split())
if B+C <= A :
  print(0)
else:
  C = B+C-A
  print(C)



Verdicts: Runtime Error

Runtime Error

Accepted
Problem: We have two bottles for holding water.  Bottle 1 can hold up to A milliliters of water, and now it contains B milliliters of water.  Bottle 2 contains C milliliters of water.  We will transfer water from Bottle 2 to Bottle 1 as much as possible.  How much amount of water will remain in Bottle 2?

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

print("{}-{}+{}".format(c,a,b))


a, b, c = input().split()

print("{}-{}+{}".format(c,a,b)

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

print(c-min(a-b,c))


Verdicts: Wrong Answer

Runtime Error

Accepted
Problem: We have two bottles for holding water.  Bottle 1 can hold up to A milliliters of water, and now it contains B milliliters of water.  Bottle 2 contains C milliliters of water.  We will transfer water from Bottle 2 to Bottle 1 as much as possible.  How much amount of water will remain in Bottle 2?

Submission attempts (in order):
a, b, c = map(int, input().split())
if a <= b + c:
  print(0)
else:
  print(c - (a - b))

a, b, c = map(int, input().split())
if c - (a - b) <= 0:
  print(0)
else:
  print(c - (a - b))

Verdicts: Wrong Answer

Accepted
Problem: We have two bottles for holding water.  Bottle 1 can hold up to A milliliters of water, and now it contains B milliliters of water.  Bottle 2 contains C milliliters of water.  We will transfer water from Bottle 2 to Bottle 1 as much as possible.  How much amount of water will remain in Bottle 2?

Submission attempts (in order):
A, B = map(int, input().split())
 
if (A+B)%2 == 0:
    print( int((A+B)/2) )
else:
    print('IMPOSSIBLE')

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

print(max(0,c-(a-b)))

Verdicts: Runtime Error

Accepted
Problem: We have two bottles for holding water.  Bottle 1 can hold up to A milliliters of water, and now it contains B milliliters of water.  Bottle 2 contains C milliliters of water.  We will transfer water from Bottle 2 to Bottle 1 as much as possible.  How much amount of water will remain in Bottle 2?

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

a, b, c = map(int,input().split())
if (a-b) <= c:
    print(c-(a-b))
else:
    print("0")

Verdicts: Wrong Answer

Accepted
Problem: We have two bottles for holding water.  Bottle 1 can hold up to A milliliters of water, and now it contains B milliliters of water.  Bottle 2 contains C milliliters of water.  We will transfer water from Bottle 2 to Bottle 1 as much as possible.  How much amount of water will remain in Bottle 2?

Submission attempts (in order):
A,B,C = map(int,input().split())
if B+C>=A:
    print(C-B)
else:
    print(0)
    

A,B,C = map(int,input().split())
if B+C>=A:
    print(B-A)
else:
    print(0)
    

A,B,C = map(int,input().split())
if B+C>=A:
    print(B+C-A)
else:
    print(0)
    

Verdicts: Wrong Answer

Wrong Answer

Accepted
Problem: We have two bottles for holding water.  Bottle 1 can hold up to A milliliters of water, and now it contains B milliliters of water.  Bottle 2 contains C milliliters of water.  We will transfer water from Bottle 2 to Bottle 1 as much as possible.  How much amount of water will remain in Bottle 2?

Submission attempts (in order):
A,B,C = map(int,input().split())

if B+C <= A:
  print(B+C)
else:
  print(A)
    

A,B,C = map(int,input().split())

if B+C <= A:
  print(0)
else:
  print(B-(A-C))
    


Verdicts: Wrong Answer

Accepted
Problem: We have two bottles for holding water.  Bottle 1 can hold up to A milliliters of water, and now it contains B milliliters of water.  Bottle 2 contains C milliliters of water.  We will transfer water from Bottle 2 to Bottle 1 as much as possible.  How much amount of water will remain in Bottle 2?

Submission attempts (in order):
import math

a, b, t = [int(x) for x in input().split()]
print(math.ceil((t + 0.5) // a) * b)


a, b, c = [int(x) for x in input().split()]

req = b + c

print((req - a) if req > a else 0)


Verdicts: Wrong Answer

Accepted
Problem: We have two bottles for holding water.  Bottle 1 can hold up to A milliliters of water, and now it contains B milliliters of water.  Bottle 2 contains C milliliters of water.  We will transfer water from Bottle 2 to Bottle 1 as much as possible.  How much amount of water will remain in Bottle 2?

Submission attempts (in order):
a,b,c=map(int,input().split)
print(c-a+b);

N=int(input())
all=0
a=5
while(a>0):
  if(10**a>N):
    a-=1
    continue
  if(a%2==1):
    all+=9*10**(a-1)
    a-=1
    continue
  else:
    all+=N+1-10**a
  
  a-=1


a,b,c=map(int,input().split())
if(c-a+b > 0):
  print(c-a+b)
else:
  print(0)

Verdicts: Runtime Error

Runtime Error

Accepted
Problem: We have two bottles for holding water.  Bottle 1 can hold up to A milliliters of water, and now it contains B milliliters of water.  Bottle 2 contains C milliliters of water.  We will transfer water from Bottle 2 to Bottle 1 as much as possible.  How much amount of water will remain in Bottle 2?

Submission attempts (in order):
N = int(input())
2 ans = 0
3 for i in range(1, N + 1):
4 if len(str(i)) % 2 == 1:
5 ans += 1
6 print(ans)


if B >= 1 and A >= B and A <= 20 and C >= 1 and C <= 20:
    capacity = A - B
    if capacity <= C:
        remaining = C - capacity
    else:
        remaining = 0
    print(remaining)

if N < 10:
    print(N)
elif N >= 10 and N < 100:
    print(9)
elif N >= 100 and N < 1000:
    print(N - 90)
elif N >= 1000 and N < 10000:
    print(909)
elif N >= 10000 and N < 100000:
    print(N-9090)
else:
    print(90909)

A,B,C = map(int, input().split())

if B >= 1 and A >= B and A <= 20 and C >= 1 and C <= 20:
    capacity = A - B
    if capacity <= C:
        remaining = C - capacity
    else:
        remaining = 0
    print(remaining)

Verdicts: Runtime Error

Runtime Error

Runtime Error

Accepted
Problem: We have two bottles for holding water.  Bottle 1 can hold up to A milliliters of water, and now it contains B milliliters of water.  Bottle 2 contains C milliliters of water.  We will transfer water from Bottle 2 to Bottle 1 as much as possible.  How much amount of water will remain in Bottle 2?

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

s = a-b

if b-s < 0:
  print(c-s)
else:
  print(0)
  


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

s = a-c

if b-s < 0:
  print(b-s)
else:
  print(0)
  


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

s = a-b

if c-s > 0:
  print(c-s)
else:
  print(0)
  


Verdicts: Wrong Answer

Wrong Answer

Accepted
Problem: We have two bottles for holding water.  Bottle 1 can hold up to A milliliters of water, and now it contains B milliliters of water.  Bottle 2 contains C milliliters of water.  We will transfer water from Bottle 2 to Bottle 1 as much as possible.  How much amount of water will remain in Bottle 2?

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

a,b,c = map(int,input().split())
if a+c => b:
    print(b)
else:
    pritn(a+c)

a,b,c = map(int,input().split())
if b+c >= a:
    print(c-a+b)
else:
    print(0)

Verdicts: Runtime Error

Runtime Error

Accepted
Problem: We have two bottles for holding water.  Bottle 1 can hold up to A milliliters of water, and now it contains B milliliters of water.  Bottle 2 contains C milliliters of water.  We will transfer water from Bottle 2 to Bottle 1 as much as possible.  How much amount of water will remain in Bottle 2?

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

print(min(c-(b-a),0))

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

print(max(c-(b-a),0))

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

print(max(c- (a - b),0))

Verdicts: Wrong Answer

Wrong Answer

Accepted
Problem: We have two bottles for holding water.  Bottle 1 can hold up to A milliliters of water, and now it contains B milliliters of water.  Bottle 2 contains C milliliters of water.  We will transfer water from Bottle 2 to Bottle 1 as much as possible.  How much amount of water will remain in Bottle 2?

Submission attempts (in order):
a,b,c = list(map(int, input().split()))
d = c-(a-b)
if d >= 0:
  print(d)
else if:
  print(0)


a,b,c = list(map(int, input().split()))
d = c-(a-b)
if d >= 0:
  print(d)
else:
  print(0)

Verdicts: Runtime Error

Accepted
Problem: We have two bottles for holding water.  Bottle 1 can hold up to A milliliters of water, and now it contains B milliliters of water.  Bottle 2 contains C milliliters of water.  We will transfer water from Bottle 2 to Bottle 1 as much as possible.  How much amount of water will remain in Bottle 2?

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

if c-(a-b)>0:
    print(c-(a-b))
else:
    print(0

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

if c-(a-b)>0:
    print(c-(a-b))
else:
    print(0)


Verdicts: Runtime Error

Accepted
Problem: We have two bottles for holding water.  Bottle 1 can hold up to A milliliters of water, and now it contains B milliliters of water.  Bottle 2 contains C milliliters of water.  We will transfer water from Bottle 2 to Bottle 1 as much as possible.  How much amount of water will remain in Bottle 2?

Submission attempts (in order):
import math
a = int(input())
data_size = int (math.log10(a) + 1)
if a == 100000:
    print(90909)
elif data_size == 5:
    print(a - 10000 + 1 + 909)
elif data_size == 4:
    print(909)
elif data_size == 3:
    print(a - 100 + 1 + 9)
elif data_size == 2:   
    print(9)
elif data_size == 1:   
    print(a - 1 + 1)

import math
a = int(input())
if ((a > 0) & (a < 100001)):
    data_size = int (math.log10(a) + 1)
    if a == 100000:
        print(90909)
    elif data_size == 5:
        print(a - 10000 + 1 + 909)
    elif data_size == 4:
        print(909)
    elif data_size == 3:
        print(a - 100 + 1 + 9)
    elif data_size == 2:   
        print(9)
    elif data_size == 1:   
        print(a - 1 + 1)

a,b,c=(int(x) for x in input().split())
if ((c - (a - b)) < 0):
    print(0)
else:
    print(c - (a - b))

Verdicts: Runtime Error

Runtime Error

Accepted
Problem: It is known that the area of a regular dodecagon inscribed in a circle of radius a is 3a^2.  Given an integer r, find the area of a regular dodecagon inscribed in a circle of radius r.

Submission attempts (in order):
import bisect

N = int(input())
A = [int(input()) for i in range(N)]

ans = 1
X = [-A[0]]
for i in range(1,N):
    if -X[ans-1] >= A[i]:
        X.append(-A[i])
        ans += 1
    else:
        pos = bisect.bisect_left(X,-A[i])
        X[pos] = -A[i]

print(ans)


r = int(input())
print(3*r**2)

Verdicts: Runtime Error

Accepted
Problem: It is known that the area of a regular dodecagon inscribed in a circle of radius a is 3a^2.  Given an integer r, find the area of a regular dodecagon inscribed in a circle of radius r.

Submission attempts (in order):
S = int(input()
print(3 * S * S)

S = int(input())
print(S * S * 3)


Verdicts: Runtime Error

Accepted
Problem: It is known that the area of a regular dodecagon inscribed in a circle of radius a is 3a^2.  Given an integer r, find the area of a regular dodecagon inscribed in a circle of radius r.

Submission attempts (in order):
n, d = map(int, input().split())
a = n // (2*d + 1) + 1
print(a)

n = int(input())
A = [int(input()) for i in range(n)]
  
for i in range(n):
  Ai = A[:]
  Ai.pop(i)
  print(max(Ai))

a = int(input())

pritn(a**3 * 2)

a = int(input())
s = a**2 * 3
print(s)

Verdicts: Runtime Error

Runtime Error

Runtime Error

Accepted
Problem: It is known that the area of a regular dodecagon inscribed in a circle of radius a is 3a^2.  Given an integer r, find the area of a regular dodecagon inscribed in a circle of radius r.

Submission attempts (in order):
r=int(input())
print(3*r^2)

r=int(input())
print(3*(r**2))

Verdicts: Wrong Answer

Accepted
Problem: It is known that the area of a regular dodecagon inscribed in a circle of radius a is 3a^2.  Given an integer r, find the area of a regular dodecagon inscribed in a circle of radius r.

Submission attempts (in order):
import sys
import math
import copy
a = int(input());
b = [int(input()) for i in range(a)];
c = 0;
f = max(b);
for i in range(a):
    if max(b) > b[i]:
        print(f);
    elif max(b) == b[i]:
        d = copy.copy(b);
        del d[i]
        e = max(d);
        print(e);


import sys
r = int(input());
ans = 3*r*r;
print(ans);


Verdicts: Runtime Error

Accepted
Problem: It is known that the area of a regular dodecagon inscribed in a circle of radius a is 3a^2.  Given an integer r, find the area of a regular dodecagon inscribed in a circle of radius r.

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

for i in range(10**6):
  if int((S+i**2)**(1/2))==(S+i**2)**(1/2):
    y=i
    x=int((S+i**2)**(1/2))
    
print("{} {} {} {} {} {}".format(0,0,x,y,y,x))

print(3*int(input())**2)

Verdicts: Runtime Error

Accepted
Problem: It is known that the area of a regular dodecagon inscribed in a circle of radius a is 3a^2.  Given an integer r, find the area of a regular dodecagon inscribed in a circle of radius r.

Submission attempts (in order):
r=map(int,input().split)
print(3r**2)

r=map(int,input().split())
print(3r**2)


r=int(input())
print(3*r**2)

Verdicts: Runtime Error

Runtime Error

Accepted
Problem: It is known that the area of a regular dodecagon inscribed in a circle of radius a is 3a^2.  Given an integer r, find the area of a regular dodecagon inscribed in a circle of radius r.

Submission attempts (in order):
print(3*(int(input()))^2)

print(3*(int(input()))**2)

Verdicts: Wrong Answer

Accepted
Problem: It is known that the area of a regular dodecagon inscribed in a circle of radius a is 3a^2.  Given an integer r, find the area of a regular dodecagon inscribed in a circle of radius r.

Submission attempts (in order):
r = input()
print(3 * a**2)

r = input()
print(3 * r**2)

r =int(input())
print(3* r**2)

Verdicts: Runtime Error

Runtime Error

Accepted
Problem: It is known that the area of a regular dodecagon inscribed in a circle of radius a is 3a^2.  Given an integer r, find the area of a regular dodecagon inscribed in a circle of radius r.

Submission attempts (in order):
a = input
a = int(a)
print(a **2 * 3)

a = int(input())
print(a**2 * 3)

Verdicts: Runtime Error

Accepted
Problem: It is known that the area of a regular dodecagon inscribed in a circle of radius a is 3a^2.  Given an integer r, find the area of a regular dodecagon inscribed in a circle of radius r.

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

S = 3*a^2

print(S)

a = int(input())

S = 3*a*a

print(S)

Verdicts: Wrong Answer

Accepted
Problem: It is known that the area of a regular dodecagon inscribed in a circle of radius a is 3a^2.  Given an integer r, find the area of a regular dodecagon inscribed in a circle of radius r.

Submission attempts (in order):
print(int(input())^2*3)

print(int(input())**2*3)

Verdicts: Wrong Answer

Accepted
Problem: It is known that the area of a regular dodecagon inscribed in a circle of radius a is 3a^2.  Given an integer r, find the area of a regular dodecagon inscribed in a circle of radius r.

Submission attempts (in order):
print(3*(int(input())**2)

print(3*(int(input())**2))

Verdicts: Runtime Error

Accepted
Problem: It is known that the area of a regular dodecagon inscribed in a circle of radius a is 3a^2.  Given an integer r, find the area of a regular dodecagon inscribed in a circle of radius r.

Submission attempts (in order):
i = input()
out = 3*i*i
print(out)

i = int(input())
out = 3*i*i
print(out)

Verdicts: Runtime Error

Accepted
Problem: It is known that the area of a regular dodecagon inscribed in a circle of radius a is 3a^2.  Given an integer r, find the area of a regular dodecagon inscribed in a circle of radius r.

Submission attempts (in order):
x,a = map(int,input().split())
if x < a:
  print(0)
else:
  print(10)

r = int(input())
print(3*r**2)

Verdicts: Runtime Error

Accepted
Problem: It is known that the area of a regular dodecagon inscribed in a circle of radius a is 3a^2.  Given an integer r, find the area of a regular dodecagon inscribed in a circle of radius r.

Submission attempts (in order):
a = r*r*3
print(a)

3*r*r

n=int(input()) 
a = n*n*3
print(a)

Verdicts: Runtime Error

Runtime Error

Accepted
Problem: It is known that the area of a regular dodecagon inscribed in a circle of radius a is 3a^2.  Given an integer r, find the area of a regular dodecagon inscribed in a circle of radius r.

Submission attempts (in order):
r=int(input())
print=(3r**2)

r=int(input())
s=3*r**2
print(s)

Verdicts: Runtime Error

Accepted
Problem: It is known that the area of a regular dodecagon inscribed in a circle of radius a is 3a^2.  Given an integer r, find the area of a regular dodecagon inscribed in a circle of radius r.

Submission attempts (in order):
n,d=input().split()
n=int(n)
d=int(d)
a=d*2+1
if n%a==0:
    print(n//a)
else:
    print(n//a+1)

r=int(input())
a=3*r**2
print(int(a))

Verdicts: Runtime Error

Accepted
Problem: It is known that the area of a regular dodecagon inscribed in a circle of radius a is 3a^2.  Given an integer r, find the area of a regular dodecagon inscribed in a circle of radius r.

Submission attempts (in order):
from bisect import bisect_left
N, M, X = map(int, input().split())
A = sorted(list(map(int, input().split())))
index = bisect_left(A, X)
print(min(len(A[:index]), len(A[index:])))

print(3*int(input())**2)

Verdicts: Runtime Error

Accepted
Problem: It is known that the area of a regular dodecagon inscribed in a circle of radius a is 3a^2.  Given an integer r, find the area of a regular dodecagon inscribed in a circle of radius r.

Submission attempts (in order):
print(3*(int(input())**2)

print(3*int(input())**2)

Verdicts: Runtime Error

Accepted
Problem: It is known that the area of a regular dodecagon inscribed in a circle of radius a is 3a^2.  Given an integer r, find the area of a regular dodecagon inscribed in a circle of radius r.

Submission attempts (in order):
# -*- coding: utf-8 -*-
print(3*int(input)**2)


# -*- coding: utf-8 -*-
print(3*int(input())**2)


Verdicts: Runtime Error

Accepted
Problem: It is known that the area of a regular dodecagon inscribed in a circle of radius a is 3a^2.  Given an integer r, find the area of a regular dodecagon inscribed in a circle of radius r.

Submission attempts (in order):
r = input()
print (3 * r ** 2)

r = int(input())
print (3 * r ** 2)


Verdicts: Runtime Error

Accepted
Problem: It is known that the area of a regular dodecagon inscribed in a circle of radius a is 3a^2.  Given an integer r, find the area of a regular dodecagon inscribed in a circle of radius r.

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

b=[-1]*N
b[N-1]=a[N-1]
for i in range(N-1)[::-1]: 
    sum_count = N // (i+1)
    if sum_count==1:
        b[i]=a[i]
    else:
        sum1=0
        for j in range(sum_count-1):
            sum1 = sum1 + b[(i+1)*(j+1)]
        sum1 = sum1 % 2
        b[i]=(a[i]+ sum1)%2

print(sum(b))
for i in range(N):
    if b[i]==1:
        print(i+1,end=" ")

N = int(input())
print(3*N*N)

Verdicts: Runtime Error

Accepted
Problem: It is known that the area of a regular dodecagon inscribed in a circle of radius a is 3a^2.  Given an integer r, find the area of a regular dodecagon inscribed in a circle of radius r.

Submission attempts (in order):
print(int(input) ** 2 * 3)

r = int(input())
print(3 * r ** 2)

Verdicts: Runtime Error

Accepted
Problem: It is known that the area of a regular dodecagon inscribed in a circle of radius a is 3a^2.  Given an integer r, find the area of a regular dodecagon inscribed in a circle of radius r.

Submission attempts (in order):
value=input().split()

print(3*value*value)

r=int(input())

print(3*(r**2))

Verdicts: Runtime Error

Accepted
Problem: It is known that the area of a regular dodecagon inscribed in a circle of radius a is 3a^2.  Given an integer r, find the area of a regular dodecagon inscribed in a circle of radius r.

Submission attempts (in order):
r = int(input())
return 3 * r ** 2 

r = int(input())
print(3 * r ** 2 )



Verdicts: Runtime Error

Accepted
Problem: It is known that the area of a regular dodecagon inscribed in a circle of radius a is 3a^2.  Given an integer r, find the area of a regular dodecagon inscribed in a circle of radius r.

Submission attempts (in order):
n,d=list(map(int, input().split()))
 
d1=d*2+1
print((n+d1-1)/d1)

n=int(input())
print(3*n*n)

Verdicts: Runtime Error

Accepted
Problem: It is known that the area of a regular dodecagon inscribed in a circle of radius a is 3a^2.  Given an integer r, find the area of a regular dodecagon inscribed in a circle of radius r.

Submission attempts (in order):
r = input()
print(3*r**2)

r = int(input())

print(3*r**2)

Verdicts: Runtime Error

Accepted
Problem: It is known that the area of a regular dodecagon inscribed in a circle of radius a is 3a^2.  Given an integer r, find the area of a regular dodecagon inscribed in a circle of radius r.

Submission attempts (in order):
n=int(input())
a=list(int,input().split(), for _ in range(0,n))
x=max(a)
y=a.index(x)
del a[y]
z=max(a)

for i in range(0,n):
  if i!=y:
    print(x)
  else:
    print(z)
  


r=int(input())
print(3*(r**2))

Verdicts: Runtime Error

Accepted
Problem: It is known that the area of a regular dodecagon inscribed in a circle of radius a is 3a^2.  Given an integer r, find the area of a regular dodecagon inscribed in a circle of radius r.

Submission attempts (in order):
a = int(input())
print(3*(a**2)

a = int(input())
print(3*(a**2))

Verdicts: Runtime Error

Accepted
Problem: It is known that the area of a regular dodecagon inscribed in a circle of radius a is 3a^2.  Given an integer r, find the area of a regular dodecagon inscribed in a circle of radius r.

Submission attempts (in order):
r = int(input())
ans = 3 * (r ^ 2)

r = int(input())
ans = 3 * (r * r)
print(ans)

Verdicts: Wrong Answer

Accepted
Problem: It is known that the area of a regular dodecagon inscribed in a circle of radius a is 3a^2.  Given an integer r, find the area of a regular dodecagon inscribed in a circle of radius r.

Submission attempts (in order):
def add(n):
    return n + 1

if __name__ == "__main__":
    n = int(input())
    buf = input().split()
    a = [int(s) for s in buf]
    
    boxes = [0] * n

    for i in reversed(range(n)):
        print(i)
        pick_boxes = boxes[i::i+1]
        pick_boxes.pop(0)
        print(pick_boxes)
        if pick_boxes:
            if (sum(pick_boxes) % 2) == a[i]:
                boxes[i] = 0
            else:
                boxes[i] = 1
        else:
            boxes[i] = a[i]

    box_name = [i for i, x in enumerate(boxes) if x == 1]
    box_name = list(map(add, box_name))
    print(len(box_name))
    if len(box_name) != 0:
        box_name = ' '.join(map(str,box_name))
        print(box_name)


def add(n):
    return n + 1

if __name__ == "__main__":
    n = int(input())
    buf = input().split()
    a = [int(s) for s in buf]
    
    boxes = [0] * n

    for i in reversed(range(n)):
        pick_boxes = boxes[i::i+1]
        pick_boxes.pop(0)
        if pick_boxes:
            if (sum(pick_boxes) % 2) == a[i]:
                boxes[i] = 0
            else:
                boxes[i] = 1
        else:
            boxes[i] = a[i]

    box_name = [i for i, x in enumerate(boxes) if x == 1]
    box_name = list(map(add, box_name))
    print(len(box_name))
    if len(box_name) != 0:
        box_name = ' '.join(map(str,box_name))
        print(box_name)


r = int(input())

print("{}".format(3*r**2))

Verdicts: Runtime Error

Runtime Error

Accepted
Problem: It is known that the area of a regular dodecagon inscribed in a circle of radius a is 3a^2.  Given an integer r, find the area of a regular dodecagon inscribed in a circle of radius r.

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

print(3*a*a)

a = int(input())

print(3*a*a)

Verdicts: Runtime Error

Accepted
Problem: It is known that the area of a regular dodecagon inscribed in a circle of radius a is 3a^2.  Given an integer r, find the area of a regular dodecagon inscribed in a circle of radius r.

Submission attempts (in order):
n = int(input())
a = [0] + list(map(int, input().split()))
for i in range(n // 2, 0, -1):
    a[i] = sum(a[i::i]) % 2
print(sum(a))
print(*(i for i, b in enumerate(a) if b))

r = int(input())
print(3*r**2)

Verdicts: Runtime Error

Accepted