TnT/Multi_CodeNet4Repair huggingface dataset:
Problem: We have two desks: A and B. Desk A has a vertical stack of N books on it, and Desk B similarly has M books on it.  It takes us A_i minutes to read the i-th book from the top on Desk A (1 \leq i \leq N), and B_i minutes to read the i-th book from the top on Desk B (1 \leq i \leq M).  Consider the following action:    * Choose a desk with a book remaining, read the topmost book on that desk, and remove it from the desk.  How many books can we read at most by repeating this action so that it takes us at most K minutes in total? We ignore the time it takes to do anything other than reading.

Submission attempts (in order):
for i in range(1, N + 1):
    A_ls[i] += A_ls[i - 1]
for i in range(1, M + 1):
    B_ls[i] += B_ls[i - 1]

b_cnt, rst = M, 0
for a_cnt in range(N + 1):
    if A_ls[a_cnt] > K:
        break
    while A_ls[a_cnt] + B_ls[b_cnt] > K:
        b_cnt -= 1
    rst = max(rst, a_cnt + b_cnt)
print(rst)

N, M, K = map(int, input().split(' '))
A_ls = [0] + list(map(int, input().split(' ')))
for i in range(N):
    A_ls[i + 1] += A_ls[i]
B_ls = [0] + list(map(int, input().split(' ')))
for i in range(M):
    B_ls[i + 1] = B_ls[i]

b_cnt, rst = M, 0 
for a_cnt in range(N + 1):
    if A_ls[a_cnt] > K:
        break
    while A_ls[a_cnt] + B_ls[b_cnt] > K and b_cnt >= 0:
        b_cnt -= 1
    rst = max(rst, a_cnt + b_cnt)
print(rst)

N, M, K = map(int, input().split(' '))
A_ls = [0] + list(map(int, input().split(' ')))
for i in range(N):
    A_ls[i + 1] += A_ls[i]
B_ls = [0] + list(map(int, input().split(' ')))
for i in range(M):
    B_ls[i + 1] += B_ls[i]
b_cnt, rst = M, 0
for a_cnt in range(N + 1):
    if A_ls[a_cnt] > K:
        break
    while A_ls[a_cnt] + B_ls[b_cnt] > K and b_cnt >= 0:
        b_cnt -= 1
    rst = max(rst, a_cnt + b_cnt)
print(rst)

Verdicts: Runtime Error

Wrong Answer

Accepted
Problem: We have two desks: A and B. Desk A has a vertical stack of N books on it, and Desk B similarly has M books on it.  It takes us A_i minutes to read the i-th book from the top on Desk A (1 \leq i \leq N), and B_i minutes to read the i-th book from the top on Desk B (1 \leq i \leq M).  Consider the following action:    * Choose a desk with a book remaining, read the topmost book on that desk, and remove it from the desk.  How many books can we read at most by repeating this action so that it takes us at most K minutes in total? We ignore the time it takes to do anything other than reading.

Submission attempts (in order):
N,M,K = map(int,input().split())
A = list(map(int,input().split()))
B = list(map(int,input().split()))
a = [0]
b = [0]
for i in range(N):
  a.append(a[i] + A[i])
for i in range(M):
  b.append(b[i] + B[i])
  
ans = 0
j = M
for i in range(N+1):
  if a[i] > K:
    break
  while b[i] > K - a[i]:
    j -= 1
  ans = max(ans,i+j)
print(ans)

N,M,K = map(int,input().split())
A = list(map(int,input().split()))
B = list(map(int,input().split()))
a = [0]
b = [0]
for i in range(N):
  a.append(a[i] + A[i])
for i in range(M):
  b.append(b[i] + B[i])
  
ans = 0
j = M
for i in range(N+1):
  if a[i] > K:
    break
  while b[j] > K - a[i]:
    j -= 1
  ans = max(ans,i+j)
print(ans)

Verdicts: Runtime Error

Accepted
Problem: We have two desks: A and B. Desk A has a vertical stack of N books on it, and Desk B similarly has M books on it.  It takes us A_i minutes to read the i-th book from the top on Desk A (1 \leq i \leq N), and B_i minutes to read the i-th book from the top on Desk B (1 \leq i \leq M).  Consider the following action:    * Choose a desk with a book remaining, read the topmost book on that desk, and remove it from the desk.  How many books can we read at most by repeating this action so that it takes us at most K minutes in total? We ignore the time it takes to do anything other than reading.

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

a, b = [0], [0]
for i in range(N):
a.append(a[i] + A[i])
for i in range(M):
b.append(b[i] + B[i])

ans, j = 0, M
for i in range(N + 1):
if a[i] > K:
break
while b[j] > K - a[i]:
j -= 1
ans = max(ans, i + j)
print(ans)

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

a, b = [0], [0]
for i in range(N):
 a.append(a[i] + A[i])
for i in range(M):
 b.append(b[i] + B[i])

ans, j = 0, M
for i in range(N + 1):
 if a[i] > K:
  break
 while b[j] > K - a[i]:
  j -= 1
 ans = max(ans, i + j)
print(ans)

Verdicts: Runtime Error

Accepted
Problem: We have two desks: A and B. Desk A has a vertical stack of N books on it, and Desk B similarly has M books on it.  It takes us A_i minutes to read the i-th book from the top on Desk A (1 \leq i \leq N), and B_i minutes to read the i-th book from the top on Desk B (1 \leq i \leq M).  Consider the following action:    * Choose a desk with a book remaining, read the topmost book on that desk, and remove it from the desk.  How many books can we read at most by repeating this action so that it takes us at most K minutes in total? We ignore the time it takes to do anything other than reading.

Submission attempts (in order):
N,M,K = map(int,input().split())
A=list(map(int,input().split()))
B=list(map(int,input().split()))
 
a, b =[0],[0]
for i in range(N):
    a.append(a[i] + A[i])
for i in range(M):
    b.append(b[i] + B[i])
 
ans, j =0,M
for i in range(N+1):
    while b[j] > K-a[i]:
        j -= 1
    ans = max(ans, i+j)
print(ans)

N,M,K = map(int,input().split())
A=list(map(int,input().split()))
B=list(map(int,input().split()))
 
a, b =[0],[0]
for i in range(N):
    a.append(a[i] + A[i])
for i in range(M):
    b.append(b[i] + B[i])
 
ans, j =0,M
for i in range(N+1):
  while b[j] > K-a[i]:
        j -= 1
  ans = max(ans, i+j)
print(ans)

N,M,K = map(int,input().split())
A=list(map(int,input().split()))
B=list(map(int,input().split()))
 
a, b =[0],[0]
for i in range(N):
    a.append(a[i] + A[i])
for i in range(M):
    b.append(b[i] + B[i])
 
ans, j =0,M
for i in range(N+1):
    if a[i] > K:
        break
    while b[j] > K-a[i]:
        j -= 1
    ans = max(ans, i+j)
print(ans)

Verdicts: Runtime Error

Runtime Error

Accepted
Problem: We have two desks: A and B. Desk A has a vertical stack of N books on it, and Desk B similarly has M books on it.  It takes us A_i minutes to read the i-th book from the top on Desk A (1 \leq i \leq N), and B_i minutes to read the i-th book from the top on Desk B (1 \leq i \leq M).  Consider the following action:    * Choose a desk with a book remaining, read the topmost book on that desk, and remove it from the desk.  How many books can we read at most by repeating this action so that it takes us at most K minutes in total? We ignore the time it takes to do anything other than reading.

Submission attempts (in order):
n,m,k = map(int, input().split())
ln = list(map(int, input().split()))
lm = list(map(int, input().split()))
def sub(l,k):
    res = 0
    for i in l:
        if k >= i:
            k -= i
            res += 1
        else:
            return k,res
    return k,res
        
def main(n,m,k,ln,lm):
    nk = sum(ln)
    mk = sum(lm)
    if k>=nk+mk:
        return n+m
    elif k>nk:
        ik, res = sub(lm, k-nk)
        res += n
        startn = n - 1
        startm = res - 1
    else:
        ik, res = sub(ln, k)
        startn = res - 1
        ik, res2 = sub(lm, ik)
        res += res2
        startm = res2
        
    ires = res
    for i in range(startn,-1,-1):
        ik += ln[i]
        ik, delta = sub(lm[startm:], ik)
        startm += delta
        ires = ires - 1 + delta
        res = max(res, ires)
    return res

main(n,m,k,ln,lm)

n,m,k = map(int, input().split())
ln = list(map(int, input().split()))
lm = list(map(int, input().split()))

def csum(l,ll):
    s = 0
    res = [0] * ll
    for i in range(ll):
        s += l[i]
        res[i] = s
    return res

csln = csum(ln,n)
cslm = csum(lm,m)

import numpy as np

lk = [k-x for x in csln if k-x >=0]
res = np.searchsorted(np.array(cslm), np.array(lk), side='right').
rmax = 0
for i,j in zip(range(1,len(lk)+1), res):
    rmax = max(rmax, i+j)
print(int(rmax))

n,m,k = map(int, input().split())
ln = list(map(int, input().split()))
lm = list(map(int, input().split()))

def csum(l,ll):
    s = 0
    res = [0] * ll
    for i in range(ll):
        s += l[i]
        res[i] = s
    return res

csln = csum(ln,n)
cslm = csum(lm,m)

import numpy as np

lk = [k] + [k-x for x in csln if k-x >=0]
res = np.searchsorted(np.array(cslm), np.array(lk), side='right')
rmax = 0
for i,j in zip(range(len(lk)+1), res):
    rmax = max(rmax, i+j)
print(rmax)

Verdicts: Wrong Answer

Runtime Error

Accepted
Problem: We have two desks: A and B. Desk A has a vertical stack of N books on it, and Desk B similarly has M books on it.  It takes us A_i minutes to read the i-th book from the top on Desk A (1 \leq i \leq N), and B_i minutes to read the i-th book from the top on Desk B (1 \leq i \leq M).  Consider the following action:    * Choose a desk with a book remaining, read the topmost book on that desk, and remove it from the desk.  How many books can we read at most by repeating this action so that it takes us at most K minutes in total? We ignore the time it takes to do anything other than reading.

Submission attempts (in order):
n,m,k=map(int,input().split())
a=list(map(int,input().split()))
b=list(map(int,input().split()))
a.insert(0,0)
b.insert(0,0)
j=1
while j<=m and b[j]+b[j-1]<=k:
  b[j]+=b[j-1]
  j+=1
  

j-=1
books=j
i=1
while i<n+1:
  while j>0 and a[i]+b[j]>k:
    j-=1
  if(j>=0) and a[i]+b[j]<=k:
    books=max(books,j+i)
    a[i]+=a[i-1]
    i+=1
  else:
    break
print(books)
  


n,m,k=map(int,input().split())
a=list(map(int,input().split()))
b=list(map(int,input().split()))
a.insert(0,0)
b.insert(0,0)
j=1
while j<=m and b[j]+b[j-1]<=k:
  b[j]+=b[j-1]
  j+=1
  

j-=1
books=j
i=1
while i<n:
  while j>0 and a[i]+b[j]>k:
    j-=1
  if(j>=0) and a[i]+b[j]<=k:
    books=max(books,j+i)
    a[i]+=a[i-1]
  else:
    break
print(books)
  

n,m,k=map(int,input().split())
a=list(map(int,input().split()))
b=list(map(int,input().split()))
a.insert(0,0)
b.insert(0,0)
j=1
while j<=m and b[j]+b[j-1]<=k:
  b[j]+=b[j-1]
  j+=1
  

j-=1
books=j
#print(books,b)

i=1
while i<n+1:
  a[i]+=a[i-1]
  while j>=0 and a[i]+b[j]>k:
    j-=1
  if(j!=-1) and a[i]+b[j]<=k:
    books=max(books,j+i)
    
    
    i+=1
    #print(books,b[:j+1],a[:i])
  else:
    break
print(books)
  

Verdicts: Wrong Answer

Wrong Answer

Accepted
Problem: We have two desks: A and B. Desk A has a vertical stack of N books on it, and Desk B similarly has M books on it.  It takes us A_i minutes to read the i-th book from the top on Desk A (1 \leq i \leq N), and B_i minutes to read the i-th book from the top on Desk B (1 \leq i \leq M).  Consider the following action:    * Choose a desk with a book remaining, read the topmost book on that desk, and remove it from the desk.  How many books can we read at most by repeating this action so that it takes us at most K minutes in total? We ignore the time it takes to do anything other than reading.

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

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


a_pre,b_pre=[0],[0]

for i in range(n):
	a_pre.append(a_pre[i]+a[i])

for i in range(m):
	b_pre.append(b_pre[i]+b[i])

ans,j=0,m
for i in range(n+1):
	if a_pre[i]>k:
		break
	while b[j]>k-a[i]:
		j-=1

	ans=max(ans,i+j)

print(ans)


n,m,k=map(int,input().split())

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


a_pre,b_pre=[0],[0]

for i in range(n):
	a_pre.append(a_pre[i]+a[i])

for i in range(m):
	b_pre.append(b_pre[i]+b[i])

ans,j=0,m
for i in range(n+1):
	if a_pre[i]>k:
		break
	while b_pre[j]>k-a_pre[i]:
		j-=1

	ans=max(ans,i+j)

print(ans)


Verdicts: Runtime Error

Accepted
Problem: We have two desks: A and B. Desk A has a vertical stack of N books on it, and Desk B similarly has M books on it.  It takes us A_i minutes to read the i-th book from the top on Desk A (1 \leq i \leq N), and B_i minutes to read the i-th book from the top on Desk B (1 \leq i \leq M).  Consider the following action:    * Choose a desk with a book remaining, read the topmost book on that desk, and remove it from the desk.  How many books can we read at most by repeating this action so that it takes us at most K minutes in total? We ignore the time it takes to do anything other than reading.

Submission attempts (in order):
n,m,k=map(int,input().split())
A=list(map(int,input().split()))
B=list(map(int,input().split()))

a,b=[0],[0]
for i in range(n):
    a.append(a[i]+A[i])
for i in range(m):
    b.append(b[i]+B[i])
print(a)
print(b)
ans,j=0,m
for i in range(n+1):
    if a[i]>k:
        break
    while b[j]>k-a[i]:
        j-=1
    ans=max(ans,i+j)
print(ans)

n,m,k=map(int,input().split())
A=list(map(int,input().split()))
B=list(map(int,input().split()))

a,b=[0],[0]
for i in range(n):
    a.append(a[i]+A[i])
for i in range(m):
    b.append(b[i]+B[i])
ans,j=0,m
for i in range(n+1):
    if a[i]>k:
        break
    while b[j]>k-a[i]:
        j-=1
    ans=max(ans,i+j)
print(ans)

Verdicts: Wrong Answer

Accepted
Problem: We have two desks: A and B. Desk A has a vertical stack of N books on it, and Desk B similarly has M books on it.  It takes us A_i minutes to read the i-th book from the top on Desk A (1 \leq i \leq N), and B_i minutes to read the i-th book from the top on Desk B (1 \leq i \leq M).  Consider the following action:    * Choose a desk with a book remaining, read the topmost book on that desk, and remove it from the desk.  How many books can we read at most by repeating this action so that it takes us at most K minutes in total? We ignore the time it takes to do anything other than reading.

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

a=0;b=0
cnt=0
while a < len(A) and b < len(B) and K > 0:
	a_ = A[a]
	b_ = B[b]
    
    if a_ > K and b_ > K:
      break
      
    if a_ < b_ and a_ <= K:
		cnt += 1
		K -= a_
		A.pop(0)
	elif a_ >= b_ and b_ <= K:
      cnt += 1
      K -= b_
      B.pop(0)
      
print(cnt)   
      

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

a=0;b=0
cnt=0
while a < len(A) and b < len(B) and K > 0:
  a_ = A[a]
  b_ = B[b]
    
  if a_ > K and b_ > K:
    break
      
  if a_ < b_ and a_ <= K:
    cnt += 1
	K -= a_
    a += 1
  elif a_ >= b_ and b_ <= K:
    cnt += 1
    K -= b_
    b += 1
      
print(cnt)  


a=0;b=0
cnt=0
while a < len(A) and b < len(B) and K > 0:
	a_ = A[a]
	b_ = B[b]
    
    if a_ > K and b_ > K:
      break
      
    if a_ < b_ and a_ <= K:
		cnt += 1
		K -= a_
        a += 1
	elif a_ >= b_ and b_ <= K:
      cnt += 1
      K -= b_
      b += 1
      
print(cnt)  

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

a=0;b=0
cnt=0
while a < len(A) and b < len(B) and K > 0:
    a_ = A[a]
    b_ = B[b]
    print(a_, b_,K)
    
    if a_ > K and b_ > K:
        break
      
    if a_ < b_ and a_ <= K:
        cnt += 1
        K -= a_
        a += 1
    elif a_ >= b_ and b_ <= K:
        cnt += 1
        K -= b_
        b += 1
    
print(cnt)

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

A = [0]
B = [0]
ans = 0

for i in range(n):
    A.append(A[i] + a[i])

for i in range(m):
    B.append(B[i] + b[i])

j = m

for i in range(n + 1):
    if A[i] > k:
        break
    while A[i] + B[j] > k:
        j -= 1
    ans = max(ans, i + j)

print(ans)

Verdicts: Runtime Error

Runtime Error

Runtime Error

Wrong Answer

Accepted
Problem: We have two desks: A and B. Desk A has a vertical stack of N books on it, and Desk B similarly has M books on it.  It takes us A_i minutes to read the i-th book from the top on Desk A (1 \leq i \leq N), and B_i minutes to read the i-th book from the top on Desk B (1 \leq i \leq M).  Consider the following action:    * Choose a desk with a book remaining, read the topmost book on that desk, and remove it from the desk.  How many books can we read at most by repeating this action so that it takes us at most K minutes in total? We ignore the time it takes to do anything other than reading.

Submission attempts (in order):
n,m,k=map(int,input().split())
A=list(map(int,input().split()))
B=list(map(int,input().split()))

a=[0]
b=[0]

for i in range(n):
    a.append(a[i]+A[i])
for i in range(m):
    b.append(b[i]+B[i])
    
c=0
idx=m

for i in range(n+1):
    while 1:
        if idx<0:
            break
        if a[i]+b[idx]>k:
            print(a[i]+b[idx])
            idx-=1
            continue
        else:
            c=max(i+idx,c)
            break
else:
    print(c)

n,m,k=map(int,input().split())
A=list(map(int,input().split()))
B=list(map(int,input().split()))

a=[0]
b=[0]

for i in range(n):
    a.append(a[i]+A[i])
for i in range(m):
    b.append(b[i]+B[i])
    
c=0
idx=m

for i in range(n+1):
    while 1:
        if idx<0:
            break
        if a[i]+b[idx]>k:
            idx-=1
            continue
        else:
            c=max(i+idx,c)
            break
else:
    print(c)

Verdicts: Wrong Answer

Accepted
Problem: We have two desks: A and B. Desk A has a vertical stack of N books on it, and Desk B similarly has M books on it.  It takes us A_i minutes to read the i-th book from the top on Desk A (1 \leq i \leq N), and B_i minutes to read the i-th book from the top on Desk B (1 \leq i \leq M).  Consider the following action:    * Choose a desk with a book remaining, read the topmost book on that desk, and remove it from the desk.  How many books can we read at most by repeating this action so that it takes us at most K minutes in total? We ignore the time it takes to do anything other than reading.

Submission attempts (in order):
# -*- coding: utf-8 -*-


def main():
    n, m, k = map(int, input().split())
    A = list(map(int, input().split()))
    B = list(map(int, input().split()))
    a = A[0]
    A2 = [a]
    b = B[0]
    B2 = [b]
    for i in range(1, n):
        a += A[i]
        A2.append(a)

    for i in range(1, m):
        b += B[i]
        B2.append(b)
    max = -2
    for ia, aa in enumerate(A2):
        for ib, bb in enumerate(B2):
            if aa + bb > k:
                break
            if ia + ib > max:

                max = ia + ib
    print(max + 2)


def answer():
    N, M, K = map(int, input().split())


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

    a, b = [0], [0]
    for i in range(N):
        a.append(a[i] + A[i])
        for i in range(M):
            b.append(b[i] + B[i])

        ans, j = 0, M
        for i in range(N + 1):
        if a[i] > K:
        break
        while b[j] > K - a[i]:
        j -= 1
        ans = max(ans, i + j)
        print(ans)
if __name__ == '__main__':
    answer()

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




#     A = list(map(int, input().split()))
#     B = list(map(int, input().split()))
#     a = A[0]
#     A2 = [a]
#     b = B[0]
#     B2 = [b]

#         a += A[i]
#         A2.append(a)
# 

#         b += B[i]
#         B2.append(b)
#     max = -2

#         for ib, bb in enumerate(B2):

#                 break
#             if ia + ib > max:
# 
#                 max = ia + ib
#     print(max + 2)
# 

def answer():
    N, M, K = map(int, input().split())


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

    a, b = [0], [0]
    for i in range(N):
        a.append(a[i] + A[i])
    for i in range(M):
        b.append(b[i] + B[i])

    ans, j = 0, M
    for i in range(N + 1):
        if a[i] > K:
            break
        while b[j] > K - a[i]:
            j -= 1
        ans = max(ans, i + j)
    print(ans)
if __name__ == '__main__':
    answer()

Verdicts: Runtime Error

Accepted
Problem: We have two desks: A and B. Desk A has a vertical stack of N books on it, and Desk B similarly has M books on it.  It takes us A_i minutes to read the i-th book from the top on Desk A (1 \leq i \leq N), and B_i minutes to read the i-th book from the top on Desk B (1 \leq i \leq M).  Consider the following action:    * Choose a desk with a book remaining, read the topmost book on that desk, and remove it from the desk.  How many books can we read at most by repeating this action so that it takes us at most K minutes in total? We ignore the time it takes to do anything other than reading.

Submission attempts (in order):
n,m,l = map(int,input().split(" "))
Ns = input().split(" ")
N = [int (n) for n in Ns]
Ms = input().split(" ")
M = [int(n) for n in Ms]
#print(n,m,l)
#print(N,M)
i = 0
j = 0
total = 0
cnt = 0
while total <= l:
  if  N[i] <= M[j]:
    total = total + N[i]
    i = i+1
    cnt = cnt +1
  else:
    total = total + M[j]
    j = j+1
    cnt = cnt +1
print(cnt)

N,M,K = list(map(int,input().split()))
A = list(map(int,input().split()))
B = list(map(int,input().split()))
a,b = [0],[0]
for i in range(N):
  a.append(a[i]+A[i])
for j in range(M):
  b.append(b[j]+B[j])
ans = 0
#print(a,b,N+1)
j = M
for i in range(N+1):
  if a[i] > K:
    break

  while K - a[i] < b[j]:
    j = j-1
  ans = max(ans,i+j)

print(ans)

Verdicts: Runtime Error

Accepted
Problem: We have two desks: A and B. Desk A has a vertical stack of N books on it, and Desk B similarly has M books on it.  It takes us A_i minutes to read the i-th book from the top on Desk A (1 \leq i \leq N), and B_i minutes to read the i-th book from the top on Desk B (1 \leq i \leq M).  Consider the following action:    * Choose a desk with a book remaining, read the topmost book on that desk, and remove it from the desk.  How many books can we read at most by repeating this action so that it takes us at most K minutes in total? We ignore the time it takes to do anything other than reading.

Submission attempts (in order):
import bisect
N,M,K=list(map(int,input().split()))
A=list(map(int,input().split()))
B=list(map(int,input().split()))
bka=[0]*N
bkb=[0]*M
bka[0]=A[0]
bkb[0]=B[0]

for i in range(1,N):
    bka[i]+=bka[i-1]+A[i]
for j in range(1,M):
    bkb[j]+=bkb[j-1]+B[j]
ans=bisect.bisect_left(bka,K)
for i in range(N):
    count=i+1
    k=K-bka[i]
    if k<0:continue
    c = bisect.bisect_left(bkb,k)
    if c!=M:
        if bkb[c]==k:
            c+=count+1
    else:c+=count

    # print(bkb,k,c,count)
    ans=max(ans,c)
print(ans)

import bisect
N,M,K=list(map(int,input().split()))
A=list(map(int,input().split()))
B=list(map(int,input().split()))
bka=[0]*N
bkb=[0]*M
bka[0]=A[0]
bkb[0]=B[0]

for i in range(1,N):
    bka[i]+=bka[i-1]+A[i]
for j in range(1,M):
    bkb[j]+=bkb[j-1]+B[j]
ans=0
c=bisect.bisect_left(bkb,K)
if c != M:
    if bkb[c] == K:
        c +=+ 1
ans = max(ans, c)
for i in range(N):
    count=i+1
    k=K-bka[i]
    if k<0:continue
    c = bisect.bisect_left(bkb,k)
    b=c
    if c!=M:
        if bkb[c]==k:
            c+=count+1
        else:c+=count
    else:c+=count
    ans=max(ans,c)
print(ans)

Verdicts: Wrong Answer

Accepted
Problem: We have two desks: A and B. Desk A has a vertical stack of N books on it, and Desk B similarly has M books on it.  It takes us A_i minutes to read the i-th book from the top on Desk A (1 \leq i \leq N), and B_i minutes to read the i-th book from the top on Desk B (1 \leq i \leq M).  Consider the following action:    * Choose a desk with a book remaining, read the topmost book on that desk, and remove it from the desk.  How many books can we read at most by repeating this action so that it takes us at most K minutes in total? We ignore the time it takes to do anything other than reading.

Submission attempts (in order):
#n = int(input())
n,m,k = [int(z) for z in input().split()]
a = [int(z) for z in input().split()]
b = [int(z) for z in input().split()]
time = 0
i = j = 0
while (time<=k and i<n and j<m):
      if (a[i]>b[j]):
            if (time+a[i]<=k):
                  time += a[i]
                  i+=1
            else:
                  break
      else:
            if (time+b[j]<=k):
                  time += b[j]
                  j+=1
            else:
                  break
while (i<n and time<=k):
      if (time+a[i]<=k):
                  time += a[i]
                  i+=1
      else:
            break
while (j<n and time<=k):
      if (time+b[j]<=k):
                  time += b[j]
                  j+=1
      else:
            break
print(i+j)


import sys
import bisect
import itertools
import collections
import fractions
import heapq
import math
from operator import mul
from functools import reduce
from functools import lru_cache


def solve():
    readline = sys.stdin.buffer.readline
    mod = 10 ** 9 + 7
    N, M, K = map(int, readline().split())
    A = list(map(int, readline().split()))
    B = list(map(int, readline().split()))


    nowtime = 0

    Asum = [0]
    Bsum = [0]
    tmp = 0
    for a in A:
        tmp += a
        Asum.append(tmp)
    tmp = 0
    for b in B:
        tmp += b
        Bsum.append(tmp)

    maxbooks = 0
    before = M
    bcount = M
    for acount in range(N+1):
        if Asum[acount] > K:
            continue
        while K < Asum[acount] + Bsum[bcount]:
            bcount -= 1
        maxbooks = max(maxbooks, acount+bcount)

    print(maxbooks)

if __name__ == '__main__':
    solve()


Verdicts: Runtime Error

Accepted
Problem: We have two desks: A and B. Desk A has a vertical stack of N books on it, and Desk B similarly has M books on it.  It takes us A_i minutes to read the i-th book from the top on Desk A (1 \leq i \leq N), and B_i minutes to read the i-th book from the top on Desk B (1 \leq i \leq M).  Consider the following action:    * Choose a desk with a book remaining, read the topmost book on that desk, and remove it from the desk.  How many books can we read at most by repeating this action so that it takes us at most K minutes in total? We ignore the time it takes to do anything other than reading.

Submission attempts (in order):
# import numpy as np
# import math
# import copy
# from collections import deque
import sys
input = sys.stdin.readline



def main():
    N,M,K = map(int,input().split())
    A = list(map(int,input().split()))
    B = list(map(int,input().split()))

    a = 0
    b = 0

    time = 0
    res = 0

    isAexist = True
    isBexist = True

    for i in range(N+M+1):
        if isAexist:
            tA = A[a]
        else:
            tA = 10 ** 9 + 1
        if isBexist:
            tB = B[b]
        else:
            tB = 10 ** 9 + 1
        if (tA <= tB) & isAexist:
            if time + tA <= K:
                time += tA
                a += 1
                res += 1
                if a == N:
                    isAexist = False
        elif (tA >= tB) & isBexist:
            if time + tB <= K:
                time += tB
                b += 1
                res += 1
                if b == M:
                    isBexist = False
        print(res,time,a,b)
        if isAexist:
            if time + A[a] > K:
                isAexist = False
        if isBexist:
            if time + B[b] > K:
                isBexist = False
        if (not isAexist) & (not isBexist):
            break

    print(res)



main()


# import numpy as np
# import math
# import copy
# from collections import deque
import sys
input = sys.stdin.readline



def main():
    N,M,K = map(int,input().split())
    A = list(map(int,input().split()))
    B = list(map(int,input().split()))

    sumA = [0] * (N+1)
    sumB = [0] * (M+1)

    for i in range(1,N+1):
        sumA[i] = sumA[i-1] + A[i-1]

    for i in range(1,M+1):
        sumB[i] = sumB[i-1] + B[i-1]

    for i in range(M,-1,-1):
        if sumB[i] <= K:
            endB = i
            break

    for i in range(N,-1,-1):
        if sumA[i] <= K:
            endA = i
            break

    res = 0

    for i in range(0,endA+1):
        for j in range(endB,-1,-1):
            if sumA[i] + sumB[j] <= K:
                endB = j
                res = max(res,i+j)
                break

    print(res)



main()


Verdicts: Wrong Answer

Accepted
Problem: We have two desks: A and B. Desk A has a vertical stack of N books on it, and Desk B similarly has M books on it.  It takes us A_i minutes to read the i-th book from the top on Desk A (1 \leq i \leq N), and B_i minutes to read the i-th book from the top on Desk B (1 \leq i \leq M).  Consider the following action:    * Choose a desk with a book remaining, read the topmost book on that desk, and remove it from the desk.  How many books can we read at most by repeating this action so that it takes us at most K minutes in total? We ignore the time it takes to do anything other than reading.

Submission attempts (in order):
n, m, k = map(int, input().split())
A = list(map(int, input().split()))
B = list(map(int, input().split()))

a_max = 0
sum = 0
for i in range(n):
  sum += A[i]
  if sum > k:
    sum -= A[i]
    break
  a_max += 1
  
ans = a_max
kotae =ans
j = 0

print(a_max, 0)
for i in range(m):
  sum += B[i]
  ans += 1
  while sum > k and j < a_max:
    sum -= A[a_max-1-j]
    ans -= 1
    j += 1
  if sum > k:
    print(kotae)
    exit()
  kotae = max(kotae, ans)
  if a_max-j == 0:
    break
  
print(kotae)
  
    
    
  

n, m, k = map(int, input().split())
A = list(map(int, input().split()))
B = list(map(int, input().split()))

a_max = 0
sum = 0
for i in range(n):
  sum += A[i]
  if sum > k:
    sum -= A[i]
    break
  a_max += 1
  
ans = a_max
kotae =ans
j = 0


for i in range(m):
  sum += B[i]
  ans += 1
  while sum > k and j < a_max:
    sum -= A[a_max-1-j]
    ans -= 1
    j += 1
  if sum > k:
    print(kotae)
    exit()
  kotae = max(kotae, ans)

  
print(kotae)
  
    
    
  

Verdicts: Wrong Answer

Accepted
Problem: We have two desks: A and B. Desk A has a vertical stack of N books on it, and Desk B similarly has M books on it.  It takes us A_i minutes to read the i-th book from the top on Desk A (1 \leq i \leq N), and B_i minutes to read the i-th book from the top on Desk B (1 \leq i \leq M).  Consider the following action:    * Choose a desk with a book remaining, read the topmost book on that desk, and remove it from the desk.  How many books can we read at most by repeating this action so that it takes us at most K minutes in total? We ignore the time it takes to do anything other than reading.

Submission attempts (in order):
N, M, K = map(int,input().split())
A = list(map(int,input().split()))
B = list(map(int,input().split()))
SUM_A=[0 for _ in range(N+1)]
for i in range(N):
    SUM_A[i+1] = A[i] + SUM_A[i]
    if SUM_A[i+1] > K:
        N=i
        break

SUM_B=[0 for _ in range(M+1)]
for i in range(M):
    SUM_B[i+1] = B[i] + SUM_B[i]
    if SUM_B[i+1] > K:
        M=i
        break

print(N,M)
cnt = 0
for i in range(N+1):
    for j in range(M+1):
        if SUM_A[i]+SUM_B[j] <= K:
            if i+j > cnt:
                cnt = i+j
        else:
            break
print(cnt)

N, M, K = map(int,input().split())
A = list(map(int,input().split()))
B = list(map(int,input().split()))
SUM_A=[0 for _ in range(N+1)]
for i in range(N):
    SUM_A[i+1] = A[i] + SUM_A[i]
    if SUM_A[i+1] > K:
        N=i
        break

SUM_B=[0 for _ in range(M+1)]
for i in range(M):
    SUM_B[i+1] = B[i] + SUM_B[i]
    if SUM_B[i+1] > K:
        M=i
        break

cnt = N
for i in range(N+1)[::-1]:
    V=K-SUM_A[i]
    j = cnt-i
    while j < M+1 and SUM_B[j] <= V:
        j += 1
    j-=1
    if i+j > cnt:
        cnt = i+j
print(cnt)

Verdicts: Wrong Answer

Accepted
Problem: We have two desks: A and B. Desk A has a vertical stack of N books on it, and Desk B similarly has M books on it.  It takes us A_i minutes to read the i-th book from the top on Desk A (1 \leq i \leq N), and B_i minutes to read the i-th book from the top on Desk B (1 \leq i \leq M).  Consider the following action:    * Choose a desk with a book remaining, read the topmost book on that desk, and remove it from the desk.  How many books can we read at most by repeating this action so that it takes us at most K minutes in total? We ignore the time it takes to do anything other than reading.

Submission attempts (in order):
import sys
input = sys.stdin.readline
from itertools import accumulate

N,M,K = map(int,input().split())
A = list(map(int,input().split()))
B = list(map(int,input().split()))
a = [0]+list(accumulate(A))
b = [0]+list(accumulate(B))
ans,j = 0,M
for i in range(N+1):
    if i > K:
        break
    while b[j] > K - a[i]:
        j -= 1
    ans = max(ans,i+j)
print(ans)

import sys
from itertools import accumulate
input = sys.stdin.readline

N,M,K=map(int,input().split())
A = list(map(int,input().split()))
B = list(map(int,input().split()))
A_acc = list(accumulate(A))
B_acc = list(accumulate(B))
a = 0
b = 0
ans = 0
deep = ""
for n,z in enumerate(zip(A_acc,B_acc)):
    i,j = z
    if i < j and i <= K:
        deep = 'A'
    elif j <= K:
        deep = 'B'
    else:
        break
    ans += 1
if deep == 'A':
    K -= A_acc[n]
    s = B
else:
    K -= B_acc[n]
    s = A
for i in s:
    if K >= i:
        ans += 1
        K -= i
    else:
        break
print(ans)

import sys
input = sys.stdin.readline
from itertools import accumulate

N,M,K = map(int,input().split())
A = list(map(int,input().split()))
B = list(map(int,input().split()))
a = [0]+list(accumulate(A))
b = [0]+list(accumulate(B))
ans,j = 0,M
for i in range(N+1):
    if a[i] > K:
        break
    while b[j] > K - a[i]:
        j -= 1
    ans = max(ans,i+j)
print(ans)

Verdicts: Runtime Error

Wrong Answer

Accepted
Problem: We have two desks: A and B. Desk A has a vertical stack of N books on it, and Desk B similarly has M books on it.  It takes us A_i minutes to read the i-th book from the top on Desk A (1 \leq i \leq N), and B_i minutes to read the i-th book from the top on Desk B (1 \leq i \leq M).  Consider the following action:    * Choose a desk with a book remaining, read the topmost book on that desk, and remove it from the desk.  How many books can we read at most by repeating this action so that it takes us at most K minutes in total? We ignore the time it takes to do anything other than reading.

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

a, b = [0], [0]
for i in range(N):
    a.append(a[i] + A[i])
for i in range(M):
    b.append(b[i] + B[i])

ans, j = 0, M
for i in range(N + 1):
    if a[i] > K:
        break
while b[j] > K - a[i]:
    j -= 1
    ans = max(ans, i + j)
print(ans)


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

a, b = [0], [0]
for i in range(N):
a.append(a[i] + A[i])
for i in range(M):
b.append(b[i] + B[i])

ans, j = 0, M
for i in range(N + 1):
    if a[i] > K:
        break
while b[j] > K - a[i]:
    j -= 1
    ans = max(ans, i + j)
print(ans)

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

a, b = [0], [0]
for i in range(N):
    a.append(a[i] + A[i])
for i in range(M):
    b.append(b[i] + B[i])

ans, j = 0, M
for i in range(N + 1):
    if a[i] > K:
        break
    while b[j] > K - a[i]:
        j -= 1
    ans = max(ans, i + j)
print(ans)

Verdicts: Runtime Error

Runtime Error

Accepted
Problem: We have two desks: A and B. Desk A has a vertical stack of N books on it, and Desk B similarly has M books on it.  It takes us A_i minutes to read the i-th book from the top on Desk A (1 \leq i \leq N), and B_i minutes to read the i-th book from the top on Desk B (1 \leq i \leq M).  Consider the following action:    * Choose a desk with a book remaining, read the topmost book on that desk, and remove it from the desk.  How many books can we read at most by repeating this action so that it takes us at most K minutes in total? We ignore the time it takes to do anything other than reading.

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

a, b = [0], [0]
for i in range(N):
a.append(a[i] + A[i])
for i in range(M):
b.append(b[i] + B[i])

ans, j = 0, M
for i in range(N + 1):
if a[i] > K:
break
while b[j] > K - a[i]:
j -= 1
ans = max(ans, i + j)
print(ans)

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

a, b = [0], [0]
for i in range(N):
  a.append(a[i] + A[i])
for i in range(M):
  b.append(b[i] + B[i])

ans, j = 0, M
for i in range(N + 1):
  if a[i] > K:
    break
  while b[j] > K - a[i]:
    j -= 1
  ans = max(ans, i + j)
print(ans)


Verdicts: Runtime Error

Accepted
Problem: We have two desks: A and B. Desk A has a vertical stack of N books on it, and Desk B similarly has M books on it.  It takes us A_i minutes to read the i-th book from the top on Desk A (1 \leq i \leq N), and B_i minutes to read the i-th book from the top on Desk B (1 \leq i \leq M).  Consider the following action:    * Choose a desk with a book remaining, read the topmost book on that desk, and remove it from the desk.  How many books can we read at most by repeating this action so that it takes us at most K minutes in total? We ignore the time it takes to do anything other than reading.

Submission attempts (in order):
import numpy as np
from collections import deque
def main():
    N, M, K = map(int, input().split())
    A = deque(list(map(int, input().split())))
    B = deque(list(map(int, input().split())))
    
    cost = K
    count = 0
    while True:
        if len(A) ==0 and len(B) == 0:
            break
        elif len(A) == 0:
            cost -= B.popleft()
        elif len(B) == 0:
            cost -= A.popleft()

        else:
            a = A[0] + A[1]
            b = B[0]
            if a < b:
                cost -= A.popleft()
            else:
                cost -= B.popleft()

        if cost < 0:
            break
        count += 1
    
    print(count)


if __name__ == '__main__':
    main()


import numpy as np
def main():
    N, M, K = map(int, input().split())
    A = list(map(int, input().split()))
    B = list(map(int, input().split()))
    
    ia = 0
    ib = 0
    cost = 0
    while True:
        if ia < len(A) and ib < len(B):
            if A[ia] < B[ia]:
                if cost + A[ia] > K:
                    break
                ia += 1
                cost += A[ia]
            else:
                if cost + B[ib] > K:
                    break
                ib += 1
                cost += B[ib]
        elif ia < len(A):
            if cost + A[ia] > K:
                break
            ia += 1
            cost += A[ia]
        elif ib < len(B):
            if cost + B[ib] > K:
                break
            ib += 1
            cost += B[ib]
        else:
            break
    
    count = ia + ib
    print(count)


if __name__ == '__main__':
    main()


from bisect import bisect_right
import numpy as np
def main():
    N, M, K = map(int, input().split())
    A = np.array([0] + list(map(int, input().split())), dtype=np.int64)
    B = np.array([0] + list(map(int, input().split())), dtype=np.int64)

    As = np.cumsum(A)
    Bs = np.cumsum(B)

    count = 0

    for i, a in enumerate(As):
        if a > K:
            break
        j = bisect_right(Bs, K-a)
        
        count = max(count, (i+j-1))

    print(count)

if __name__ == '__main__':
    main()

Verdicts: Runtime Error

Runtime Error

Accepted
Problem: We have two desks: A and B. Desk A has a vertical stack of N books on it, and Desk B similarly has M books on it.  It takes us A_i minutes to read the i-th book from the top on Desk A (1 \leq i \leq N), and B_i minutes to read the i-th book from the top on Desk B (1 \leq i \leq M).  Consider the following action:    * Choose a desk with a book remaining, read the topmost book on that desk, and remove it from the desk.  How many books can we read at most by repeating this action so that it takes us at most K minutes in total? We ignore the time it takes to do anything other than reading.

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

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

a_, b_ = [0], [0]

for i in range(n):
    a_.append(a_[i] + a[i])
for j in range(m):
    b_.append(b_[j] + b[j])

ans = 0
for i in range(n+1):
    
    idx2 = m
    while a_[i] + b_[idx2] > k:
        idx2 -= 1
    
    ans = max(ans, i + idx2)
print(ans)

n, m, k = map(int, input().split())

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

a_, b_ = [0], [0]

for i in range(n):
    a_.append(a_[i] + a[i])
for i in range(m):
    b_.append(b_[i] + b[i])

ans = 0
j = m

for i in range(n+1):
    
    if a_[i] > k:
        break
    while b_[j] > k - a_[i]:
        j -= 1
    ans = max(ans, i + j)
print(ans)

Verdicts: Runtime Error

Accepted
Problem: We have two desks: A and B. Desk A has a vertical stack of N books on it, and Desk B similarly has M books on it.  It takes us A_i minutes to read the i-th book from the top on Desk A (1 \leq i \leq N), and B_i minutes to read the i-th book from the top on Desk B (1 \leq i \leq M).  Consider the following action:    * Choose a desk with a book remaining, read the topmost book on that desk, and remove it from the desk.  How many books can we read at most by repeating this action so that it takes us at most K minutes in total? We ignore the time it takes to do anything other than reading.

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

A = [0], B = [0]
for i in range(n):
    A.append(A[i] + a[i])
for i in range(m):
    B.append(B[i] + b[i])

ans, j = 0, m
for i in range(n+1):
    if A[i] > k:
        break
    while B[j] > k - A[i]:
        j -= 1
    ans = max(ans, i+j)
print(ans)

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

A, B = [0], [0]
for i in range(n):
    A.append(A[i] + a[i])
for i in range(m):
    B.append(B[i] + b[i])

ans, j = 0, m
for i in range(n+1):
    if A[i] > k:
        break
    while B[j] > k - A[i]:
        j -= 1
    ans = max(ans, i+j)
print(ans)


Verdicts: Runtime Error

Accepted
Problem: We have two desks: A and B. Desk A has a vertical stack of N books on it, and Desk B similarly has M books on it.  It takes us A_i minutes to read the i-th book from the top on Desk A (1 \leq i \leq N), and B_i minutes to read the i-th book from the top on Desk B (1 \leq i \leq M).  Consider the following action:    * Choose a desk with a book remaining, read the topmost book on that desk, and remove it from the desk.  How many books can we read at most by repeating this action so that it takes us at most K minutes in total? We ignore the time it takes to do anything other than reading.

Submission attempts (in order):
from itertools import accumulate
from bisect import bisect_right
N, M, K = map(int, input().split())
A = [0] + list(map(int, input().split()))
B = list(map(int, input().split()))
SA = list(accumulate(A))
SB = list(accumulate(B))

ans = 0
for i in range(len(SA)):
    if tmp := K-SA[i] < 0:
        break
    ans = max(ans, bisect_right(SB, tmp))
print(ans)

from itertools import accumulate
from bisect import bisect_right
N, M, K = map(int, input().split())
A = [0] + list(map(int, input().split()))
B = list(map(int, input().split()))
SA = list(accumulate(A))
SB = list(accumulate(B))

ans = 0
for i in range(len(SA)):
    ans = max(ans, bisect_right(SB, K-SA[i]))
print(ans)

from itertools import accumulate
from bisect import bisect_right
N, M, K = map(int, input().split())
A = [0] + list(map(int, input().split()))
B = list(map(int, input().split()))
SA = list(accumulate(A))
SB = list(accumulate(B))

ans = 0
for i in range(len(SA)):
    if tmp := K-SA[i] < 0:
        break
    ans = max(ans, bisect_right(SB, tmp)+i)
print(ans)

from itertools import accumulate
from bisect import bisect_right
N, M, K = map(int, input().split())
A = [0] + list(map(int, input().split()))
B = list(map(int, input().split()))
SA = list(accumulate(A))
SB = list(accumulate(B))

ans = 0
for n in range(len(SA)):
    tmp = K-SA[n]
    if tmp < 0:
        break
    m = bisect_right(SB, tmp)
    ans = max(ans, n+m)
print(ans)

Verdicts: Wrong Answer

Wrong Answer

Wrong Answer

Accepted
Problem: We have two desks: A and B. Desk A has a vertical stack of N books on it, and Desk B similarly has M books on it.  It takes us A_i minutes to read the i-th book from the top on Desk A (1 \leq i \leq N), and B_i minutes to read the i-th book from the top on Desk B (1 \leq i \leq M).  Consider the following action:    * Choose a desk with a book remaining, read the topmost book on that desk, and remove it from the desk.  How many books can we read at most by repeating this action so that it takes us at most K minutes in total? We ignore the time it takes to do anything other than reading.

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

import numpy as np
import bisect

ra=[0]+list(np.cumsum(a))
rb=[0]+list(np.cumsum(b))

ans=0
for i in range(n+1):    
    capa = k - ra[i]
    if ra[i]>0:
        break
    j=bisect.bisect_right(rb,capa)


    ans=max(ans,i-1+j)
            
print(ans)    

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

import numpy as np
import bisect

ra=[0]+list(np.cumsum(a))
rb=[0]+list(np.cumsum(b))

ans=0
for i in range(n+1):    
    capa = k - ra[i]
    if ra[i]>0:
        break
    j=bisect.bisect_right(rb,capa)


    ans=max(ans,i-1+j)
            
print(ans)    


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

import numpy as np
import bisect

ra=[0]+list(np.cumsum(a))
rb=[0]+list(np.cumsum(b))

ans=0
for i in range(n+1):    
    capa = k - ra[i]
    if ra[i]>k:
        break
    j=bisect.bisect_right(rb,capa)


    ans=max(ans,i-1+j)
            
print(ans)    


Verdicts: Wrong Answer

Wrong Answer

Accepted
Problem: We have two desks: A and B. Desk A has a vertical stack of N books on it, and Desk B similarly has M books on it.  It takes us A_i minutes to read the i-th book from the top on Desk A (1 \leq i \leq N), and B_i minutes to read the i-th book from the top on Desk B (1 \leq i \leq M).  Consider the following action:    * Choose a desk with a book remaining, read the topmost book on that desk, and remove it from the desk.  How many books can we read at most by repeating this action so that it takes us at most K minutes in total? We ignore the time it takes to do anything other than reading.

Submission attempts (in order):

N, M, K = map(int,input().split())

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


a, b = [0], [0]

for i in range(N):
    a.append(a[i] + A[i])
for i in range(M):
    b.append(b[i] + B[i])


ans, j = 0, M





for i in range(N+1):
    if a[i] > K:
        break
    # print(a[i])
    while b[j] > K -a[i]:
        
        j -= 1
        print(b[j])
    
    ans = max(ans, i + j)
print(ans)





N, M, K = map(int,input().split())

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


a, b = [0], [0]

for i in range(N):
    a.append(a[i] + A[i])
for i in range(M):
    b.append(b[i] + B[i])


ans, j = 0, M





for i in range(N+1):
    if a[i] > K:
        break
    # print(a[i])
    while b[j] > K -a[i]:
        
        j -= 1
        # print(b[j])
    
    ans = max(ans, i + j)
print(ans)


Verdicts: Wrong Answer

Accepted