problem_id
stringlengths 6
6
| user_id
stringlengths 10
10
| time_limit
float64 1k
8k
| memory_limit
float64 262k
1.05M
| problem_description
stringlengths 48
1.55k
| codes
stringlengths 35
98.9k
| status
stringlengths 28
1.7k
| submission_ids
stringlengths 28
1.41k
| memories
stringlengths 13
808
| cpu_times
stringlengths 11
610
| code_sizes
stringlengths 7
505
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|---|---|---|---|---|---|---|---|---|---|---|
p02623 | u310709512 | 2,000 | 1,048,576 | 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. | ['N, M, K = map(int, input().split())\nA = list(map(int, input().split()))\nB = list(map(int, input().split()))\n\ntotalA = [0]\ntotalB = [0]\nsumA = 0\nfor i in range(N):\n sumA += A[0]\n totalA.append(sumA)\nsumB = 0\nfor i in range(M):\n sumB += B[0]\n totalB.append(sumB)\n\niA = N\niB = 0\ncountMax = 0\nwhile iA >= 0:\n if totalA[iA] + totalB[iB] > K:\n iA -= 1\n else:\n if iB < M:\n countMax = max(countMax, iA + iB)\n iB += 1\n elif iA == N and iB == M:\n countMax = max(countMax, iA + iB)\n break\n\nprint(countMax)\n\n', 'N, M, K = map(int, input().split())\nA = list(map(int, input().split()))\nB = list(map(int, input().split()))\n\ntotalA = [0]\ntotalB = [0]\nsumA = 0\nfor i in range(N):\n sumA += A[i]\n totalA.append(sumA)\nsumB = 0\nfor i in range(M):\n sumB += B[i]\n totalB.append(sumB)\n\niB = M\ncountMax = 0\nfor iA in range(N+1):\n if totalA[iA]>K:\n break\n else:\n while totalA[iA] + totalB[iB] > K:\n iB -= 1\n countMax = max(countMax,iA+iB)\n\nprint(countMax)\n\n'] | ['Runtime Error', 'Accepted'] | ['s352302446', 's547748760'] | [8860.0, 47456.0] | [25.0, 282.0] | [567, 457] |
p02623 | u311379832 | 2,000 | 1,048,576 | 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. | ['N, M, K = map(int, input().split())\na = list(map(int, input().split()))\nb = list(map(int, input().split()))\narui = [0] * (N + 1)\nbrui = [0] * (M + 1)\nfor i in range(N):\n arui.append(a[i] + arui[i])\nfor i in range(M):\n brui.append(b[i] + brui[i])\n\nans = 0\nfor i in range(N + 1):\n if arui[i] > K:\n break\n bcnt = M\n while brui[bcnt] + arui[i] > K:\n bcnt -= 1\n ans = max(ans, i + bcnt)\n\nprint(ans)', 'N, M, K = map(int, input().split())\na = list(map(int, input().split()))\nb = list(map(int, input().split()))\narui = [0] * (N + 1)\nbrui = [0] * (M + 1)\nfor i in range(N):\n arui[i + 1] = a[i] + arui[i]\nfor i in range(M):\n brui[i + 1] = b[i] + brui[i]\n\nans = 0\nbcnt = M\nfor i in range(N + 1):\n if arui[i] > K:\n break\n\n while brui[bcnt] + arui[i] > K:\n bcnt -= 1\n ans = max(ans, i + bcnt)\n\nprint(ans)\n'] | ['Wrong Answer', 'Accepted'] | ['s401607950', 's937476151'] | [48776.0, 47544.0] | [281.0, 300.0] | [425, 425] |
p02623 | u312078744 | 2,000 | 1,048,576 | 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. | ['import itertools\nimport numpy as np \nn, m, k = map(int, input().split())\na = list(map(int, input().split()))\nb = list(map(int, input().split()))\na = [0] + a\nb = [0] + b\nacc = list(itertools.accumulate(a))\nbcc = list(itertools.accumulate(b))\n\nacc = np.array(acc)\n#print(acc)\nbcc = np.array(bcc)\n\nans = 0\n\nfor i, val in enumerate(acc):\n if (val > k):\n break\n else:\n rest = k - val\n print(rest)\n for j, val2 in enumerate(bcc):\n if (val2 <= rest):\n ans = max(ans, j + i)\n else:\n break\n\nprint(ans)\n ', 'import itertools\nimport numpy as np \nn, m, k = map(int, input().split())\na = list(map(int, input().split()))\nb = list(map(int, input().split()))\nacc = list(itertools.accumulate(a))\nbcc = list(itertools.accumulate(b))\n\nacc = np.array(acc)\n#print(acc)\nbcc = np.array(bcc)\n\nans = 0\n\nfor i in acc:\n if (i > k):\n continue\n else:\n rest = k - i\n for j in bcc:\n if (j <= rest):\n ans = max(ans, j + i)\n else:\n break\n\nprint(ans)\n ', 'import itertools\nimport numpy as np \nimport sys\nread = sys.stdin.buffer.read\nreadline = sys.stdin.buffer.readline\nreadlines = sys.stdin.buffer.readlines\n# -------------------\nn, m, k = map(int, readline().split())\nprint(n)\na = list(map(int, readline().split()))\nb = list(map(int, readline().split()))\na = [0] + a\nb = [0] + b\nacc = list(itertools.accumulate(a))\nbcc = list(itertools.accumulate(b))\n\n#acc = np.array(acc)\n#print(acc)\n#bcc = np.array(bcc)\n\nans = 0\n\nfor i, val in enumerate(acc):\n if (val > k):\n break\n else:\n rest = k - val\n #print(rest)\n for j, val2 in enumerate(bcc):\n if (val2 <= rest):\n ans = max(ans, j + i)\n else:\n break\n\nprint(ans)\n \n', 'import itertools\nimport numpy as np \nimport sys\nread = sys.stdin.buffer.read\nreadline = sys.stdin.buffer.readline\nreadlines = sys.stdin.buffer.readlines\n# -------------------\nn, m, k = map(int, readline().split())\n#print(n)\na = list(map(int, readline().split()))\nb = list(map(int, readline().split()))\na = [0] + a\nb = [0] + b\nacc = list(itertools.accumulate(a))\nbcc = list(itertools.accumulate(b))\n\n#acc = np.array(acc)\n#print(acc)\n#bcc = np.array(bcc)\n\nans = 0\nj = m\nfor i, val in enumerate(acc):\n if (val > k):\n break\n else:\n rest = k - val\n #print(rest)\n #j = m\n \n while (bcc[j] > rest):\n j -= 1\n\n ans = max(ans, j + i)\n\n\nprint(ans)'] | ['Wrong Answer', 'Wrong Answer', 'Wrong Answer', 'Accepted'] | ['s353669769', 's525862088', 's751243880', 's673337853'] | [66984.0, 66732.0, 65396.0, 65448.0] | [2207.0, 2207.0, 2207.0, 323.0] | [533, 456, 693, 730] |
p02623 | u312158169 | 2,000 | 1,048,576 | 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. | ['n,m,k = map(int,input().split())\n\na = [int(x) for x in input().split()]\nb = [int(x) for x in input().split()]\n\nif k < a[0] and k < b[0]:\n print(0)\n exit()\n\nif sum(a) + sum(b) <= k:\n print(n+m)\n exit()\n\na_temp = a[0]\nb_temp = b[0]\ntemp = 0\ncount = 0\nhour = 0\ni = 0\nj = 0\n\nwhile k > hour:\n \n count += 1\n if a_temp <= b_temp:\n hour += a_temp\n i += 1\n if i == len(a):\n a_temp = 10**10\n else:\n a_temp = a[i]\n else:\n hour += b_temp\n j += 1\n if j == len(b):\n b_temp = 10**10\n else:\n b_temp = b[i]\n\nprint(count)\n', 'n,m,k = map(int,input().split())\n\nA = [int(x) for x in input().split()]\nB = [int(x) for x in input().split()]\n\na, b = [0], [0]\nfor i in range(n):\n a.append(a[i] + A[i])\nfor i in range(m):\n b.append(b[i] + B[i])\n\nans, j = 0, m\nfor i in range(n + 1):\n if a[i] > k:\n break\n while b[j] > k - a[i]:\n j -= 1\n ans = max(ans, i + j)\n\nprint(ans)\n'] | ['Runtime Error', 'Accepted'] | ['s562826980', 's924067925'] | [42224.0, 47356.0] | [256.0, 314.0] | [628, 368] |
p02623 | u315600877 | 2,000 | 1,048,576 | 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. | ['N,M,K=map(int,input("入力してください").split())\ncostA=list(map(int,input().split()))\ncostB=list(map(int,input().split()))\ncostA=[0]+costA\ncostB=[0]+costB\n\nfor i in range(N):\n costA[i+1]+=costA[i]\n \nfor i in range(M):\n costB[i+1]+=costB[i]\n\nprint(costA)\nprint(costB)\n\ncostB.append(float("inf"))\n\ndef upper_bound(list1,K):\n low=0\n high=len(list1)-1 #index\n while high-low!=1:\n mid=(high+low)//2\n if list1[mid]>K:\n high=mid\n else:\n low=mid\n return high \n\nnumber=[]\nfor i in range(N+1): \n time=K-costA[i]\n number.append(i+upper_bound(costB,time)-1)\n print(i)\n print(upper_bound(costB,time))\n print(number)\n \nprint(max(number))\n', 'N,M,K=map(int,input("入力してください").split())\ncostA=list(map(int,input().split()))\ncostB=list(map(int,input().split()))\ncostA=[0]+costA\ncostB=[0]+costB\n\nfor i in range(N):\n costA[i+1]+=costA[i]\n \nfor i in range(M):\n costB[i+1]+=costB[i]\n\n\n\ncostB.append(float("inf"))\n\ndef upper_bound(list1,K):\n low=0\n high=len(list1)-1 #index\n while high-low!=1:\n mid=(high+low)//2\n if list1[mid]>K:\n high=mid\n else:\n low=mid\n return high \n\nnumber=[]\nfor i in range(N+1): \n time=K-costA[i]\n if upper_bound(costB,time)!=1:\n number.append(i+upper_bound(costB,time)-1)\n else:\n number.append(0)\n"""print(i)\n print(upper_bound(costB,time))\n print(number)"""\n \nprint(max(number))\n\n\n\n\n', 'N,M,K=map(int,input("入力してください").split())\ncostA=list(map(int,input().split()))\ncostB=list(map(int,input().split()))\ncostA=[0]+costA\ncostB=[0]+costB\n\nfor i in range(N):\n costA[i+1]+=costA[i]\n \nfor i in range(M):\n costB[i+1]+=costB[i]\n\nprint(costA)\nprint(costB)\n\ncostB.append(float("inf"))\n\ndef upper_bound(list1,K):\n low=0\n high=len(list1)-1 #index\n while high-low!=1:\n mid=(high+low)//2\n if list1[mid]>K:\n high=mid\n else:\n low=mid\n return high \n\nnumber=[]\nfor i in range(N+1): \n time=K-costA[i]\n if upper_bound(costB,time)!=1:\n number.append(i+upper_bound(costB,time)-1)\n else:\n number.append(0)\n print(i)\n print(upper_bound(costB,time))\n print(number)\n \nprint(max(number))', 'N,M,K=map(int,input().split())\ncostA=list(map(int,input().split()))\ncostB=list(map(int,input().split()))\ncostA=[0]+costA\ncostB=[0]+costB\n\nfor i in range(N):\n costA[i+1]+=costA[i]\n \nfor i in range(M):\n costB[i+1]+=costB[i]\n\n\n\n\n\ndef upper_bound(list1,K):\n low=0\n high=len(list1) #index\n while high-low!=1:\n mid=(high+low)//2\n if list1[mid]>K:\n high=mid\n else:\n low=mid\n return low \n\nnumber=[]\nfor i in range(N+1): \n time=K-costA[i]\n if upper_bound(costB,time)!=1:\n number.append(i+upper_bound(costB,time)-1)\n else:\n number.append(0)\n"""print(i)\n print(upper_bound(costB,time))\n print(number)"""\n \nprint(max(number))\n\n\n\n\n', 'N,M,K=map(int,input().split())\ncostA=list(map(int,input().split()))\ncostB=list(map(int,input().split()))\ncostA=[0]+costA\ncostB=[0]+costB\n\nfor i in range(N):\n costA[i+1]+=costA[i]\n \nfor i in range(M):\n costB[i+1]+=costB[i]\n\n\n\n""\n\ndef upper_bound(list1,K):\n low=0\n high=len(list1) #index\n while high-low!=1:\n mid=(high+low)//2\n if list1[mid]>K:\n high=mid\n else:\n low=mid\n return low \n\nnumber=[]\n#print(costA)\n\nfor i in range(N+1): \n time=K-costA[i]\n if time<0:\n break\n #print(i,upper_bound(costB,time))\n number.append(i+upper_bound(costB,time))\n\n"""print(i)\n print(upper_bound(costB,time))\n print(number)"""\n \nprint(max(number))\n\n\n\n\n'] | ['Runtime Error', 'Wrong Answer', 'Runtime Error', 'Wrong Answer', 'Accepted'] | ['s052467878', 's354411764', 's542146545', 's741544635', 's361321379'] | [164156.0, 40112.0, 160244.0, 40200.0, 40156.0] | [2091.0, 1138.0, 2361.0, 1128.0, 659.0] | [813, 2037, 886, 1982, 2004] |
p02623 | u317779196 | 2,000 | 1,048,576 | 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. | ["'''ABC172 C'''\nimport numpy as np\nn,m,k = map(int,input().split())\na = np.array([int(i) for i in input().split()], dtype='int64')\nb = np.array([int(i) for i in input().split()], dtype='int64')\n\na_c = np.zeros(n+1,dtype='int64')\na_c[1:] = np.cumsum(a)\n\nb_c = np.zeros(m+1,dtype='int64')\nb_c[1:] = np.cumsum(b)\n\na_c = a_c[a_c <= k]\nb_c = b_c[b_c <= k]\nans = 0\nfor i,ai in enumerate(a_c):\n n = np.searchsorted(b_c, k - ai, side = 'right')\n ans = max(i + n, ans)\nprint(ans)", "import numpy as np\nn,m,k = map(int,input().split())\na = np.array([int(i) for i in input().split()], dtype='int64')\nb = np.array([int(i) for i in input().split()], dtype='int64')\n\n\na_c = np.zeros(n+1,dtype='int64')\na_c[1:] = np.cumsum(a)\n\nb_c = np.zeros(m+1,dtype='int64')\nb_c[1:] = np.cumsum(b)\n\na_c = a_c[a_c <= k]\nb_c = b_c[b_c <= k]\n\nans_l = np.searchsorted(b_c, k - a_c, side='right') - 1\nans_l += np.arange(len(a_c))\nprint(ans_l.max())"] | ['Wrong Answer', 'Accepted'] | ['s474547178', 's690126984'] | [53440.0, 53460.0] | [783.0, 237.0] | [475, 499] |
p02623 | u319589470 | 2,000 | 1,048,576 | 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. | ['#!/usr/bin/env python3\nimport sys\nsys.setrecursionlimit(10**7)\nimport bisect\nimport heapq\nimport itertools\nimport math\nfrom collections import Counter, defaultdict, deque\nfrom copy import deepcopy\nfrom decimal import Decimal\nfrom math import gcd\nfrom operator import add, itemgetter, mul, xor\ndef cmb(n,r,mod):\n bunshi=1\n bunbo=1\n for i in range(r):\n bunbo = bunbo*(i+1)%mod\n bunshi = bunshi*(n-i)%mod\n return (bunshi*pow(bunbo,mod-2,mod))%mod\nmod = 10**9+7\ndef I(): return int(input())\ndef LI(): return list(map(int,input().split()))\ndef MI(): return map(int,input().split())\ndef LLI(n): return [list(map(int, input().split())) for _ in range(n)]\n\n\n\n\n\n\n\nn,m,k = MI()\na = deque(LI())\nb = deque(LI())\nans = 0\nwhile k > 0 and len(a) + len(b) > 0:\n if len(a) > 0:\n book = a.popleft()\n if len(a) == 0:\n book = 10**10\n if len(b) > 0:\n book2 = b.popleft()\n if len(b) == 0:\n book2 = 10**10\n if book < book2:\n k -= book\n if k > 0:\n ans += 1\n if book > book2\n k -= book2\n if k > 0:\n ans += 1\nprint(ans)', '#!/usr/bin/env python3\nimport sys\nsys.setrecursionlimit(10**7)\nimport bisect\nimport heapq\nimport itertools\nimport math\nfrom collections import Counter, defaultdict, deque\nfrom copy import deepcopy\nfrom decimal import Decimal\nfrom math import gcd\nfrom operator import add, itemgetter, mul, xor\ndef cmb(n,r,mod):\n bunshi=1\n bunbo=1\n for i in range(r):\n bunbo = bunbo*(i+1)%mod\n bunshi = bunshi*(n-i)%mod\n return (bunshi*pow(bunbo,mod-2,mod))%mod\nmod = 10**9+7\ndef I(): return int(input())\ndef LI(): return list(map(int,input().split()))\ndef MI(): return map(int,input().split())\ndef LLI(n): return [list(map(int, input().split())) for _ in range(n)]\n\n\n\n\n\n\n\nn,m,k = MI()\na = LI()\nb = LI()\na_cum = [0]\nb_cum = [0]\nans = 0\nfor i in range(1,n+1):\n a_cum.append(a[i-1]+a_cum[i-1])\nfor i in range(1,m+1):\n b_cum.append(b[i-1]+b_cum[i-1])\nfor i in range(m+1):\n K = k-b_cum[i]\n if K >= 0:\n \n j = bisect.bisect_right(a_cum, K) - 1\n ans = max(ans,i+j)\nprint(ans)\n\n'] | ['Runtime Error', 'Accepted'] | ['s068268642', 's654150904'] | [8892.0, 48428.0] | [27.0, 361.0] | [1627, 1573] |
p02623 | u323045245 | 2,000 | 1,048,576 | 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. | ['n,m,k=map(int, input().split())\na=list(map(int,input().split()))\nb=list(map(int,input().split()))\nA=[0]*(n+1)\nB=[0]*(m+1)\nc = 0\n\nfor i in range(n):\n A[i+1] = A[i] + a[i]\nfor i in range(m):\n B[i+1] = B[i] + b[i]\n#\nfor i in range(n+1):\n if A[i]>k:\n bookA=i-1\n timeA = A[i-1]\n break\nelse:\n bookA = n\n timeA = A[n]\n \nfor i in range(m+1):\n if B[i] >k:\n bookB = i-1\n timeB = B[i-1]\n break\nelse:\n bookB = m\n timeB = B[m]\n#\n"""\nprint(bookA,timeA,)\nprint(bookB,timeB)\nif bookA == 0 and bookB == 0:\n print(0)\n exit()\n"""\nif bookA < bookB:\n time = timeB\n book = bookB\n for i in range(n+1):\n if k-time < A[i]:\n break\n c = i\n else:\n c = n\n book += c \n\nelif bookA > bookB:\n book = bookA\n time = timeA\n for i in range(m+1):\n if k-time < B[i]:\n break\n c = i\n else:\n c = m\n book += c\n\nelse:\n if timeA > timeB:\n time = timeB\n book = bookB\n for i in range(n+1):\n if k-time < A[i]:\n break\n c = i\n book += c\n elif timeB > timeA:\n time = timeA\n book = bookA\n for i in range(m+1):\n if k-time < B[i]:\n break\n c = i\n book += c\n print(c)\n else:\n if A[bookA-1] > B[bookB-1]:\n book = bookB-1\n for i in range(n+1):\n if A[i] > k:\n break\n c = i\n book += c\n elif A[bookA-1] < B[bookB-1]:\n book = bookA-1\n for i in range(m+1):\n if B[i] > k:\n break\n c = i\n book += c\n else:\n book = bookA\n for i in range(m+1):\n if k-timeA < B[i]:\n book += i-1\n else:\n book += m\nif book < 0:\n print(0)\nelse:\n print(book)', 'n,m,k=map(int, input().split())\nA=list(map(int,input().split()))\nB=list(map(int,input().split()))\n\n\ncumA = [0]*(n+1)\nfor i in range(n):\n cumA[i+1] = cumA[i] + A[i]\n\ncumB = [0]*(m+1)\nfor i in range(m):\n cumB[i+1] = cumB[i] + B[i]\n\nans = 0\nj = m\n\nfor i in range(n+1):\n if cumA[i]>k: \n break\n while cumB[j] > k-cumA[i] and j>0: \n j -= 1\n ans = max(ans,i+j)\nprint(ans) '] | ['Wrong Answer', 'Accepted'] | ['s067002085', 's057624721'] | [47552.0, 47328.0] | [247.0, 295.0] | [1966, 570] |
p02623 | u324090406 | 2,000 | 1,048,576 | 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. | ['n, m, k = list(map(int, input().split()))\na = list(map(int, input().split()))\nb = list(map(int, input().split()))\nai = -1\nbi = -1\nanswer = 0\nt_sum = 0\nif sum(a) + sum(b) <= k:\n print(n + m)\nelif sum(a) <= k:\n t_sum = sum(a)\n ai = n-1\n while True:\n bi += 1\n if t_sum + b[bi] > k:\n break\n t_sum += b[bi] \n bi -= 1\n answer = ai + bi + 2\n while ai >= 0:\n t_sum -= a[ai]\n ai -= 1\n while bi <= m:\n bi += 1\n if t_sum + b[bi] > k:\n break\n t_sum += b[bi]\n bi -= 1\n if ai + bi +2 > answer:\n answer = ai + bi +2\n print(answer)\nelse:\n for i in range(n):\n t_sum += a[i]\n if t_sum > k:\n ai = i-1\n break\n while True:\n bi += 1\n if t_sum + b[bi] > k:\n break\n t_sum += b[bi] \n bi -= 1\n answer = ai + bi + 2\n while ai >= 0:\n t_sum -= a[ai]\n ai -= 1\n while bi <= m:\n bi += 1\n if t_sum + b[bi] > k:\n break\n t_sum += b[bi]\n bi -= 1\n if ai + bi +2 > answer:\n answer = ai + bi +2\n print(answer)', 'n, m, k = list(map(int, input().split()))\na = list(map(int, input().split()))\nb = list(map(int, input().split()))\nai = -1\nbi = -1\nanswer = 0\nt_sum = 0\nif sum(a) + sum(b) <= k:\n print(n + m)\nelif sum(a) <= k:\n t_sum = sum(a)\n ai = n-1\n while True:\n bi += 1\n if bi == m:\n break\n if t_sum + b[bi] > k:\n break\n t_sum += b[bi] \n bi -= 1\n answer = ai + bi + 2\n while ai >= 0:\n t_sum -= a[ai]\n ai -= 1\n while bi <= m-1:\n bi += 1\n if bi == m:\n break\n if t_sum + b[bi] > k:\n break\n t_sum += b[bi]\n bi -= 1\n if ai + bi +2 > answer:\n answer = ai + bi +2\n print(answer)\nelse:\n while True:\n ai += 1\n if t_sum + a[ai] > k:\n break\n t_sum += a[ai]\n ai -= 1\n while True:\n bi += 1\n if bi == m:\n break\n if t_sum + b[bi] > k:\n break\n t_sum += b[bi]\n bi -= 1\n answer = ai + bi + 2\n while ai >= 0:\n t_sum -= a[ai]\n ai -= 1\n while bi <= m-1:\n bi += 1\n if bi == m:\n break\n if t_sum + b[bi] > k:\n break\n t_sum += b[bi]\n bi -= 1\n if ai + bi +2 > answer:\n answer = ai + bi +2\n print(answer)'] | ['Runtime Error', 'Accepted'] | ['s264858768', 's454605852'] | [40700.0, 41736.0] | [297.0, 330.0] | [1200, 1380] |
p02623 | u325660636 | 2,000 | 1,048,576 | 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. | ['N, M, K = map(int, input().split())\na_l = list(map(int, input().split()))\nb_l = list(map(int, input().split()))\n\ni_l, j_l = [0], [0]\n\nbest0 = M\nans = 0\n\nfor i in range(N):\n i_l.append(i_l[i] + a_l[i])\n\nfor i in range(M):\n j_l.append(j_l[i] + b_l[i])\n\n# for j in range(M):\n\n# if total > K:\n\n# break\n# elif total == K:\n\n# else:\n# continue\n\nfor k in range(N+1):\n if i_l[i] > K:\n break\n while i_l[k] + j_l[best0] > K:\n best0 -= 1\n ans = max(ans, k+best0)\nprint(ans)\n ', 'N, M, K = map(int, input().split())\na_l = list(map(int, input().split()))\nb_l = list(map(int, input().split()))\n\ni_l, j_l = [0], [0]\n\nbest0 = M\nans = 0\n\nfor i in range(N):\n i_l.append(i_l[i] + a_l[i])\n\nfor i in range(M):\n j_l.append(j_l[i] + b_l[i])\n\n# for j in range(M):\n\n# if total > K:\n\n# break\n# elif total == K:\n\n# else:\n# continue\n\nfor k in range(N+1):\n if i_l[k] > K:\n break\n while i_l[k] + j_l[best0] > K:\n best0 -= 1\n ans = max(ans, k+best0)\nprint(ans)\n '] | ['Runtime Error', 'Accepted'] | ['s356010868', 's842355749'] | [47480.0, 47600.0] | [281.0, 284.0] | [584, 584] |
p02623 | u328755070 | 2,000 | 1,048,576 | 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. | ['N, M, K = list(map(int, input().split()))\nA = list(map(int, input().split()))\nB = list(map(int, input().split()))\n\ntime = 0\ncount = -1\nsaidaicho = len(A) + len(B)\nwhile time <= K:\n time += max(A[0], B[0])\n count += 1\n\n if A[0] > B[0]:\n A.pop(0)\n else:\n B.pop(0)\n\n if len(A) == 0:\n A.append(0)\n elif len(B) == 0:\n B.append(0)\n\n if count == saidaicho:\n break\nprint(count)\n', 'N, M, K = list(map(int, input().split()))\nA = list(map(int, input().split()))\nB = list(map(int, input().split()))\n\nruiA = [0]\nruiB = [0]\nfor i in range(N):\n ruiA.append(ruiA[-1] + A[i])\nfor i in range(M):\n ruiB.append(ruiB[-1] + B[i])\n\nMAX = 0\ni = 0\nfor j in range(M, -1, -1):\n if ruiA[i] + ruiB[j] <= K:\n MAX = max(MAX, i+j)\n best = j\n break\n \nfor i in range(1, N+1):\n for j in range(best, -1, -1):\n if ruiA[i] + ruiB[j] <= K:\n MAX = max(MAX, i+j)\n break\n\n\nprint(MAX)\n', 'N, M, K = list(map(int, input().split()))\nA = list(map(int, input().split()))\nB = list(map(int, input().split()))\n\nruiA = [0]\nruiB = [0]\nfor i in range(N):\n ruiA.append(ruiA[-1] + A[i])\nfor i in range(M):\n ruiB.append(ruiB[-1] + B[i])\n\nMAX = 0\nj = M\n\nfor i in range(N+1):\n if ruiA[i] > K:\n break\n while ruiA[i] + ruiB[j] > K:\n j -= 1\n MAX = max(MAX, i+j)\n\nprint(MAX)\n'] | ['Wrong Answer', 'Runtime Error', 'Accepted'] | ['s417361044', 's994845875', 's116065680'] | [40548.0, 47492.0, 47408.0] | [2206.0, 304.0, 299.0] | [426, 526, 396] |
p02623 | u331105860 | 2,000 | 1,048,576 | 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. | ['N, M, K = map(int, input().split())\nA = [int(x) for x in input().split()]\nB = [int(x) for x in input().split()]\n\ntotal = 0\nmaxb = 0\nfor i in range(N):\n if total + A[i] <= K:\n total += A[i]\n maxb += 1\n else:\n break\n\ni = maxb - 1\nfor j in range(M):\n\n if total + B[j] <= K:\n total += B[j]\n else:\n if i == -1:\n break\n total -= A[i]\n i -= 1\n j -= 1\n\n if maxb < i + j + 2:\n maxb = i + j + 2\nprint(maxb)\n', 'N, M, K = map(int, input().split())\nA = [int(x) for x in input().split()]\nB = [int(x) for x in input().split()]\n\ntotal = 0\nAsum = sum(A)\nnA = len(A)\nBsum = 0\nans = 0\nfor nB in range(len(B) + 1):\n if nB != 0:\n Bsum += B[nB-1]\n if Bsum > K:\n break\n while Asum + Bsum > K:\n nA -= 1\n Asum -= A[nA]\n ans = max(nA + nB, ans)\n\nprint(ans)\n'] | ['Wrong Answer', 'Accepted'] | ['s843556736', 's389901503'] | [42476.0, 40684.0] | [234.0, 238.0] | [486, 371] |
p02623 | u343490140 | 2,000 | 1,048,576 | 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. | ['n, m, k = map(int, input().split())\nA = list(map(int, input().split()))\nB = list(map(int, input().split()))\na = [0]\nb = [0]\nfor i in range(n):\n a.append(a[i] + A[i])\nfor i in range(m):\n b.append(b[i] + B[i])\nans = 0\nj = m\nfor i in range(n + 1):\n if a[0] > k:\n break\n while b[j] > k - a[i]:\n j -= 1\n ans = max(ans, i + j)\nprint(ans)\n', 'n, m, k = map(int, input().split())\nA = list(map(int, input().split()))\nB = list(map(int, input().split()))\na = [0]\nb = [0]\nfor i in range(n):\n a.append(a[i] + A[i])\nfor i in range(m):\n b.append(b[i] + B[i])\nans = 0\nj = m\nfor i in range(n + 1):\n if a[i] > k:\n break\n while b[j] > k - a[i]:\n j -= 1\n ans = max(ans, i + j)\nprint(ans)\n'] | ['Runtime Error', 'Accepted'] | ['s808740023', 's977388778'] | [47428.0, 47560.0] | [311.0, 290.0] | [361, 361] |
p02623 | u345621867 | 2,000 | 1,048,576 | 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. | ['N, M, K = map(int,input().split())\nA = list(map(int,input().split()))\nB = list(map(int,input().split()))\nAsum = [0]\nBsum = [0]\na = 0\nb = 0\nfor i in range(N):\n a += A[i]\n Asum.append(a)\nfor i in range(M):\n b += B[i]\n Bsum.append(b)\nAsum.append(0)\nBsum.append(0)\nres, j = 0, 0\nfor i in range(N+1):\n if Asum[i] > K:\n break\n while Asum[i] + Bsum[j] > K:\n j -= 1\n res = max(res,i+j)\nprint(res)', 'N, M, K = map(int,input().split())\nA = list(map(int,input().split()))\nB = list(map(int,input().split()))\nAsum = []\nBsum = []\nfor i in range(N):\n a = 0\n a += A[i]\n Asum.append(a)\nfor i in range(M):\n b = 0\n b += B[i]\n Bsum.append(b)\nAsum.append(0)\nBsum.append(0)\nres = 0\nfor i in range(-1,N-1):\n for j in range(-1,M-1):\n if Asum[i] + Bsum[j] <= K:\n res = max(res,(i+j+2))\n else:\n break\nprint(res)', 'N, M, K = map(int,input().split())\nA = list(map(int,input().split()))\nB = list(map(int,input().split()))\nAsum = [0]\nBsum = [0]\na = 0\nb = 0\nfor i in range(N):\n a += A[i]\n Asum.append(a)\nfor i in range(M):\n b += B[i]\n Bsum.append(b)\nAsum.append(0)\nBsum.append(0)\nres, j = 0, M\nfor i in range(N+1):\n if Asum[i] > K:\n break\n while Asum[i] + Bsum[j] > K:\n j -= 1\n res = max(res,i+j)\nprint(res)'] | ['Wrong Answer', 'Wrong Answer', 'Accepted'] | ['s015909569', 's794357984', 's259528070'] | [47464.0, 41140.0, 47444.0] | [260.0, 2207.0, 288.0] | [427, 449, 423] |
p02623 | u346395915 | 2,000 | 1,048,576 | 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. | ['n,m,k = map(int, input().split())\na_li = list(map(int, input().split()))\nb_li = list(map(int, input().split()))\n\na_li = [0] + a_li\nb_li = [0] + b_li\n\ntime = 0\nans = 0\na = 0\ntemp = 0\n\na_wa = [0] * (n+10)\nb_wa = [0] * (m+10)\n\ns = 0\nfor i in range(n+1):\n s += a_li[i]\n a_wa[i] = s\n\ns = 0\nfor i in range(m+1):\n s += b_li[i]\n b_wa[i] = s\n \n\nfor i in range(n+1):\n for j in range(min(ans-temp, m), m+1):\n time = a_wa[i] + b_wa[j]\n if time <= k:\n ans = max(ans, i+j)\n temp = i\n else:\n break\n \nprint(ans)', 'n,m,k = map(int, input().split())\na_li = list(map(int, input().split()))\nb_li = list(map(int, input().split()))\n\na_li = [0] + a_li\nb_li = [0] + b_li\n\ntime = 0\nans = 0\na = 0\ntemp = 0\n\na_wa = [0] * (n+10)\nb_wa = [0] * (m+10)\n\ns = 0\nfor i in range(n+1):\n s += a_li[i]\n a_wa[i] = s\n\ns = 0\nfor i in range(m+1):\n s += b_li[i]\n b_wa[i] = s\n \n\nfor i in range(n+1):\n for j in range(max(min(temp-i, m), 0), m+1):\n time = a_wa[i] + b_wa[j]\n if time <= k:\n ans = max(ans, i+j)\n temp = ans\n else:\n break\n \nprint(ans)\n'] | ['Wrong Answer', 'Accepted'] | ['s380632982', 's728737354'] | [48316.0, 47100.0] | [428.0, 530.0] | [578, 587] |
p02623 | u347640436 | 2,000 | 1,048,576 | 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. | ['from itertools import accumulate\nfrom bisect import bisect_left\n\nN, M, K = map(int, input().split())\nA = list(map(int, input().split()))\nB = list(map(int, input().split()))\n\na = [0] + list(accumulate(A))\nb = [0] + list(accumulate(B))\n\nresult = 0\nfor i in range(N):\n t = K - a[i]\n if t < 0:\n break\n j = bisect_left(b, t)\n if b[j] == t:\n result = max(result, i + j + 1)\n else:\n result = max(result, i + j)\nprint(result)\n', '\nfrom itertools import accumulate\nfrom bisect import bisect_right\n\nN, M, K = map(int, input().split())\nA = list(map(int, input().split()))\nB = list(map(int, input().split()))\n\na = [0] + list(accumulate(A))\nb = [0] + list(accumulate(B))\n\nresult = 0\nfor i in range(N + 1):\n if a[i] > K:\n break\n j = bisect_right(b, K - a[i])\n result = max(result, i + j - 1)\nprint(result)\n'] | ['Runtime Error', 'Accepted'] | ['s182324519', 's578015460'] | [50504.0, 50584.0] | [288.0, 266.0] | [454, 411] |
p02623 | u349384236 | 2,000 | 1,048,576 | 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. | [' N, M, K = map(int, input().split())\n A = list(map(int, input().split()))\n B = list(map(int, input().split()))\n a, b = [0], [0]\n for i in range(N):\n a.append(a[i] + A[i])\n for i in range(M):\n b.append(b[i] + B[i])\n\n ans, j = 0, M\nfor i in range(N + 1):\n if a[i] > K:\n break\n while b[j] > K - a[i]:\n\tj -= 1\nans = max(ans, i + j)\nprint(ans)\n', 'n,m,k=map(int,input().split())\na=list(map(int,input().split()))\nb=list(map(int,input().split()))\nA=[0]\nB=[0]\nfor i in range(n):\n A.append(a[i]+A[i])\nfor j in range(m):\n B.append(b[j]+B[j])\n\ni=0\nj=m\nans=0\ntmp=0\nwhile(i<n+1 and j>=0):\n if(k>=A[i]):\n tmp=k-A[i]\n if(tmp-B[j]>=0):\n ans=max(ans,i+j)\n i+=1\n else:\n j-=1\n else:\n break\nprint(ans)\n \n \n\n \n'] | ['Runtime Error', 'Accepted'] | ['s985267820', 's167534419'] | [8848.0, 47336.0] | [25.0, 359.0] | [352, 440] |
p02623 | u350247040 | 2,000 | 1,048,576 | 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. | ['n,m,k=map(int,input().split())\nA=list(map(int,input().split()))\nB=list(map(int,input().split()))\ncnt=0\nwhile cnt==n+m:\n if len(B)==0:\n k-=A[0]\n del A[0]\n cnt+=1\n elif len(A)==0:\n k-=B[0]\n del B[0]\n cnt+=1\n elif A[0]<=B[0] and k>A[0]:\n k-=A[0]\n del A[0]\n cnt+=1\n elif B[0]<A[0] and k>B[0]:\n k-=B[0]\n del B[0]\n cnt+=1\n else:\n break\nprint(cnt)', 'n,m,k=map(int,input().split())\na=list(map(int,input().split()))\nb=list(map(int,input().split()))\n\nb2=[]\nb2.append(0)\nfor i in range(1,m+1):\n b2.append(b2[i-1]+b[i-1])\na.insert(0,0)\ncnt=0\na_sum=0\nfor i in range(n+1):\n j=m\n a_sum+=a[i]\n while True:\n if k>=a_sum+b2[j]:\n cnt=max(cnt,i+j)\n m=j\n break\n j-=1\n if j <0:\n break\nprint(cnt)'] | ['Wrong Answer', 'Accepted'] | ['s773245202', 's074162675'] | [39768.0, 40696.0] | [109.0, 273.0] | [384, 363] |
p02623 | u352676541 | 2,000 | 1,048,576 | 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. | ['#N,M,K = map(int,input().split())\n#A = list(map(int,input().split()))\n#B = list(map(int,input().split()))\n\nN,M,K =map(int,"5 7 100".split())\nA = [10,10,20,30,3]\nB = [3,9,30,20,1,7,20]\n\n\nA_min = [0]\nfor i in range(N):\n A_min.append(A_min[i]+A[i])\nB_min = [0]\nfor i in range(M):\n B_min.append(B_min[i]+B[i])\n\n\nans = 0\nB_len = M\nfor i in range(len(A_min)):\n if A_min[i] > K:\n break\n while B_min[B_len] > K - A_min[i]:\n B_len -= 1\n ans = max(ans,i+B_len)\nprint(ans)', 'N,M,K = map(int,input().split())\nA = list(map(int,input().split()))\nB = list(map(int,input().split()))\n\n\nA_min = [0]\nfor i in range(N):\n A_min.append(A_min[i]+A[i])\nB_min = [0]\nfor i in range(M):\n B_min.append(B_min[i]+B[i])\n\n\nans = 0\nB_len = len(B_min)\nfor i in range(len(A_min)):\n if A_min[i] > K:\n break\n while true:\n if B_min[B_len] > K - A_min[i]:\n break\n ans = max(ans,i+B_len)\n\nprint(ans)', 'N,M,K = map(int,input().split())\nA = list(map(int,input().split()))\nB = list(map(int,input().split()))\n\n\nA_min = [0]\nfor i in range(N):\n A_min.append(A_min[i]+A[i])\nB_min = [0]\nfor i in range(M):\n B_min.append(B_min[i]+B[i])\n\n\nans = 0\nB_len = len(B_min)\nfor i in range(len(A_min)):\n if Amin[i] > K:\n break\n while true:\n if B_min[B_len] > K - A_min[i]:\n break\n ans = max(ans,i+B_len)\n\nprint(ans)', 'N,M,K = map(int,input().split())\nA = list(map(int,input().split()))\nB = list(map(int,input().split()))\n\n\nA_min = [0]\nfor i in range(N):\n A_min.append(A_min[i]+A[i])\nB_min = [0]\nfor i in range(M):\n B_min.append(B_min[i]+B[i])\n\n\nans = 0\nB_len = M\nfor i in range(len(A_min)):\n print("i=",str(i))\n if A_min[i] > K:\n break\n while B_min[B_len] > K - A_min[i]:\n print("B_len=",str(B_len))\n print("A_min=",str(A_min[i]))\n print("B_min=",str(B_min[B_len]))\n ans = max(ans,i+B_len)\n B_len -= 1\nprint(ans)', 'N,M,K = map(int,input().split())\nA = list(map(int,input().split()))\nB = list(map(int,input().split()))\n\n\nA_min = [0]\nfor i in range(N):\n A_min.append(A_min[i]+A[i])\nB_min = [0]\nfor i in range(M):\n B_min.append(B_min[i]+B[i])\n\n\nans = 0\nB_len = M\nfor i in range(len(A_min)):\n if A_min[i] > K:\n break\n while B_min[B_len] > K - A_min[i]:\n B_len -= 1\n ans = max(ans,i+B_len)\nprint(ans)'] | ['Wrong Answer', 'Runtime Error', 'Runtime Error', 'Wrong Answer', 'Accepted'] | ['s009872893', 's372798342', 's776861787', 's815132959', 's138481657'] | [9216.0, 47372.0, 47368.0, 51952.0, 47500.0] | [30.0, 191.0, 184.0, 747.0, 281.0] | [618, 559, 558, 663, 537] |
p02623 | u353099601 | 2,000 | 1,048,576 | 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. | ['from collections import deque\n\nN, M, K = map(int, input().split())\nA = deque(list(map(int, input().split())))\nB = deque(list(map(int, input().split())))\n\nif sum(A) + sum(B) < K:\n print(N + M)\n exit()\n\nans = 0\n\nwhile K > 0:\n if len(A) != 0:\n # if len(B) == 0 or A[0] <= B[0]:\n if len(B) == 0 or sum(A) <= sum(B):\n if A[0] <= K:\n K -= A.popleft()\n ans += 1\n continue\n if len(B) != 0:\n # if len(A) == 0 or A[0] >= B[0]:\n if len(A) == 0 or sum(A) >= sum(B):\n if B[0] <= K:\n K -= B.popleft()\n ans += 1\n continue\n\nprint(ans)\n', 'N, M, K = map(int, input().split())\nA = list(map(int, input().split()))\nB = list(map(int, input().split()))\n\n\nA_sum, B_sum = [0], [0]\nfor i, a in enumerate(A):\n A_sum.append(A_sum[i] + a)\nfor i, b in enumerate(B):\n B_sum.append(B_sum[i] + b)\n\nans = 0\nj = M\n\n\n\n\n\nfor i in range(N+1):\n if A_sum[i] > K:\n break\n while B_sum[j] > K - A_sum[i]:\n j -= 1\n ans = max(ans, i+j)\n\nprint(ans)\n'] | ['Time Limit Exceeded', 'Accepted'] | ['s090652836', 's491631577'] | [42648.0, 48824.0] | [2206.0, 278.0] | [671, 775] |
p02623 | u357120030 | 2,000 | 1,048,576 | 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. | ['from functools import lru_cache\n\nri = lambda S: [int(v) for v in S.split()]\n\n\ndef rii():\n return ri(input())\n\n\nN, M, K = rii()\nA = rii()\nB = rii()\n\ni = j = 0\n\n\n@lru_cache(None)\ndef dp(i, j, k):\n ki = kj = mki = mkj = 0\n\n if i < N:\n Ki = k - A[i] if A[i] <= k else 0\n mki = 1 + dp(i + 1, j, Ki) if Ki >= 0 else 0\n if j < M:\n kj = k - B[j] if B[j] <= k else 0\n mkj = 1 + dp(i, j + 1, kj) if kj >= 0 else 0\n\n return max(mki, mkj)\n\n\nprint(dp(0, 0, K))\n#print(dp.cache_info())\n', 'ri = lambda S: [int(v) for v in S.split()]\n\n\ndef rii():\n return ri(input())\n\n\nN, M, K = rii()\nA = rii()\nB = rii()\n\ni = j = 0\n\n\ndef dp(i, j, k):\n ki = kj = mki = mkj = 0\n\n if i < N:\n Ki = k - A[i] if A[i] <= k else 0\n mki = 1 + dp(i + 1, j, Ki) if Ki >= 0 else 0\n if j < M:\n kj = k - B[j] if B[j] <= k else 0\n mkj = 1 + dp(i, j + 1, kj) if kj >= 0 else 0\n\n return max(mki, mkj)\n\n\nprint(dp(0, 0, K))\n', 'ri = lambda S: [int(v) for v in S.split()]\n\n\ndef rii():\n return ri(input())\n\n\nN, M, K = rii()\nA = rii()\nB = rii()\n\ni = j = 0\n\n\ndef dp(i, j, k):\n print(i, j, k)\n ki = kj = mki = mkj = 0\n\n if i < N:\n Ki = k - A[i] if A[i] <= k else 0\n mki = 1 + dp(i + 1, j, Ki) if Ki >= 0 else 0\n if j < M:\n kj = k - B[j] if B[j] <= k else 0\n mkj = 1 + dp(i, j + 1, kj) if kj >= 0 else 0\n\n return max(mki, mkj)\n\n\nprint(dp(0, 0, K))\n', 'from itertools import accumulate\nfrom bisect import bisect_left\n\nri = lambda S: [int(v) for v in S.split()]\n\n\ndef rii():\n return ri(input())\n\n\nN, M, K = rii()\nA = rii()\nB = rii()\n\ni = j = 0\n\npA = [0] + list(accumulate(A))\npB = [0] + list(accumulate(B))\n\nans = 0\nfor i, a in enumerate(pA):\n br = 0\n if K >= a:\n r = K - a\n br = i\n \n if r:\n j = bisect_left(pB, r)\n if j != len(pB):\n if pB[j] > r:\n j -= 1\n br += j\n \n ans = max(ans, br)\n\nprint(ans)'] | ['Runtime Error', 'Runtime Error', 'Runtime Error', 'Accepted'] | ['s152082375', 's207488474', 's694686679', 's302435288'] | [43332.0, 42748.0, 42892.0, 50764.0] | [124.0, 125.0, 125.0, 328.0] | [515, 441, 460, 530] |
p02623 | u357751375 | 2,000 | 1,048,576 | 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. | ['\nn,m,k = map(int,input().split())\na = list(map(int,input().split()))\nb = list(map(int,input().split()))\np = 0\nz = 0\ni = 0\nj = 0\n\nwhile p < k:\n if a[i] < b[j]:\n p += a[i]\n z += 1\n i += 1\n elif a[i] > b[j]:\n p += b[j]\n z += 1\n j += 1\n else:\n if a[i+1] < b[j+1]:\n p += a[i]\n z += 1\n i += 1\n elif:\n p += b[j]\n z += 1\n j += 1\n\nprint(z)', 'n,m,k = map(int,input().split())\na = list(map(int,input().split()))\nb = list(map(int,input().split()))\n\nx = 0\nfor i in range(n):\n x += a[i]\n a[i] = x\n\nx = 0\nfor i in range(m):\n x += b[i]\n b[i] = x\n\nx = 0\nfor i in range(n):\n x = a[i]\n if x > k:\n x = a[i-1]\n ans = i\n j = i-2\n break\n if i + 1 == n:\n ans = n\n j = i-1\n\ny = 0\nfor i in range(m):\n y = b[i]\n if x + y > k:\n while x + y > k and j >= 0:\n x = b[j]\n j -= 1\n if x + y <= k:\n ans = max(ans,i+j+2)\n\nprint(ans)', '\nn,m,k = map(int,input().split())\na = list(map(int,input().split()))\nb = list(map(int,input().split()))\np = 0\ni = 0\nj = 0\nwhile p < k:\n if a[i] < b[j]:\n p += 1\n i += 1\n elif a[i] > b[j]:\n p += 1\n j += 1\n else:\n if a[i+1] < b[j+1]:\n p += 1\n i += 1\n else:\n p += 1\n j += 1\n\nprint(p)', 'n,m,k = map(int,input().split())\na = list(map(int,input().split()))\nb = list(map(int,input().split()))\nc = [0]*(n+1)\nd = [0]*(m+1)\nfor i in range(n):\n c[i+1] = a[i]+c[i]\nfor i in range(m):\n d[i+1] = b[i]+d[i]\nfor i in range(m+1)[::-1]:\n if d[i] <= k:\n x = i\n break\nans = x\nfor i in range(1,n+1):\n while c[i] + d[x] > k:\n if x >= 1:\n x -= 1\n else:\n break\n if c[i] + d[x] <= k:\n ans = max(ans,i+x)\nprint(ans)'] | ['Runtime Error', 'Runtime Error', 'Runtime Error', 'Accepted'] | ['s166204442', 's466171406', 's726417325', 's952820078'] | [9068.0, 40464.0, 42288.0, 47344.0] | [29.0, 349.0, 212.0, 309.0] | [487, 570, 400, 478] |
p02623 | u358859892 | 2,000 | 1,048,576 | 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. | ['import bisect\nn, m, k = map(int, input().split())\na = list(map(int, input().split()))\nb = list(map(int, input().split()))\n\nfor i in range(1, n):\n a[i] += a[i-1]\nfor i in range(1, m):\n b[i] += b[i-1]\n\nI = bisect.bisect_right(a, k)\nans = bisect.bisect_right(b, k)\nli = list(range(I))\nli = li[::-1]\nfor i in li:\n tmp = bisect.bisect_right(b, k-a[i])\n if a[i] + b[tmp-1] <= k && tmp != 0:\n ans = max(ans, i+tmp+1)\n\nprint(ans)', 'import bisect\nn, m, k = map(int, input().split())\na = list(map(int, input().split()))\nb = list(map(int, input().split()))\n\na2 = [0] * (n + 1)\nb2 = [0] * (m + 1)\nfor i in range(0, n):\n a2[i+1] = a[i] + a2[i]\nfor i in range(0, m):\n b2[i+1] = b[i] + b2[i]\n\nans = 0\nfor i in range(n+1):\n tmp = bisect.bisect_right(b2, k-a2[i]) - 1\n if a2[i] + b2[tmp] <= k and tmp >= 0:\n ans = max(ans, i+tmp)\n\nprint(ans)'] | ['Runtime Error', 'Accepted'] | ['s492929197', 's049868800'] | [9052.0, 47616.0] | [27.0, 344.0] | [440, 419] |
p02623 | u359358631 | 2,000 | 1,048,576 | 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. | ['def main():\n N, M, K = map(int, input().split())\n A = list(map(int, input().split()))\n B = list(map(int, input().split()))\n\n a, b = [0], [0]\n for i in range(N):\n a.append(a[i] + A[i])\n for i in range(M):\n a.append(b[i] + B[i])\n\n ans, j = 0, M\n for i in range(N + 1):\n if a[i] > K:\n break\n\n while a[i] + b[j] > K:\n j -= 1\n\n ans = max(ans, i + j)\n\n print(ans)\n\n\nif __name__ == "__main__":\n main()\n', 'def main():\n N, M, K = map(int, input().split())\n A = list(map(int, input().split()))\n B = list(map(int, input().split()))\n\n a, b = [0], [0]\n for i in range(N):\n a.append(a[i] + A[i])\n for i in range(M):\n b.append(b[i] + B[i])\n\n ans, j = 0, M\n for i in range(N + 1):\n if a[i] > K:\n break\n\n while a[i] + b[j] > K:\n j -= 1\n\n ans = max(ans, i + j)\n\n print(ans)\n\n\nif __name__ == "__main__":\n main()\n'] | ['Runtime Error', 'Accepted'] | ['s656972231', 's666798604'] | [40596.0, 47452.0] | [140.0, 216.0] | [482, 482] |
p02623 | u363074342 | 2,000 | 1,048,576 | 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. | ['n, M, k = map(int,input().split())\na_l = list(map(int,input().split()))\nb_l = list(map(int,input().split()))\nread_a = [0]*n\nread_b = [0]*M\ntime = 0\nans = 0\nmemo = 0\nfor j in range(n):\n if time + a_l[j] <= k:\n time += a_l[j]\n read_a[j] = 1\n memo = j\n ans = j+1\n else:\n break\n\nkotae = [ans]\n\n\ndef dfs(i,a,t,m):\n global kotae\n if i < M:\n if b_l[i] + t <= k:\n tt = t + b_l[i]\n aa = a + 1\n kotae.append(aa)\n ii = i+1\n mm = m\n #print(aa)\n dfs(ii,aa,tt,mm)\n \n else:\n if m < 0:\n continue\n tt = t - a_l[m]\n mm = m - 1\n aa = a - 1\n ii = i\n #print(aa)\n dfs(ii,aa,tt,mm)\n #print(aa)\n\n\ndfs(0,ans,time,memo)\n\n\nprint(max(kotae))\n \n\n\n\n', 'import sys\nsys.setrecursionlimit(10**6)\nn, M, k = map(int,input().split())\na_l = list(map(int,input().split()))\nb_l = list(map(int,input().split()))\nread_a = [0]*n\nread_b = [0]*M\ntime = 0\nans = 0\nmemo = -1\nfor j in range(n):\n if time + a_l[j] <= k:\n time += a_l[j]\n read_a[j] = 1\n memo = j\n ans = j+1\n else:\n break\n\nkotae = [ans]\n\n\ndef dfs(i,a,t,m):\n global kotae\n if i < M:\n if b_l[i] + t <= k:\n tt = t + b_l[i]\n aa = a + 1\n kotae.append(aa)\n ii = i+1\n mm = m\n #print(aa)\n dfs(ii,aa,tt,mm)\n \n else:\n if m >= 0:\n tt = t - a_l[m]\n mm = m - 1\n aa = a - 1\n ii = i\n dfs(ii,aa,tt,mm)\n \n\n\ndfs(0,ans,time,memo)\n\n\nprint(kotae)\n \n\n\n\n', 'n, M, k = map(int,input().split())\na_l = list(map(int,input().split()))\nb_l = list(map(int,input().split()))\nread_a = [0]*n\nread_b = [0]*M\ntime = 0\nans = 0\nmemo = 0\nfor j in range(n):\n if time + a_l[j] <= k:\n time += a_l[j]\n read_a[j] = 1\n memo = j\n ans = j+1\n else:\n break\n\nkotae = [ans]\n\n\ndef dfs(i,a,t,m):\n global kotae\n if i < M:\n if b_l[i] + t <= k:\n tt = t + b_l[i]\n aa = a + 1\n kotae.append(aa)\n ii = i+1\n mm = m\n #print(aa)\n dfs(ii,aa,tt,mm)\n \n else:\n if m < 0:\n continue\n tt = t - a_l[m]\n mm = m - 1\n aa = a - 1\n ii = i\n #print(aa)\n dfs(ii,aa,tt,mm)\n #print(aa)\n\n\ndfs(0,ans,time,memo)\n\n\nprint(max(kotae))\n \n\n\n\n', 'import sys\nsys.setrecursionlimit(10**6)\nn, M, k = map(int,input().split())\na_l = list(map(int,input().split()))\nb_l = list(map(int,input().split()))\nread_a = [0]*n\nread_b = [0]*M\ntime = 0\nans = 0\nmemo = -1\nfor j in range(n):\n if time + a_l[j] <= k:\n time += a_l[j]\n read_a[j] = 1\n memo = j\n ans = j+1\n else:\n break\n\nkotae = [ans]\n\n\ndef dfs(i,a,t,m):\n global kotae\n if i < M:\n if b_l[i] + t <= k:\n tt = t + b_l[i]\n aa = a + 1\n kotae.append(aa)\n ii = i+1\n mm = m\n #print(aa)\n dfs(ii,aa,tt,mm)\n \n else:\n if m >= 0:\n tt = t - a_l[m]\n mm = m - 1\n aa = a - 1\n ii = i\n dfs(ii,aa,tt,mm)\n \n\n\ndfs(0,ans,time,memo)\n\n\nprint(max(kotae))\n \n\n\n\n'] | ['Runtime Error', 'Wrong Answer', 'Runtime Error', 'Accepted'] | ['s131271905', 's342952216', 's393769352', 's485462455'] | [9124.0, 408808.0, 9124.0, 405596.0] | [28.0, 630.0, 31.0, 614.0] | [874, 873, 874, 878] |
p02623 | u363599046 | 2,000 | 1,048,576 | 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. | ['n, m, k = map(int, input().split())\na = list(map(int, input().split()))\nb = list(map(int, input().split()))\n\np, q = [0], [0]\nfor i in range(n):\n\tp.append(p[i] + a[i])\nfor i in range(m):\n\tq.append(q[i] + b[i])\n\nj, ans = m, 0\nfor i in range(N+1):\n\tif a[i] > k:\n\t\tbreak\n\n\twhile b[j] > k - a[i]:\n\t\tj -= 1\n\n\tans = max(ans, i+j)\n\nprint(ans)', 'n, m, k = map(int, input().split())\na = list(map(int, input().split()))\nb = list(map(int, input().split()))\n\np, q = [0], [0]\nfor i in range(n):\n\tp.append(p[i] + a[i])\nfor i in range(m):\n\tq.append(q[i] + b[i])\n\nj, ans = m, 0\nfor i in range(n+1):\n\tif p[i] > k:\n\t\tbreak\n\n\twhile q[j] > k - p[i]:\n\t\tj -= 1\n\n\tans = max(ans, i+j)\n\nprint(ans)'] | ['Runtime Error', 'Accepted'] | ['s965204331', 's402497931'] | [47496.0, 47416.0] | [191.0, 291.0] | [334, 334] |
p02623 | u364027015 | 2,000 | 1,048,576 | 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. | ['N,M,K=map(int,input().split())\nA=list(map(int,input().split()))\nB=list(map(int,input().split()))\nc=0\nimport numpy as np\nA=np.array(A).cumsum()\nB=np.array(B).cumsum()\nfor i in range(N):\n d=K-A[i]\n if d>=0:\n for j in reversed(range(M)):\n if d>=B[j]:\n if c<i+j+2:\n c=i+j+2\n M=j+1\n break\n if d<B[0]:\n c=i+1\n else:\n for j in reverse(range(M)):\n if K-B[j]>=0:\n if c<j+1:\n c=j+1\n break\n break\nprint(c)', 'N,M,K=map(int,input().split())\nA=list(map(int,input().split()))\nB=list(map(int,input().split()))\nc=0\nA.insert(0,0)\nB.insert(0,0)\nimport numpy as np\nA=np.array(A).cumsum()\nB=np.array(B).cumsum()\n\nfor i in range(N+1):\n d=K-A[i]\n for j in reversed(range(M+1)):\n if d-B[j]>=0:\n if c<i+j:\n c=i+j\n M=j\n break\nprint(c)'] | ['Runtime Error', 'Accepted'] | ['s908758063', 's426603356'] | [45900.0, 45860.0] | [542.0, 552.0] | [477, 338] |
p02623 | u364386647 | 2,000 | 1,048,576 | 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. | ['def resolve():\n N, M, K = list(map(int, input().split()))\n A = list(map(int, input().split()))\n B = list(map(int, input().split()))\n\n read = deque()\n\n while sum(read) <= K:\n if len(A) == 0 and len(B) == 0:\n break\n\n if len(A) == 0:\n if sum(read) + B[0] <= K:\n read.append(B[0])\n B.pop(0)\n continue\n\n if len(B) == 0:\n if sum(read) + A[0] <= K:\n read.append(A[0])\n A.pop(0)\n continue\n\n if A[0] < B[0]:\n if sum(read) + A[0] <= K:\n read.append(A[0])\n A.pop(0)\n else:\n if sum(read) + B[0] <= K:\n read.append(B[0])\n\n B.pop(0)\n\n print(len(read))\n\n\nresolve()', 'def resolve():\n N, M, K = list(map(int, input().split()))\n A = list(map(int, input().split()))\n B = list(map(int, input().split()))\n\n ans = 10 ** 10\n t = 0\n\n for i in range(M):\n t += B[i]\n j = M\n ans = 0\n for i in range(N+1):\n while j > 0 and t > K:\n j -= 1\n t -= B[j]\n \n if t > K:\n break\n \n ans = max(ans, i+j)\n \n if i == N:\n break\n t += A[i]\n \n print(ans)\n\nresolve()'] | ['Runtime Error', 'Accepted'] | ['s977441030', 's500005896'] | [40096.0, 40404.0] | [111.0, 185.0] | [783, 508] |
p02623 | u364555831 | 2,000 | 1,048,576 | 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. | ['N, M, K = map(int, input().split())\nA = list(map(int, input().split()))\nB = list(map(int, input().split()))\n\nsum_A = [0] * (N + 1)\nsum_B = [0] * (M + 1)\n\nfor a in range(1, N + 1):\n sum_A[a] = sum_A[a - 1] + A[a - 1]\nfor b in range(1, M + 1):\n sum_B[b] = sum_B[b - 1] + B[b - 1]\n\ncount = 0\n\nprint(sum_A)\nprint(sum_B)\nlast_index = M\n\nfor a in range(len(sum_A)):\n print("----------------------")\n print("a", a)\n if sum_A[a] >= K:\n count = max(count, a)\n break\n now = sum_A[a]\n rest = K - now\n print("last_index", last_index)\n \n\n while True:\n if last_index < 0:\n able = 0\n break\n if sum_B[last_index] > rest:\n last_index -= 1\n else:\n able = last_index\n break\n\n print("now", now, "rest", rest, "able", able)\n print("今回の最大countは", "a + able = ", a + able)\n count = max(count, a + able)\n print("これまでの最大count", count)\n\nprint(count)\n', 'N, M, K = map(int, input().split())\nA = list(map(int, input().split()))\nB = list(map(int, input().split()))\n\nsum_A = [0] * (N + 1)\nsum_B = [0] * (M + 1)\n\nfor a in range(1, N + 1):\n sum_A[a] = sum_A[a - 1] + A[a - 1]\nfor b in range(1, M + 1):\n sum_B[b] = sum_B[b - 1] + B[b - 1]\n\ncount = 0\n\n# print(sum_A)\n\nlast_index = M\n\nfor a in range(len(sum_A)):\n # print("----------------------")\n # print("a", a)\n if sum_A[a] > K:\n count = max(count, a - 1)\n break\n elif sum_A[a] == K:\n count = max(count, a)\n now = sum_A[a]\n rest = K - now\n \n \n\n while True:\n if last_index < 0:\n able = 0\n break\n if sum_B[last_index] > rest:\n last_index -= 1\n else:\n able = last_index\n break\n\n \n \n count = max(count, a + able)\n \n\nprint(count)\n'] | ['Wrong Answer', 'Accepted'] | ['s286296166', 's375606528'] | [74760.0, 47488.0] | [1106.0, 329.0] | [1047, 1120] |
p02623 | u365077014 | 2,000 | 1,048,576 | 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. | ['#include<bits/stdc++.h>\n#define INF 2000000000\n#define MOD 1000000007\n#define EPS (1e-10)\n\nusing namespace std;\n\nint main(int argc, char *argv[]) {\n\n\tlong N,M; cin >> N >> M;\n\tlong K; cin >> K;\n\t\n\tvector<long> A(N+1,0);\n\tfor (int i = 1; i <= N; i++) {\n\t\tint a; cin >> a;\n\t\tA[i] = A[i-1] + a;\n\t}\n\n\tvector<long> B(M+1,0);\n\tfor (int i = 1; i <= M; i++) {\n\t\tint b; cin >> b;\n\t\tB[i] = B[i-1] + b;\n\t}\n\n\tlong ans = 0;\n\tlong best = M;\n\tfor (int i = 1; i <= N; i++) {\n\t\tif (A[i] > K) {\n\t\t\tbreak;\t\n\t\t}\n\t\t\n\t\twhile (1 < best and B[best] > K - A[i]) {\n\t\t\tbest--;\t\t\n\t\t}\n\t\tans = max(ans, i+best);\n\n\t}\n\tcout << ans << endl;\n\n\n\n\treturn 0;\n}\n', 'N, M, K = map(int, input().split())\nA = list(map(int, input().split()))\nB = list(map(int, input().split()))\n\na, b = [0], [0]\nfor i in range(N):\n\ta.append(a[i] + A[i])\nfor i in range(M):\n\tb.append(b[i] + B[i])\n \nans, j = 0, M\nfor i in range(N + 1):\n\tif a[i] > K:\n\t\tbreak\n\twhile b[j] > K - a[i]:\n\t\tj -= 1\n\tans = max(ans, i + j)\nprint(ans)'] | ['Runtime Error', 'Accepted'] | ['s976899559', 's984852369'] | [8876.0, 47364.0] | [24.0, 279.0] | [624, 341] |
p02623 | u366996583 | 2,000 | 1,048,576 | 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. | ['N,M,K=map(int,input().split())\nimport numpy as np\na=np.array(list(map(int,input().split())))\nb=np.array(list(map(int,input().split())))\na_s=a.cumsum()\nb_s=b.cumsum()\nan=0\nans=0\nwhile an<N and a_s[an]<=K:\n bn=0\n while bn<M and a_s[an]+b_s[bn]<=K:\n bn+=1\n if an+bn+1>ans:\n ans=an+bn+1\n an+=1\nprint(ans)', 'N,M,K=map(int,input().split())\na=list(map(int,input().split()))\nb=list(map(int,input().split()))\na_s=[0]\nb_s=[0]\nfor i in range(N):\n if a_s[i]+a[i]>K:\n break\n a_s.append(a[i]+a_s[i])\nfor i in range(M):\n if b_s[i]+b[i]>K:\n break\n b_s.append(b[i]+b_s[i])\nbn=len(b_s)-1\nans=0\nfor i in range(len(a_s)):\n while a_s[i]+b_s[bn]>K:\n bn-=1\n ans=max(ans,i+bn)\n \nprint(ans)'] | ['Runtime Error', 'Accepted'] | ['s782453138', 's518968680'] | [8940.0, 40748.0] | [24.0, 328.0] | [312, 378] |
p02623 | u370576244 | 2,000 | 1,048,576 | 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. | ['N,M,K = map(int,input().split())\nA=list(map(int,input().split()))\nB=list(map(int,input().split()))\n\nr0=0\nrr = 0\nt0=0\ni = 0\nj = 0\nflag = 0\n\nwhile t0<=K:\n t0 = t0 + A[i]\n if t0<=K:\n r0 = r0 + 1\n print(t0,r0)\n if i==N-1:\n break\n i = i + 1\n\nif t0>K:\n t0 = t0 - A[i]\n\nwhile t0<=K:\n t0 = t0 + B[j]\n if t0<=K:\n r0 = r0 + 1\n print(t0,r0)\n if j==M-1:\n flag = 1\n break\n j = j + 1\n\nif t0>K:\n t0 = t0 - B[j]\n\nrr = r0\nKk = K\ntk = t0\nrk = r0\n\nfor k in range(i):\n if flag == 1:\n break\n Kk = Kk + A[i-k] - tk\n print(Kk)\n tk = 0\n ik = 0\n rk = rk - 1\n while tk<=Kk:\n tk = tk + B[j]\n if tk<=Kk:\n rk = rk + 1\n print(tk,rk)\n if j == M-1:\n flag = 1\n break\n j = j+1\n rr = max(rk,rr)\n\n\nprint(rr)', 'N,M,K = map(int,input().split())\nA=list(map(int,input().split()))\nB=list(map(int,input().split()))\n\na,b = [0],[0]\nfor i in range(N):\n a.append(a[i]+A[i])\nfor i in range(M):\n b.append(b[i]+B[i])\n\nAns,j = 0,M\nfor i in range(N+1):\n if a[i]>K:\n break\n while b[j]>K-a[i]:\n j = j-1\n Ans = max(Ans,i+j)\n\nprint(Ans)'] | ['Wrong Answer', 'Accepted'] | ['s503112356', 's581329476'] | [41076.0, 47436.0] | [588.0, 288.0] | [897, 336] |
p02623 | u370721525 | 2,000 | 1,048,576 | 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. | ["N, M, K = map(int, input().split())\nA = list(map(int, input().split()))\nB = list(map(int, input().split()))\n\ndef doku(li, K, time):\n cnt = 0\n tmp = 0\n for i in range(len(li)):\n if time+li[i] <= K:\n tmp += li[i]\n cnt += 1\n else:\n return cnt\n return cnt\n\nans = 0\ntime = 0\nwhile time+min(A[0], B[0]) <= K:\n plus = min(A[0], B[0])\n time += plus\n ans += 1\n if A[0] <= B[0]:\n A.pop(0)\n if not A:\n ans += doku(B, K, time)\n print('A')\n break;\n else:\n B.pop(0)\n if not B:\n ans += doku(A, K, time)\n print('B')\n break;\n print(A, B, time, ans)\n \nprint(ans)", 'N, M, K = map(int, input().split())\nA = list(map(int, input().split()))\nB = list(map(int, input().split()))\n\na = [0]\nb = [0]\nfor i in range(N):\n a.append(A[i] + a[i])\nfor i in range(M):\n b.append(B[i] + b[i])\n \nans = 0\nj = M\nfor i in range(N+1):\n if a[i] > K:\n break;\n while b[j] > K-a[i]:\n j -= 1\n ans = max(ans, i+j)\n \nprint(ans)'] | ['Runtime Error', 'Accepted'] | ['s602102478', 's562343345'] | [158524.0, 47456.0] | [2369.0, 300.0] | [620, 344] |
p02623 | u374935093 | 2,000 | 1,048,576 | 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. | ['import numpy as np\nN,M,K = map(int,input().split())\nA = list(map(int,input().split()))\nB = list(map(int,input().split()))\nans = 0\nA_array = np.array(A)\nB_array = np.array(B)\nA_cum = np.cumsum(A_array)\nB_cum = np.cumsum(B_array)\nfor i in range(len(A_cum)):\n k = A_cum[i]\n if k >K:\n break\n a = 0\n b = len(B_cum)-1\n print(B_check)\n while b - a != 0:\n c = int(round((a+b)/2))\n if B_check[c] < 0:\n b = c\n else:\n a = c\n if b-a == 1:\n if B_check[b] < 0:\n b = a\n else:\n a = b\n check = i+b+2\n if check>ans:\n ans = check\nprint(ans)', 'import bisect\nN,M,K = map(int,input().split())\nA = list(map(int,input().split()))\nB = list(map(int,input().split()))\nans = 0\nA_cum = [0]\nB_cum = [0]\nA_cum[0] = A[0]\nfor j in range(len(A)-1):\n A_cum.append(A_cum[j]+ A[j+1])\nB_cum[0] = B[0]\nfor l in range(len(B)-1):\n B_cum.append(B_cum[l] + B[l+1])\nfor i in range(len(A_cum)):\n k = A_cum[i]\n if k >K:\n break\n b = bisect.bisect(B_cum,K-k)\n check = i+b+1\n if check>ans:\n ans = check\ncheck = bisect.bisect(B_cum,K)\nif check>ans:\n ans = check \nprint(ans)'] | ['Runtime Error', 'Accepted'] | ['s414669920', 's653909832'] | [58420.0, 49068.0] | [223.0, 301.0] | [661, 540] |
p02623 | u377834804 | 2,000 | 1,048,576 | 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. | ['N, M, K = map(int, input().split())\nA = list(map(int, input().split()))\nB = list(map(int, input().split()))\n\nmmax = 0\npre = False\nnum = 0\nans = 0\nans2 = 0\nfor i in range(N+1):\n if i == 0:\n ans = 0\n else:\n ans += A[i-1]\n if ans > K:\n break\n for j in range(M+1):\n if j == 0:\n ans2 = ans\n else:\n ans2 += B[j-1]\n if ans2 > K:\n break\n else:\n mmax = max(mmax, i+j)\n print(i, j, ans2, mmax)\n\nprint(mmax)', 'N, M, K = map(int, input().split())\nA = list(map(int, input().split()))\nB = list(map(int, input().split()))\n\na = [0]\nb = [0]\nfor i in range(N):\n a.appned(a[i]+A[i])\nfor i in range(M):\n b.append(b[i]+B[i])\n\nans, j = 0, M\nfor i in range(len(A)):\n if a[i] > K:\n break\n while B[j] > K - a[i]:\n j -= 1\n ans = max(ans, i+j)\nprint(ans)', 'N, M, K = map(int, input().split())\nA = list(map(int, input().split()))\nB = list(map(int, input().split()))\n \npa, pb = 0, 0\n \nans = 0\nfa, fb = False, False\nwhile 1:\n if pa >= N and pb < M:\n if K - B[pb] >= 0:\n ans += 1\n pb += 1\n else:\n break\n elif pb >= M and pa < N:\n if K - A[pa] >= 0:\n ans += 1\n pa += 1\n else:\n break\n elif pa >= N and pb >= M:\n break\n elif A[pa] < B[pb]:\n if K - A[pa] >= 0:\n ans += 1\n pa += 1\n else:\n break\n else:\n if K - B[pb] >= 0:\n ans += 1\n pb += 1\n else:\n break\nprint(ans)', 'N, M, K = map(int, input().split())\nA = list(map(int, input().split()))\nB = list(map(int, input().split()))\n\npa, pb = 0, 0\n\nans = 0\nwhile 1:\n if A[pa] < B[pb]:\n if K - A[pa] >= 0:\n ans += 1\n pa += 1\n else:\n break\n else:\n if K - B[pb] >= 0:\n ans += 1\n pb += 1\n else:\n break', 'N, M, K = map(int, input().split())\nA = list(map(int, input().split()))\nB = list(map(int, input().split()))\n \na = [0]\nb = [0]\nfor i in range(N):\n a.append(a[i]+A[i])\nfor i in range(M):\n b.append(b[i]+B[i])\n\nans, j = 0, M\nfor i in range(N+1):\n if a[i] > K:\n break\n while b[j] > K - a[i]:\n j -= 1\n ans = max(ans, i+j)\nprint(ans)'] | ['Wrong Answer', 'Runtime Error', 'Wrong Answer', 'Runtime Error', 'Accepted'] | ['s040076614', 's043819332', 's578979115', 's596897794', 's001635819'] | [60448.0, 40516.0, 39900.0, 40440.0, 47320.0] | [2567.0, 114.0, 252.0, 190.0, 284.0] | [444, 339, 591, 316, 337] |
p02623 | u380408685 | 2,000 | 1,048,576 | 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. | ['from sys import stdin\nimport sys\n\nl = stdin.readline().rstrip().split(\' \')\na = stdin.readline().rstrip().split(\' \')\nb = stdin.readline().rstrip().split(\' \')\nk = int(l[2])\nn = int(l[0])\nm = int(l[1])\nia = 0\nib = 0\nb_sum = 0\ncount = 0\na_sum = 0\nres = []\n\nfor ib in range(m):\n b_sum += int(b[ib])\n if b_sum <= k:\n count += 1\n else:\n break\nres.append(count)\ncount = 0\nfor ia in range(n):\n a_sum = a_sum + int(a[ia])\n if k < a_sum:\n #print("ok")\n res.append(count)\n continue\n else:\n count = ia + 1\n b_sum = a_sum\n for ib in range(m):\n b_sum += int(b[ib])\n if b_sum <= k:\n count += 1\n else:\n break\n res.append(count)\nprint(res)', "from sys import stdin\nimport sys\n\nl = stdin.readline().rstrip().split(' ')\na = stdin.readline().rstrip().split(' ')\nb = stdin.readline().rstrip().split(' ')\nk = int(l[2])\nn = int(l[0])\nm = int(l[1])\nia = 0\nib = 0\nsum = 0\ncount = 0\na_sum = 0\nres = []\nb_list = []\nfor bt in b:\n sum += int(bt)\n if sum <= k:\n count += 1\n b_list.append(sum)\n else:\n break\n#print(b_list)\nres.append(count)\ncount = 0\nia = 1\nfor at in a:\n a_sum += int(at)\n if k < a_sum:\n print(max(res))\n sys.exit()\n count = ia\n sum = a_sum\n for bsum in b_list:\n sum = a_sum + bsum\n if k < sum:\n print(max(res))\n sys.exit()\n else:\n count += 1\n ia += 1\n res.append(count)\nprint(max(res))", "from sys import stdin\nimport sys\n\nl = stdin.readline().rstrip().split(' ')\nA = list(map(int, input().split()))\nB = list(map(int, input().split()))\nk = int(l[2])\nn = int(l[0])\nm = int(l[1])\n\nsum = 0\na = [0]\nfor i in range(n):\n sum += A[i]\n a.append(sum)\nsum = 0\n\nb = [0]\nfor i in range(m):\n sum += B[i]\n b.append(sum)\n#print(b)\n\nans = 0\nj = m\nfor i in range(n+1):\n if a[i] > k:\n break\n while b[j] > k - a[i]:\n j -= 1\n ans = max(ans, i+j)\nprint(ans)"] | ['Wrong Answer', 'Wrong Answer', 'Accepted'] | ['s596274643', 's736984543', 's070940948'] | [42060.0, 46840.0, 47268.0] | [2206.0, 2207.0, 297.0] | [733, 765, 482] |
p02623 | u382639013 | 2,000 | 1,048,576 | 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. | ['N, M, K = list(map(int, input().split()))\nA = list(map(int, input().split()))\nB = list(map(int, input().split()))\n\nans = 0\n\nif ab_min > K:\n ans = 0\nelse:\n while True:\n K = K - min(A[0],B[0])\n if K < 0:\n break\n ans += 1\n if min(A[0],B[0])==A[0]:\n A.pop(0)\n if A == []:\n A.append(10**9+1)\n else:\n B.pop(0)\n if B == []:\n B.append(10**9+1)\nprint(ans)', 'n, m, k = map(int, input().split())\nA = list(map(int, input().split()))\nB = list(map(int, input().split()))\na = [0]\nb = [0]\nfor i in range(n):\n a.append(a[i] + A[i])\nfor i in range(m):\n b.append(b[i] + B[i])\n\nans = 0 \nfor i in range(n + 1):\n if a[i] > k:\n break\n while a[i] + b[m] > k:\n m -= 1\n ans = max(ans, i + m)\nprint(ans)'] | ['Runtime Error', 'Accepted'] | ['s564347165', 's564723510'] | [40548.0, 47508.0] | [113.0, 298.0] | [472, 357] |
p02623 | u385677760 | 2,000 | 1,048,576 | 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. | ['N, M, K = map(int, input().split())\nA = list(map(int, input().split()))\nB = list(map(int, input().split()))\na, b = [0], [0]\nfor i in range(N):\na.append(a[i] + A[i])\nfor i in range(M):\nb.append(b[i] + B[i])\nans, j = 0, M\nfor i in range(N + 1):\nif a[i] > K:\nbreak\nwhile b[j] > K - a[i]:\nj -= 1\nans = max(ans, i + j)\nprint(ans)', 'N,M,K = (int(x) for x in input().split())\nA = list(map(int, input().split()))\nB = list(map(int, input().split()))\ni = 0\nj = 0\ntime = 0\ncount = 0\nbuff = -1\nwhile time < K and count-buff == 1 and i < N or j < M:\n buff = count\n if i >= N or j >= M:\n if i >= N:\n if K-time>B[j]:\n time += B[j]\n j += 1\n count += 1\n elif j >= M:\n if K-time>A[i]:\n time += A[i]\n i += 1\n count += 1\n else:\n if A[i] < B[j]:\n if K-time>A[i]:\n time += A[i]\n i += 1\n count += 1\n elif A[i] >= B[j]:\n if K-time>B[j]:\n time += B[j]\n j += 1\n count += 1\n\nprint(count)', 'N,M,K = (int(x) for x in input().split())\nA = list(map(int, input().split()))\nB = list(map(int, input().split()))\ni = 0\nj = 0\ntime = 0\ncount = 0\nbuff = 0\nwhile time < K and count-buff ==1 and i < N or j < M:\n buff = count\n if i >= N or j >= M:\n if i >= N:\n time += B[j]\n j += 1\n count += 1\n elif j >= M:\n time += A[i]\n i += 1\n count += 1\n else:\n if A[i] < B[j]:\n time += A[i]\n i += 1\n count += 1\n elif A[i] >= B[j]:\n time += B[j]\n j += 1\n count += 1\n\nprint(count)', 'N,M,K = (int(x) for x in input().split())\nA = list(map(int, input().split()))\nB = list(map(int, input().split()))\ni = 0\nj = 0\ntime = 0\ncount = 0\nbuff = -1\nwhile time < K and count-buff == 1 and i < N or j < M:\n buff = count\n if i >= N or j >= M:\n if i >= N:\n time += B[j]\n j += 1\n count += 1\n elif j >= M:\n time += A[i]\n i += 1\n count += 1\n else:\n if A[i] < B[j]:\n time += A[i]\n i += 1\n count += 1\n elif A[i] >= B[j]:\n time += B[j]\n j += 1\n count += 1\n\nprint(count)', 'N,M,K = (int(x) for x in input().split())\nA = list(map(int, input.split()))\nB = list(map(int, input.split()))\ni,j = 0\ntime = 0\nwhile time < K and i < N-1 or j < M-1:\n if A[i] < B[j]:\n time += A[i]\n i += 1\n elif A[i] > B[j]:\n time += B[j]\n j += 1\nallbooks = i + j\nprint(allbooks)', 'N, M, K = map(int, input().split())\nA = list(map(int, input().split()))\nB = list(map(int, input().split()))\na, b = [0], [0]\nfor i in range(N):\n a.append(a[i] + A[i])\nfor i in range(M):\n b.append(b[i] + B[i])\nans, j = 0, M\nfor i in range(N + 1):\n if a[i] > K:\n break\n while b[j] > K - a[i]:\n j -= 1\n ans = max(ans, i + j)\nprint(ans)'] | ['Runtime Error', 'Wrong Answer', 'Wrong Answer', 'Wrong Answer', 'Runtime Error', 'Accepted'] | ['s084458505', 's139056827', 's232927579', 's576569074', 's604299024', 's879053928'] | [9004.0, 40056.0, 42212.0, 41912.0, 9164.0, 47600.0] | [26.0, 2206.0, 286.0, 286.0, 24.0, 279.0] | [324, 794, 632, 634, 312, 360] |
p02623 | u388497015 | 2,000 | 1,048,576 | 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. | ['import bisect\nn, m, k = list(map(int, input().split()))\na = list(map(int, input().split()))\nb = list(map(int, input().split()))\na_acc = [0]\nb_acc = [0]\nnow = 0\nfor i in range(n):\n now += a[i]\n a_acc.append(now)\n\nnow = 0\nfor i in range(m):\n now += b[i]\n b_acc.append(now)\n\nans = 0\nfor i in range(n+1):\n t = k - a_acc[i]\n if t < 0:\n break\n ind = bisect.bisect_left(b_acc, t)\n ans = max(ans, i+ind)\n\nprint(ans)\n', 'import bisect\nn, m, k = list(map(int, input().split()))\na = list(map(int, input().split()))\nb = list(map(int, input().split()))\na_acc = [0]\nb_acc = [0]\nnow = 0\nfor i in range(n):\n now += a[i]\n a_acc.append(now)\n\nnow = 0\nfor i in range(m):\n now += b[i]\n b_acc.append(now)\n\nans = 0\nfor i in range(n+1):\n t = k - a_acc[i]\n if t < 0:\n break\n ind = bisect.bisect_left(b_acc, t)\n if ind == m+1:\n ans = max(ans, i+m)\n ind -= 1\n if b_acc[ind] == t:\n ans = max(ans, i+ind)\n else:\n ans = max(ans, i+ind-1)\n\nprint(ans)\n'] | ['Wrong Answer', 'Accepted'] | ['s563245710', 's325479859'] | [47368.0, 47396.0] | [310.0, 398.0] | [419, 533] |
p02623 | u388971072 | 2,000 | 1,048,576 | 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. | ['N,M,K = map(int,input().split())\nA = list(map(int,input().split()))\nB = list(map(int,input().split()))\ntime = 0\ncount = 0\nAin = 0\nBin = 0\nlenA = len(A)\nlenB = len(B)\nif(min(min(A),min(B))>K):\n print(0)\n exit()\nwhile time<=K:\n if((lenA!=Ain) and (lenB!=Bin)):\n if(A[Ain]<B[Bin]):\n time+= A[Ain]\n Ain+=1\n count +=1\n elif(A[Ain]>B[Bin]):\n time += B[Bin]\n Bin+=1\n count+=1\n if((lenA==Ain) and (lenB==Bin)):\n break\n elif(lenA==Ain):\n time += B[Bin]\n Bin += 1\n count += 1\n elif(lenB==Bin):\n time += A[Ain]\n Ain += 1\n count += 1\nif(time == K):\n print(count)\nelse:\n print(count-1)', 'N,M,K = map(int,input().split())\nA = list(map(int,input().split()))\nB = list(map(int,input().split()))\ncount = 0\nAin = 0\nBin = 0\ncheck = False\nif(min(min(A),min(B))>K):\n print(0)\nwhile K>=0:\n if(Ain != N and Bin !=M): \n if(A[Ain] < B[Bin]):\n K-=A[0]\n Ain += 1\n else:\n K-=B[0]\n Bin += 1\n \n elif((Ain == N) and (Bin == M)):\n check = True\n break\n if(Ain==N):\n K -= B[Bin]\n Bin += 1\n elif(Bin==M):\n K -= A[Ain]\n Ain += 1\nif(K==0 or check):\n print(Ain+Bin)\nelse:\n print(Ain+Bin-1)', 'N,M,K = map(int,input().split())\nA = list(map(int,input().split()))\nB = list(map(int,input().split()))\nA_sum = [0]\nB_sum = [0]\nA_sum.extend(A)\nB_sum.extend(B)\nfor i in range(1,N+1):\n A_sum[i]+=A_sum[i-1]\nfor i in range(1,M+1):\n B_sum[i]+=B_sum[i-1]\nans,j=0,M\nfor i in range(N+1):\n if A_sum[i]>K:\n break\n while B_sum[j]+A_sum[i]>K:\n j-=1\n ans = max(ans,i+j)\nprint(ans)'] | ['Wrong Answer', 'Wrong Answer', 'Accepted'] | ['s767777577', 's960878285', 's374799861'] | [40448.0, 40584.0, 49128.0] | [2206.0, 243.0, 277.0] | [725, 603, 396] |
p02623 | u391589398 | 2,000 | 1,048,576 | 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. | ['n, m, k = map(int, input().split())\nA = tuple(map(int, input().split()))\nB = tuple(map(int, input().split()))\nimport numpy as np\nfrom itertools import accumulate\nAS = np.array([0] + list(accumulate(A)))\nBS = np.array([0] + list(accumulate(B)))\n\nans = 0\nfor i in range(n+1):\n if AS[i] > k:\n break\n for j in range(ans-i-1, m+1):\n if AS[i] + BS[j] > k:\n break\n ans = max(ans, i+j)\n \nprint(ans)\n', 'n, m, k = map(int, input().split())\nA = tuple(map(int, input().split()))\nB = tuple(map(int, input().split()))\n\nfrom itertools import accumulate\nAS = [0] + list(accumulate(A))\nBS = [0] + list(accumulate(B))\n\nfrom bisect import bisect\nans = 0\nfor i in range(n+1):\n if AS[i] > k:\n break\n \n # ans = max(ans, kb+i)\n ans = max(ans, bisect(BS, k - AS[i])-1+i)\n\nprint(ans)\n'] | ['Wrong Answer', 'Accepted'] | ['s310565898', 's355311193'] | [56820.0, 50684.0] | [986.0, 283.0] | [436, 414] |
p02623 | u392441504 | 2,000 | 1,048,576 | 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. | ['N,M,K = map(int, input().split())\n \nA = [list(map(int, input().split())) for _ in range(N)]\n\nB = [list(map(int, input().split())) for _ in range(M)]\n\nsum=0\ncount = 0\nfor i in range(N+M):\n if A[0] >= B[0]:\n temp = B.pop(0)\n if sum+temp > K:\n break\n print(count)\n else:\n sum +=temp\n count += 1\n\n else:\n temp = A.pop(0)\n if sum + temp > K:\n break\n print(count)\n else:\n sum += temp\n count += 1\n\n\n\n\n\n\n', 'N,M,K = map(int, input().split())\n\n \nA = list(map(int, input().split()))\n\nB = list(map(int, input().split()))\nsum=0\ncount = 0\nfor i in range(N+M):\n if A[0] >= B[0]:\n temp = B.pop(0)\n if sum+int(temp) > K:\n break\n print(count)\n else:\n sum +=temp\n count += 1\n\n else:\n temp = A.pop(0)\n if sum + int(temp) > K:\n break\n print(count)\n else:\n sum += temp\n count += 1\n\n\n\n\n\n\n', 'N,M,K = map(int, input().split())\n \nA = [list(map(int, input().split())) for _ in range(N)]\n\nB = [list(map(int, input().split())) for _ in range(M)]\n\nsum=0\ncount = 0\nfor i in range(N+M):\n if A[0] >= B[0]:\n temp = B.pop(0)\n if sum+temp > K:\n break\n print(count)\n else:\n sum +=temp\n count += 1\n\n else:\n temp = A.pop(0)\n if sum + temp > K:\n break\n print(count)\n else:\n sum += temp\n count += 1\n\n\n\n\n\n\n', 'n,m,k = map(int, input().split())\nA = list(map(int, input().split()))\nB = list(map(int, input().split()))\n\n\na = [0]\nb = [0]\n\nfor i in range(n):\n a.append(a[i]+ A[i])\nfor i in range(m):\n b.append(b[i] + B[i])\n# print(a)\n# print(b)\n# print(a[3])\nans=0\nj = m\nfor i in range(n+1):\n if a[i] > k:\n break\n while b[j] > k - a[i]:\n j -= 1\n ans = max(ans,i+j)\nprint(ans)'] | ['Runtime Error', 'Runtime Error', 'Runtime Error', 'Accepted'] | ['s146569395', 's290617951', 's386493924', 's591750438'] | [40700.0, 40556.0, 40536.0, 47360.0] | [105.0, 2206.0, 107.0, 298.0] | [551, 521, 551, 389] |
p02623 | u394731058 | 2,000 | 1,048,576 | 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. | ['n ,m, k = map(int, input().split())\na = list(map(int, input().split()))\nb = list(map(int, input().split()))\nans = 0\nwhile k > 0:\n i, j = 0\n if a[i] <= b[j]:\n k -= a[i]\n if k >= 0\n i += 1\n ans += 1\n else:\n k -= b[j]\n if k >= 0\n j += 1\n ans += 1\nprint(ans)', 'n ,m, k = map(int, input().split())\na = list(map(int, input().split()))\nb = list(map(int, input().split()))\nans = 0\nwhile k > 0:\n i, j = 0\n if a[i] <= b[j]:\n k -= a[i]\n if k >= 0\n i += 1\n ans += 1\n if i >= n:\n for o in range(j, m):\n if k >= b[o]:\n k -= b[o]\n ans += 1\n else:\n k -= b[o]\n break\n else:\n k -= b[j]\n if k >= 0\n j += 1\n ans += 1\n if j >= m:\n for p in range(i, n):\n if k >= a[p]:\n k -= a[p]\n ans += 1\n else:\n k -= a[p]\n break\nprint(ans)', 'n, m, k = map(int, input().split())\na = list(map(int, input().split()))\nb = list(map(int, input().split()))\n\na_cs, b_cs = [0], [0]\nfor i in range(n):\n a_cs.append(a_cs[i] + a[i])\nfor i in range(m):\n b_cs.append(b_cs[i] + b[i])\n\nans, j = 0, m\nfor i in range(n+1):\n if a_cs[i] > k:\n break\n while b_cs[j] > k - a_cs[i]:\n j -= 1\n ans = min(ans, i + j)\nprint(ans)', 'n, m, k = map(int, input().split())\na = list(map(int, input().split()))\nb = list(map(int, input().split()))\n\na_cs, b_cs = [0], [0]\nfor i in range(n):\n a_cs.append(a_cs[i] + a[i])\nfor i in range(m):\n b_cs.append(b_cs[i] + b[i])\n\nans, j = 0, m\nfor i in range(n+1):\n if a_cs[i] > k:\n break\n while b_cs[j] > k - a_cs[i]:\n j -= 1\n ans = max(ans, i + j)\nprint(ans)'] | ['Runtime Error', 'Runtime Error', 'Wrong Answer', 'Accepted'] | ['s245628729', 's480091243', 's979805771', 's834444322'] | [9032.0, 8952.0, 47492.0, 47552.0] | [24.0, 25.0, 282.0, 288.0] | [289, 629, 369, 369] |
p02623 | u395894569 | 2,000 | 1,048,576 | 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. | ['from collections import deque\n\nn,m,k = map(int, input().split())\na = [int(_) for _ in input().split()]\nb = [int(_) for _ in input().split()]\n\nx = deque(a)\ny = deque(b)\n\nans=0\nfor i in range(n + m):\n print(x, y)\n try:\n temp = x.popleft() if x[0] < y[0] else y.popleft()\n except:\n if len(x) == 0:\n temp = y.popleft()\n elif len(y) == 0:\n temp=x.popleft()\n print(temp)\n\n k -= int(temp)\n if k >= 0:\n ans += 1\n else:\n break\nprint(ans)', 'n,m,k = map(int, input().split())\na = [int(_) for _ in input().split()]\nb = [int(_) for _ in input().split()]\n\nt = sum(b)\nj = m\nans = 0\nfor i in range(n+1):\n while j > 0 and t > k:\n j -= 1\n t -= b[j]\n if t > k: break\n ans = max(ans, i + j)\n if i==n:break\n t+=a[i]\n\nprint(ans)'] | ['Runtime Error', 'Accepted'] | ['s714026732', 's686189652'] | [162748.0, 40660.0] | [2384.0, 249.0] | [514, 304] |
p02623 | u396210538 | 2,000 | 1,048,576 | 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. | ['from sys import stdin\nimport sys\nimport math\nimport random\nfrom decimal import Decimal\n\n\nN, M, K = [int(x) for x in stdin.readline().rstrip().split()]\nA = list(map(int, input().split()))\nA1 = [0]\nfor i, v in enumerate(A):\n x = A1[i]+v\n A1.append(x)\nB = list(map(int, input().split()))\nB1 = [0]\nfor i, v in enumerate(B):\n x = B1[i]+v\n B1.append(x)\nc1 = 0\nc2 = 0\nans = 0\ntemp = 0\nprint(A1, B1)\nfor i in range(len(A1)):\n if A1[i] > K:\n break\n for k in range(len(B1)):\n if B1[k] > K:\n break\n if A1[i]+B1[k] <= K:\n if i+k > ans:\n ans = i+k\nprint(ans)', 'from sys import stdin\nimport sys\nimport math\nimport random\nfrom decimal import Decimal\n\n\nN, M, K = [int(x) for x in stdin.readline().rstrip().split()]\nA = list(map(int, input().split()))\nA1 = [0]\nfor i, v in enumerate(A):\n x = A1[i]+v\n A1.append(x)\nB = list(map(int, input().split()))\nB1 = [0]\nfor i, v in enumerate(B):\n x = B1[i]+v\n B1.append(x)\nc1 = 0\nc2 = 0\nans = 0\ntemp = M\n\nfor i in range(N+1):\n if A1[i] > K:\n break\n while B1[temp] > K-A1[i]:\n temp -= 1\n ans = max(ans, i+temp)\n # for k in range(temp, 0, -1):\n # if A1[i]+B1[k] <= K:\n # ans = max(ans, i+k)\n # temp = k\n # break\n# print(c1, c2)\n\nprint(ans)\n'] | ['Wrong Answer', 'Accepted'] | ['s118280687', 's103320962'] | [57068.0, 51832.0] | [2214.0, 304.0] | [621, 690] |
p02623 | u396211450 | 2,000 | 1,048,576 | 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. | ['n,m,k=map(int,input().split())\nA=list(map(int,input().split()))\nB=list(map(int,input().split()))\na=[0]\nb=[0]\nfor i in range(N):\n\ta.append(a[i] + A[i])\nfor i in range(M):\n\tb.append(b[i] + B[i])\nans, j = 0, M\nfor i in range(N + 1):\n\tif a[i] > K:\n\t\tbreak\n\twhile b[j] > K - a[i]:\n\t\tj -= 1\n\tans = max(ans, i + j)\nprint(ans)\n', 'N,M,K=map(int,input().split())\nA=list(map(int,input().split()))\nB=list(map(int,input().split()))\na=[0]\nb=[0]\nfor i in range(N):\n\ta.append(a[i] + A[i])\nfor i in range(M):\n\tb.append(b[i] + B[i])\nans, j = 0, M\nfor i in range(N + 1):\n\tif a[i] > K:\n\t\tbreak\n\twhile b[j] > K - a[i]:\n\t\tj -= 1\n\tans = max(ans, i + j)\nprint(ans)\n'] | ['Runtime Error', 'Accepted'] | ['s739137565', 's366792583'] | [40700.0, 47364.0] | [112.0, 281.0] | [319, 319] |
p02623 | u398383435 | 2,000 | 1,048,576 | 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. | ['N,M,K = input().split()\n\nN = int(N)\nM = int(M)\nK = int(K)\n\nA = list(input().split())\nA = [int(c) for c in A]\n\nSum_A = [0]\nele_A = 0\na = 0\n\nwhile Sum_A[-1] + A[a] <= K:\n ele_A += A[a]\n a += 1\n Sum_A.append(ele_A)\n if a == len(A):\n break\n\nB = list(input().split())\nB = [int(c) for c in B]\n\nSum_B = [0]\nele_B = 0\nb = 0\n\nwhile Sum_B[-1] + B[b] <= K:\n ele_B += B[b]\n b += 1\n Sum_B.append(ele_B)\n if b == len(B):\n break\n\nh = len(Sum_A) - 1\ni = 0\nj = 0\n\nwhile h > 0:\n while Sum_A[h] + Sum_B[i] <= K:\n j = max(j,h + i)\n i += 1\n if i == len(Sum_B) - 1:\n break\n if i == len(Sum_B) - 1:\n j = max(j,h + i)\n break\n h -= 1\n\nprint(j)', 'N,M,K = input().split()\n\nN = int(N)\nM = int(M)\nK = int(K)\n\nA = list(input().split())\nA = [int(c) for c in A]\n\nSum_A = [0]\nele_A = 0\na = 0\n\nwhile Sum_A[-1] + A[a] <= K:\n ele_A += A[a]\n a += 1\n Sum_A.append(ele_A)\n if a == len(A):\n break\n\nB = list(input().split())\nB = [int(c) for c in B]\n\nSum_B = [0]\nele_B = 0\nb = 0\n\nwhile Sum_B[-1] + B[b] <= K:\n ele_B += B[b]\n b += 1\n Sum_B.append(ele_B)\n if b == len(B):\n break\n\nh = len(Sum_A) - 1\ni = 0\nj = 0\n\nwhile h >= 0:\n while Sum_A[h] + Sum_B[i] <= K:\n j = max(j,h + i)\n if i == len(Sum_B) - 1:\n break\n i += 1\n if h == 0:\n break\n h -= 1\n\nprint(j)'] | ['Wrong Answer', 'Accepted'] | ['s325214086', 's151032396'] | [47764.0, 48528.0] | [422.0, 454.0] | [711, 674] |
p02623 | u399155892 | 2,000 | 1,048,576 | 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. | ["#!/usr/bin/env python\nfrom collections import deque\n\ndef main():\n n, m, k = map(int, input().split())\n a_lst = list(map(int, input().split()))\n b_lst = list(map(int, input().split()))\n if n < m:\n a_lst, b_lst = b_lst, a_lst\n n, m = m, n\n\n a_time = [0]\n b_time = []\n\n t = 0\n for a in a_lst:\n t += a\n if t > k:\n break\n a_time.append(t)\n t = 0\n for b in b_lst:\n t += b\n if t > k:\n break\n b_time.append(t)\n\n a_que = deque(a_time)\n b_que = deque(b_time)\n cnt = len(a_que)\n c_max = cnt - 1\n while c_max <= cnt and a_que and b_que:\n cnt -= 1\n a = a_que.pop()\n b = b_que.popleft()\n while a + b <= k:\n cnt += 1\n c_max = cnt\n if b_que:\n b = b_que.popleft()\n else:\n break\n\n print(c_max)\n\n\nif __name__ == '__main__':\n main()\n", "#!/usr/bin/env python\nfrom collections import deque\n\ndef main():\n n, m, k = map(int, input().split())\n a_lst = list(map(int, input().split()))\n b_lst = list(map(int, input().split()))\n if n < m:\n a_lst, b_lst = b_lst, a_lst\n n, m = m, n\n\n a_time = [0]\n b_time = [0]\n\n t = 0\n for a in a_lst:\n t += a\n if t > k:\n break\n a_time.append(t)\n t = 0\n for b in b_lst:\n t += b\n if t > k:\n break\n b_time.append(t)\n\n ret = 0\n j = len(b_time) - 1\n for i, a in enumerate(a_time):\n while a + b_time[j] > k:\n j -= 1\n ret = max(ret, i+j)\n\n print(ret)\n\nif __name__ == '__main__':\n main()\n"] | ['Wrong Answer', 'Accepted'] | ['s244436795', 's366645174'] | [44176.0, 42184.0] | [178.0, 220.0] | [944, 718] |
p02623 | u401810884 | 2,000 | 1,048,576 | 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. | ['import sys\nsys.setrecursionlimit(2147483647)\n\nN, M, K=map(int,input().split())\nA=list(map(int,input().split()))\nB=list(map(int,input().split()))\nAsum = []\nBsum = []\ncount = 0\nfor p in A:\n count += p\n Asum.append(count)\ncount = 0\nfor p in B:\n count += p\n Bsum.append(count)\nbest = 0\nBlast = M-1\nfor t in range(N):\n if Asum(t) > K:\n break\n if best < t+1:\n best=t+1\n for s in reversed(range(0,Blast+1)):\n if Asum[t]+Bsum[s]<=K:\n if best < t+1+s+1:\n best=t+1+s+1\n Blast = s\n break\nprint(best)\n\n', 'import sys\nsys.setrecursionlimit(2147483647)\n\nN, M, K=map(int,input().split())\nA=list(map(int,input().split()))\nB=list(map(int,input().split()))\nAsum = [0]\nBsum = [0]\ncount = 0\nfor p in A:\n count += p\n Asum.append(count)\ncount = 0\nfor p in B:\n count += p\n Bsum.append(count)\nbest = 0\nBlast = M-1\nfor t in range(N+1):\n if Asum[t] > K:\n break\n if best < t+1:\n best=t+1\n for s in reversed(range(0,Blast+1)):\n if Asum[t]+Bsum[s]<=K:\n if best < t+1+s+1:\n best=t+1+s+1\n Blast = s\n break\nprint(best)\n\n', 'import sys\nsys.setrecursionlimit(2147483647)\n\nN, M, K=map(int,input().split())\nA=list(map(int,input().split()))\nB=list(map(int,input().split()))\nAsum = [0]\nBsum = [0]\ncount = 0\nfor p in A:\n count += p\n Asum.append(count)\ncount = 0\nfor p in B:\n count += p\n Bsum.append(count)\nbest = 0\nBlast = M-1\nfor t in range(N+1):\n if Asum[t] > K:\n break\n if best < t:\n best=t\n for s in reversed(range(0,Blast+1)):\n if Asum[t]+Bsum[s]<=K:\n if best < t+s+1:\n best=t+s+1\n Blast = s\n break\nprint(best)\n\n', 'import sys\nsys.setrecursionlimit(2147483647)\n\nN, M, K=map(int,input().split())\nA=list(map(int,input().split()))\nB=list(map(int,input().split()))\nAsum = [0]\nBsum = [0]\ncount = 0\nfor p in A:\n count += p\n Asum.append(count)\ncount = 0\nfor p in B:\n count += p\n Bsum.append(count)\nbest = 0\nBlast = M\nfor t in range(N+1):\n if Asum[t] > K:\n break\n if best < t:\n best=t\n for s in reversed(range(0,Blast+1)):\n if Asum[t]+Bsum[s]<=K:\n if best < t+s:\n best=t+s\n Blast = s\n break\nprint(best)\n\n'] | ['Runtime Error', 'Wrong Answer', 'Wrong Answer', 'Accepted'] | ['s120867920', 's144162738', 's363190917', 's033854230'] | [47144.0, 47384.0, 47504.0, 47180.0] | [196.0, 352.0, 341.0, 334.0] | [580, 584, 576, 570] |
p02623 | u405276711 | 2,000 | 1,048,576 | 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. | ['na,nb,k = map(int, input().split())\nta = map(int, input().split())\ntb = map(int, input().split())\n\ndef get_bcnt( usek,tb=tb ):\n t = 0\n cnt = 0\n try:\n while True:\n if t + tb[cnt] > usek:\n return cnt\n else:\n t += tb[cnt]\n cnt += 1\n except:\n return len(tb)\n\nsumta = 0\nreadbook = [get_bcnt(k)] \nfor i in range(na):\n sumta += ta[i]\n if sumta > k:\n break\n if ta[i] <= k:\n readbook.append( i+1 + get_bcnt( k - sumta ) )\n\nprint( max(readbook) )\n ', 'na,nb,k = map(int, input().split())\nta = list(map(int, input().split()))\ntb = list(map(int, input().split()))\n\n\ndef get_bcnt( usek,tb=tb ):\n tb_sum = [ sum(tb[:i]) + tt for i, tt in enumerate(tb)]\n tb_sum_tf = [ i <= usek for i in tb_sum]\n \n if False in tb_sum_tf:\n return tb_sum_tf.index(False)\n else:\n return len(tb)\n \n \nreadbook = [ i+1 + get_bcnt( k - sum( ta[:i] ) ) for i in range(na) if sum( ta[:i] ) <= k ]\nreadbook.append( get_bcnt(k) )', 'na,nb,k = map(int, input().split())\nta = map(int, input().split())\ntb = map(int, input().split())\n\ndef get_bcnt( usek,tb=tb ):\n t = 0\n cnt = 0\n try:\n while True:\n if t + tb[cnt] > usek:\n return cnt\n else:\n t += tb[cnt]\n cnt += 1\n except:\n return len(tb)\n\nsumta = 0\nreadbook = [get_bcnt(k)] \nfor i in range(na):\n sumta += ta[i]\n if sumta > k:\n break\n if ta[i] <= k:\n readbook.append( i+1 + get_bcnt( k - sumta ) )\n\n\n\n\nprint( max(readbook) )\n ', 'na,nb,k = map(int, input().split())\nta = list(map(int, input().split()))\ntb = list(map(int, input().split()))\n \ndef get_bcnt( usek,tb=tb ):\n t = 0\n cnt = 0\n try:\n while True:\n if t + tb[cnt] > usek:\n return cnt\n else:\n t += tb[cnt]\n cnt += 1\n except:\n return len(tb)\n \nsumta = 0\nreadbook = [ i+1 + get_bcnt( k - sum( ta[:i] ) ) ) for i in range(na) if sum( ta[:i] ) <= k ]\nreadbook.append( get_bcnt(k) )', 'na,nb,k = map(int, input().split())\nta = list(map(int, input().split()))\ntb = list(map(int, input().split()))\n\na = [0]\nfor i in range(na):\n a.append( a[i] + ta[i] )\nb = [0]\nfor i in range(nb):\n b.append( b[i] + tb[i] )\n\nans, j=0, nb\nfor i in range(na + 1):\n if a[i] > k:\n break\n while True:\n if a[i] + b[j] > k:\n j -= 1\n else:\n break\n ans = max(ans, i+j)\n \nprint(ans)'] | ['Runtime Error', 'Wrong Answer', 'Runtime Error', 'Runtime Error', 'Accepted'] | ['s033426176', 's223608923', 's285402887', 's600318327', 's329926238'] | [40776.0, 40528.0, 40584.0, 8884.0, 47388.0] | [75.0, 2207.0, 80.0, 25.0, 302.0] | [557, 478, 572, 496, 394] |
p02623 | u405660020 | 2,000 | 1,048,576 | 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. | ['n,m,k=map(int, input().split())\na=list(map(int, input().split()))\nb=list(map(int, input().split()))\n\nfrom itertools import accumulate\na_acc=list(accumulate(a))\nb_acc=list(accumulate(b))\n\ncnt=0\n\nimport bisect\nfor i in range(n):\n nokori=k-a_acc[i]\n if nokori<b[0]:\n break\n b_i=bisect.bisect_right(b_acc,nokori)\n cnt=max(cnt,(i+1)+b_i)\n print(nokori,i,b_i)\n\nprint(cnt)\n', 'n,m,k=map(int, input().split())\na=list(map(int, input().split()))\nb=list(map(int, input().split()))\n\nfrom itertools import accumulate\na_acc=list(accumulate(a))\nb_acc=list(accumulate(b))\na_acc.insert(0,0)\nb_acc.insert(0,0)\n\ncnt=0\n\nimport bisect\nfor i in range(n+1):\n nokori=k-a_acc[i]\n if nokori<0:\n break\n b_i=bisect.bisect_right(b_acc,nokori)-1\n cnt=max(cnt,i+b_i)\n \n\nprint(cnt)'] | ['Wrong Answer', 'Accepted'] | ['s054370223', 's097303214'] | [47480.0, 47336.0] | [454.0, 277.0] | [388, 422] |
p02623 | u406355300 | 2,000 | 1,048,576 | 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. | ['N, M, K = map(int, input().split())\na = list(map(int, input().split()))\nb = list(map(int, input().split()))\nmin = 0\ncnt = 0\nfor i,k in enumerate(b):\n min += k\nj = i\nfor i in range(N):\n while(j > 0 and min > K):\n j -= 1\n min -= b[j]\n if(min > K):\n break\n cnt = max(cnt, i + j)\n if(i == N):\n break\n min += a[i]\nprint(cnt)\n', 'N, M, K = map(int, input().split())\na = list(map(int, input().split()))\nb = list(map(int, input().split()))\nmin = 0\ncnt = 0\nfor k in b:\n min += k\nj = M\nfor i in range(N+1):\n while(j > 0 and min > K):\n j -= 1\n min -= b[j]\n if(min > K):\n break\n cnt = max(cnt, i + j)\n if(i == N):\n break\n min += a[i]\nprint(cnt)\n'] | ['Wrong Answer', 'Accepted'] | ['s509508239', 's095713288'] | [40420.0, 40508.0] | [248.0, 245.0] | [366, 355] |
p02623 | u408262366 | 2,000 | 1,048,576 | 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. | ['n,m,k = map(int,input().split())\n\na = list(map(int,input().split()))\nb = list(map(int,input().split()))\n\nans = 0\ncost = 0\nb_i = 0\nif not (a[0] >= k and b[0] >= k):\n while 1:\n if ans >= n:\n if ans-n>=m:\n break\n _next = b[b_i]\n b_i += 1\n else:\n _next = a[ans]\n if (cost + _next) <= k:\n ans += 1\n cost += _next\n else:\n b_i -= 1\n break\n\n max_ans = ans\n if b_i == 0 or b_i == -1:\n for i in range(ans-1,-1,-1): \n ans -= 1\n cost -= a[i]\n while 1:\n if b_i >= m:\n break\n _next = b[b_i]\n if cost + _next > k:\n break\n cost += _next\n ans += 1\n b_i += 1\n max_ans = max(max_ans, ans)\n else:\n b_i += 1\n for i in range(n-1,-1,-1):\n ans -= 1\n cost -= a[i]\n while cost < k and b_i < m:\n ans += 1\n cost += b[b_i]\n b_i += 1\n max_ans = max(max_ans, ans)\n\n if a[0] <= k:\n print(max_ans)\n else:\n ans = 0\n cost = 0\n _next = 0\n while 1:\n if ans >= m:\n break\n _next = b[ans]\n if cost + _next > k:\n break\n cost += _next\n ans += 1\n print(ans)\nelse:\n print(0)', 'n,m,k = map(int,input().split())\n\na = list(map(int,input().split()))\nb = list(map(int,input().split()))\n\n\n\n\n\nans = 0\ncost = 0\nb_i = 0\nwhile cost <= k:\n ans += 1\n if ans >= n:\n cost += b[n-ans]\n b_i += 1\n else:\n cost += a[ans]\n\nmax_ans = ans\nif b_i == 0:\n for i in range(ans-1,-1,-1):\n ans -= 1\n cost -= a[i]\n while cost <= k and b_i < m:\n ans += 1\n cost += b[b_i]\n b_i += 1\n max_ans = max(max_ans, ans)\nelse:\n for i in range(n-1,-1,-1):\n ans -= 1\n cost -= a[i]\n while cost <= k and b_i < m:\n ans += 1\n cost += b[b_i]\n b_i += 1\n max_ans = max(max_ans, ans)\n \nif not a[0] >= k and b[0] >= k:\n print(max_ans)\nelse:\n print(0)', 'n,m,k = map(int,input().split())\n\na = list(map(int,input().split()))\nb = list(map(int,input().split()))\n\n\n\n\n\nans = 0\ncost = a[0]\nb_i = 0\nif not (a[0] >= k and b[0] >= k):\n while cost <= k:\n ans += 1\n if ans >= n:\n if ans-n>=m:\n break\n cost += b[n-ans]\n b_i += 1\n else:\n cost += a[ans]\n\n max_ans = ans\n if b_i == 0:\n for i in range(ans-1,-1,-1):\n ans -= 1\n cost -= a[i]\n while cost <= k and b_i < m:\n ans += 1\n cost += b[b_i]\n b_i += 1\n max_ans = max(max_ans, ans)\n else:\n for i in range(n-1,-1,-1):\n ans -= 1\n cost -= a[i]\n while cost <= k and b_i < m:\n ans += 1\n cost += b[b_i]\n b_i += 1\n max_ans = max(max_ans, ans)\n\n print(max_ans)\nelse:\n print(0)', 'N, M, K=map(int,input().split())\nA=list(map(int,input().split()))\nB=list(map(int,input().split()))\n\na, b=[0], [0]\nfor i in range(N):\n a.append(a[i]+A[i])\nfor i in range(M):\n b.append(b[i]+B[i])\n \nans, j=0, M\nfor i in range(N+1):\n if a[i]>K:\n break\n while b[j]>K-a[i]:\n j-=1\n ans=max(ans, i+j)\n \nprint(ans)'] | ['Wrong Answer', 'Runtime Error', 'Wrong Answer', 'Accepted'] | ['s388515193', 's759757239', 's912949761', 's878624876'] | [41040.0, 42544.0, 39988.0, 47452.0] | [324.0, 289.0, 298.0, 308.0] | [1193, 834, 920, 322] |
p02623 | u408375121 | 2,000 | 1,048,576 | 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. | ['n, m, k = map(int, input().split())\na = list(map(int, input().split()))\nb = list(map(int, input().split()))\ncum_a = []\ntotal = 0\nfor i in range(n):\n total += a[i]\n if total > k:\n break\n cum_a.append(total)\n \ncum_b = []\ntotal = 0\nfor j in range(m):\n total += b[j]\n if total > k:\n break\n cum_b.append(total)\nlim_a = len(cum_a)\nlim_b = len(cum_b)\nidx = 0\nans = lim_a\nfor l in range(1, lim_a + 1):\n while True:\n if k >= cum_a[-l] + cum_b[idx]:\n ans = lim_a - l + idx + 2\n else:\n idx += 1\n idx += 1\nprint(ans)\n', 'n, m, k = map(int, input().split())\na = list(map(int, input().split()))\nb = list(map(int, input().split()))\ncum_a = []\ntotal = 0\nfor i in range(n):\n total += a[i]\n if total > k:\n break\n cum_a.append(total)\n \ncum_b = []\ntotal = 0\nfor j in range(m):\n total += b[j]\n if total > k:\n break\n cum_b.append(total)\nlim_a = len(cum_a)\nlim_b = len(cum_b)\nidx = 0\nans = lim_a\nfor l in range(1, lim_a + 1):\n for idx in range(lim_b):\n if k >= cum_a[-l] + cum[idx]:\n ans = max(ans, lim_a - l + idx + 2)\nprint(ans)\n', 'n, m, k = map(int, input().split())\na = list(map(int, input().split()))\nb = list(map(int, input().split()))\ncum_a = []\ntotal = 0\nfor i in range(n):\n total += a[i]\n if total > k:\n break\n cum_a.append(total)\n \ncum_b = []\ntotal = 0\nfor j in range(m):\n total += b[j]\n if total > k:\n break\n cum_b.append(total)\nlim_a = len(cum_a)\nlim_b = len(cum_b)\nidx = 0\nans = lim_a\nfor l in range(1, lim_a + 1):\n while idx < lim_b:\n if k >= cum_a[-l] + cum_b[idx]:\n ans = max(ans, lim_a - l + idx + 2)\n else:\n break\n idx += 1\nans = max(ans, lim_b)\nprint(ans)\n'] | ['Runtime Error', 'Runtime Error', 'Accepted'] | ['s095580879', 's435529368', 's767673446'] | [40736.0, 40604.0, 40632.0] | [238.0, 196.0, 312.0] | [538, 521, 574] |
p02623 | u408793494 | 2,000 | 1,048,576 | 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. | ['n, m, k = map(int, input().split())\na = list(map(int, input().split()))\nb = list(map(int, input().split()))\n\np_time = 0\ntime = 0\nwhile time < k:\n if len(a) > 0 and len(b) > 0:\n if a[0] >= b[0]:\n time += a[0]\n a.remove(a[0])\n else:\n time += b[0]\n b.remove(b[0])\n elif len(a) <= 0 and len(b) <= 0:\n break\n elif len(a) <= 0 < len(b):\n time += b[0]\n b.remove(b[0])\n elif len(b) <= 0 < len(a):\n time += a[0]\n a.remove(a[0])\n if time <= k:\n p_time += 1\n\nprint(p_time)\n', 'n, m, k = map(int, input().split())\nA = list(map(int, input().split()))\nB = list(map(int, input().split()))\n\na, b = [0], [0]\nfor i in range(n):\n a.append(a[i] + A[i])\nfor j in range(m):\n b.append(b[j] + B[j])\n\nans, j = 0, m\nfor i in range(n + 1):\n if a[i] > k:\n break\n while b[j] > k - a[i]:\n j -= 1\n ans = max(ans, i + j)\nprint(ans)\n\n'] | ['Wrong Answer', 'Accepted'] | ['s288709008', 's196315324'] | [40440.0, 47540.0] | [2206.0, 284.0] | [577, 364] |
p02623 | u414980766 | 2,000 | 1,048,576 | 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. | ['\n#include <algorithm>\n\n\n\nint main()\n{\n long long n,m,k; std::cin >> n >> m >> k;\n // long long a[n+1], b[m+1];\n std::vector<long long> a{n+1}, b{m+1};\n a[0]=0; b[0]=0;\n for (int i=1;i<n+1;i++)\n std::cin >> a[i];\n for (int i=1; i<m+1; i++)\n std::cin >> b[i];\n \n for (int i=1; i<n+1; i++)\n a[i]+=a[i-1];\n for (int i=1; i<m+1; i++)\n b[i]+=b[i-1];\n\n int ans=0, r=m;\n for (int i=0; i<n+1; i++)\n {\n if(a[i]>k) break;\n while (a[i]+b[r]>k)\n r--;\n ans = std::max({ans, i+r});\n }\n \n std::cout << ans;\n\n return 0;\n}', 'import sys\nread = sys.stdin.buffer.read\nreadline = sys.stdin.buffer.readline\nreadlines = sys.stdin.buffer.readlines\n\nn,m,k = map(int, readline().split())\na = list(map(int, readline().split()))\nb = list(map(int, readline().split()))\n\nsa = [0]*(n+1)\nsb = [0]*(m+1)\n\nfor i in range(n):\n sa[i+1] = sa[i]+a[i]\nfor i in range(m):\n sb[i+1] = sb[i]+b[i]\n\nisb=m\nans = 0\nfor isa in range(n+1):\n if sa[isa]>k:\n break\n while isb>0 and sa[isa]+sb[isb]>K:\n isb-=1\n ans = max(ans, isa+isb)\n \nprint(ans)', 'import sys\nread = sys.stdin.buffer.read\nreadline = sys.stdin.buffer.readline\nreadlines = sys.stdin.buffer.readlines\n\nn,m,k = map(int, readline().split())\na = list(map(int, readline().split()))\nb = list(map(int, readline().split()))\n\nsa = [0]*(n+1)\nsb = [0]*(m+1)\n\nfor i in range(n):\n sa[i+1] = sa[i]+a[i]\nfor i in range(m):\n sb[i+1] = sb[i]+b[i]\n\nisb=m\nans = 0\nfor isa in range(n+1):\n if sa[isa]>k:\n break\n #while isb>0 and sa[isa]+sb[isb]>k:\n while sa[isa]+sb[isb]>k:\n isb-=1\n ans = max(ans, isa+isb)\n \nprint(ans)'] | ['Runtime Error', 'Runtime Error', 'Accepted'] | ['s318214925', 's670322336', 's373651099'] | [8896.0, 48168.0, 48384.0] | [26.0, 192.0, 283.0] | [651, 519, 549] |
p02623 | u416011173 | 2,000 | 1,048,576 | 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. | ['# -*- coding: utf-8 -*-\n\nN, M, K = list(map(int, input().split()))\nA = list(map(int, input().split()))\nB = list(map(int, input().split()))\n\n\na, b = [0], [0]\nfor i in range(N):\n a.append(a[i] + A[i])\nfor j in range(M):\n b.append(b[i] + B[i])\n\nans, j = 0, M\nfor i in range(N + 1):\n if a[i] > K:\n break\n while b[j] > K - a[i]:\n j -= 1\n ans = max(ans, i + j)\n\n\nprint(ans)\n', '# -*- coding: utf-8 -*-\n\ndef get_input() -> tuple:\n \n N, M, K = list(map(int, input().split()))\n A = list(map(int, input().split()))\n B = list(map(int, input().split()))\n\n return N, M, K, A, B\n\n\ndef main(N: int, M: int, K: int, A: list, B: list) -> None:\n \n \n a, b = [0], [0]\n for i in range(N):\n a.append(a[i] + A[i])\n for j in range(M):\n b.append(b[j] + B[j])\n\n ans, j = 0, M\n for i in range(N + 1):\n if a[i] > K:\n break\n while b[j] > K - a[i]:\n j -= 1\n ans = max(ans, i + j)\n\n \n print(ans)\n\n\nif __name__ == "__main__":\n \n N, M, K, A, B = get_input()\n\n \n main(N, M, K, A, B)\n'] | ['Runtime Error', 'Accepted'] | ['s326586538', 's221877862'] | [40484.0, 47152.0] | [158.0, 229.0] | [448, 1172] |
p02623 | u416758623 | 2,000 | 1,048,576 | 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. | ['\nimport sys\npin = sys.stdin.readline\npout = sys.stdout.write\nperr = sys.stderr.write\n\nN, M, K = map(int, pin().split())\nA = list(map(int, pin().split()))\nB = list(map(int, pin().split()))\na = [0]\nb = [0]\nfor i in range(N):\n a.append(a[i] + A[i])\nfor i in range(M):\n b.append(b[i] + B[i])\nans = 0\nprint(a)\nprint(b)\nfor i in range(N + 1):\n \n if a[i] > K:\n break\n \n while b[M] > K - a[i]:\n M -= 1\n ans = max(ans, M + i)\nprint(ans)\n', '\nimport sys\npin = sys.stdin.readline\npout = sys.stdout.write\nperr = sys.stderr.write\n\nN, M, K = map(int, pin().split())\nA = list(map(int, pin().split()))\nB = list(map(int, pin().split()))\na = [0]\nb = [0]\nfor i in range(N):\n a.append(a[i] + A[i])\nfor i in range(M):\n b.append(b[i] + B[i])\nans = 0\n\nfor i in range(N + 1):\n \n if a[i] > K:\n break\n \n while b[M] > K - a[i]:\n M -= 1\n ans = max(ans, M + i)\nprint(ans)\n'] | ['Wrong Answer', 'Accepted'] | ['s176250721', 's256471892'] | [55648.0, 49236.0] | [335.0, 278.0] | [566, 549] |
p02623 | u418149936 | 2,000 | 1,048,576 | 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. | ['for i in range(1, N + 1):\n A_ls[i] += A_ls[i - 1]\nfor i in range(1, M + 1):\n B_ls[i] += B_ls[i - 1]\n\nb_cnt, rst = M, 0\nfor a_cnt in range(N + 1):\n if A_ls[a_cnt] > K:\n break\n while A_ls[a_cnt] + B_ls[b_cnt] > K:\n b_cnt -= 1\n rst = max(rst, a_cnt + b_cnt)\nprint(rst)', "N, M, K = map(int, input().split(' '))\nA_ls = [0] + list(map(int, input().split(' ')))\nfor i in range(N):\n A_ls[i + 1] += A_ls[i]\nB_ls = [0] + list(map(int, input().split(' ')))\nfor i in range(M):\n B_ls[i + 1] = B_ls[i]\n\nb_cnt, rst = M, 0 \nfor a_cnt in range(N + 1):\n if A_ls[a_cnt] > K:\n break\n while A_ls[a_cnt] + B_ls[b_cnt] > K and b_cnt >= 0:\n b_cnt -= 1\n rst = max(rst, a_cnt + b_cnt)\nprint(rst)", "N, M, K = map(int, input().split(' '))\nA_ls = [0] + list(map(int, input().split(' ')))\nfor i in range(N):\n A_ls[i + 1] += A_ls[i]\nB_ls = [0] + list(map(int, input().split(' ')))\nfor i in range(M):\n B_ls[i + 1] += B_ls[i]\nb_cnt, rst = M, 0\nfor a_cnt in range(N + 1):\n if A_ls[a_cnt] > K:\n break\n while A_ls[a_cnt] + B_ls[b_cnt] > K and b_cnt >= 0:\n b_cnt -= 1\n rst = max(rst, a_cnt + b_cnt)\nprint(rst)"] | ['Runtime Error', 'Wrong Answer', 'Accepted'] | ['s531424687', 's650926755', 's981236402'] | [9072.0, 45168.0, 45208.0] | [27.0, 232.0, 293.0] | [294, 430, 429] |
p02623 | u425834921 | 2,000 | 1,048,576 | 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. | ['N,M,K = map(int,input().split())\nA = list(map(int,input().split()))\nB = list(map(int,input().split()))\na = [0]\nb = [0]\nfor i in range(N):\n a.append(a[i] + A[i])\nfor i in range(M):\n b.append(b[i] + B[i])\n \nans = 0\nj = M\nfor i in range(N+1):\n if a[i] > K:\n break\n while b[i] > K - a[i]:\n j -= 1\n ans = max(ans,i+j)\nprint(ans)', 'N,M,K = map(int,input().split())\nA = list(map(int,input().split()))\nB = list(map(int,input().split()))\na = [0]\nb = [0]\nfor i in range(N):\n a.append(a[i] + A[i])\nfor i in range(M):\n b.append(b[i] + B[i])\n \nans = 0\nj = M\nfor i in range(N+1):\n if a[i] > K:\n break\n while b[j] > K - a[i]:\n j -= 1\n ans = max(ans,i+j)\nprint(ans)'] | ['Runtime Error', 'Accepted'] | ['s725976794', 's456356589'] | [47652.0, 47280.0] | [2207.0, 289.0] | [335, 335] |
p02623 | u430536466 | 2,000 | 1,048,576 | 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. | ['N, M, K = map(int, input().split())\nA = list(map(int, input().split()))\nB = list(map(int, input().split()))\n\na, b = [0], [0]\nfor i in range(N):\na.append(a[i] + A[i])\nfor i in range(M):\nb.append(b[i] + B[i])\n\nans, j = 0, M\nfor i in range(N + 1):\nif a[i] > K:\nbreak\nwhile b[j] > K - a[i]:\nj -= 1\nans = max(ans, i + j)\nprint(ans)', 'N, M, K = map(int, input().split())\nA = list(map(int, input().split()))\nB = list(map(int, input().split()))\n\na, b = [0], [0]\nfor i in range(N):\n a.append(a[i] + A[i])\nfor i in range(M):\n b.append(b[i] + B[i])\n\nans, j = 0, M\nfor i in range(N + 1):\n if a[i] > K:\n break\n while b[j] > K - a[i]:\n j -= 1\n ans = max(ans, i + j)\nprint(ans)'] | ['Runtime Error', 'Accepted'] | ['s098965553', 's607615755'] | [8924.0, 47428.0] | [21.0, 281.0] | [326, 335] |
p02623 | u430726059 | 2,000 | 1,048,576 | 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. | ['N,M,K = map(int,input().split())\nA=list(map(int,input().split()))\nB=list(map(int,input().split()))\n \na, b =[0],[0]\nfor i in range(N):\n a.append(a[i] + A[i])\nfor i in range(M):\n b.append(b[i] + B[i])\n \nans, j =0,M\nfor i in range(N+1):\n while b[j] > K-a[i]:\n j -= 1\n ans = max(ans, i+j)\nprint(ans)', 'N,M,K = map(int,input().split())\nA=list(map(int,input().split()))\nB=list(map(int,input().split()))\n \na, b =[0],[0]\nfor i in range(N):\n a.append(a[i] + A[i])\nfor i in range(M):\n b.append(b[i] + B[i])\n \nans, j =0,M\nfor i in range(N+1):\n while b[j] > K-a[i]:\n j -= 1\n ans = max(ans, i+j)\nprint(ans)', 'N,M,K = map(int,input().split())\nA=list(map(int,input().split()))\nB=list(map(int,input().split()))\n \na, b =[0],[0]\nfor i in range(N):\n a.append(a[i] + A[i])\nfor i in range(M):\n b.append(b[i] + B[i])\n \nans, j =0,M\nfor i in range(N+1):\n if a[i] > K:\n break\n while b[j] > K-a[i]:\n j -= 1\n ans = max(ans, i+j)\nprint(ans)'] | ['Runtime Error', 'Runtime Error', 'Accepted'] | ['s176333968', 's816104122', 's982211002'] | [47496.0, 47428.0, 47368.0] | [319.0, 306.0, 312.0] | [314, 310, 345] |
p02623 | u435480265 | 2,000 | 1,048,576 | 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. | ['n,m,k = map(int, input().split())\nln = list(map(int, input().split()))\nlm = list(map(int, input().split()))\ndef sub(l,k):\n res = 0\n for i in l:\n if k >= i:\n k -= i\n res += 1\n else:\n return k,res\n return k,res\n \ndef main(n,m,k,ln,lm):\n nk = sum(ln)\n mk = sum(lm)\n if k>=nk+mk:\n return n+m\n elif k>nk:\n ik, res = sub(lm, k-nk)\n res += n\n startn = n - 1\n startm = res - 1\n else:\n ik, res = sub(ln, k)\n startn = res - 1\n ik, res2 = sub(lm, ik)\n res += res2\n startm = res2\n \n ires = res\n for i in range(startn,-1,-1):\n ik += ln[i]\n ik, delta = sub(lm[startm:], ik)\n startm += delta\n ires = ires - 1 + delta\n res = max(res, ires)\n return res\n\nmain(n,m,k,ln,lm)', "n,m,k = map(int, input().split())\nln = list(map(int, input().split()))\nlm = list(map(int, input().split()))\n\ndef csum(l,ll):\n s = 0\n res = [0] * ll\n for i in range(ll):\n s += l[i]\n res[i] = s\n return res\n\ncsln = csum(ln,n)\ncslm = csum(lm,m)\n\nimport numpy as np\n\nlk = [k-x for x in csln if k-x >=0]\nres = np.searchsorted(np.array(cslm), np.array(lk), side='right').\nrmax = 0\nfor i,j in zip(range(1,len(lk)+1), res):\n rmax = max(rmax, i+j)\nprint(int(rmax))", "n,m,k = map(int, input().split())\nln = list(map(int, input().split()))\nlm = list(map(int, input().split()))\n\ndef csum(l,ll):\n s = 0\n res = [0] * ll\n for i in range(ll):\n s += l[i]\n res[i] = s\n return res\n\ncsln = csum(ln,n)\ncslm = csum(lm,m)\n\nimport numpy as np\n\nlk = [k] + [k-x for x in csln if k-x >=0]\nres = np.searchsorted(np.array(cslm), np.array(lk), side='right')\nrmax = 0\nfor i,j in zip(range(len(lk)+1), res):\n rmax = max(rmax, i+j)\nprint(rmax)"] | ['Wrong Answer', 'Runtime Error', 'Accepted'] | ['s183246923', 's695470160', 's612764472'] | [40604.0, 8872.0, 70628.0] | [2206.0, 26.0, 427.0] | [853, 483, 481] |
p02623 | u436519884 | 2,000 | 1,048,576 | 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. | ['n,m,k=map(int,input().split())\na=list(map(int,input().split()))\nb=list(map(int,input().split()))\na.insert(0,0)\nb.insert(0,0)\nj=1\nwhile j<=m and b[j]+b[j-1]<=k:\n b[j]+=b[j-1]\n j+=1\n \n\nj-=1\nbooks=j\ni=1\nwhile i<n+1:\n while j>0 and a[i]+b[j]>k:\n j-=1\n if(j>=0) and a[i]+b[j]<=k:\n books=max(books,j+i)\n a[i]+=a[i-1]\n i+=1\n else:\n break\nprint(books)\n \n', 'n,m,k=map(int,input().split())\na=list(map(int,input().split()))\nb=list(map(int,input().split()))\na.insert(0,0)\nb.insert(0,0)\nj=1\nwhile j<=m and b[j]+b[j-1]<=k:\n b[j]+=b[j-1]\n j+=1\n \n\nj-=1\nbooks=j\ni=1\nwhile i<n:\n while j>0 and a[i]+b[j]>k:\n j-=1\n if(j>=0) and a[i]+b[j]<=k:\n books=max(books,j+i)\n a[i]+=a[i-1]\n else:\n break\nprint(books)\n ', 'n,m,k=map(int,input().split())\na=list(map(int,input().split()))\nb=list(map(int,input().split()))\na.insert(0,0)\nb.insert(0,0)\nj=1\nwhile j<=m and b[j]+b[j-1]<=k:\n b[j]+=b[j-1]\n j+=1\n \n\nj-=1\nbooks=j\n#print(books,b)\n\ni=1\nwhile i<n+1:\n a[i]+=a[i-1]\n while j>=0 and a[i]+b[j]>k:\n j-=1\n if(j!=-1) and a[i]+b[j]<=k:\n books=max(books,j+i)\n \n \n i+=1\n #print(books,b[:j+1],a[:i])\n else:\n break\nprint(books)\n '] | ['Wrong Answer', 'Wrong Answer', 'Accepted'] | ['s734253249', 's957491945', 's715177035'] | [40496.0, 42204.0, 42168.0] | [326.0, 2206.0, 361.0] | [368, 356, 426] |
p02623 | u439676056 | 2,000 | 1,048,576 | 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. | ['import sys\nimport numpy as np\ninput = sys.stdin.readline\n\nN, M, K = map(int, input().split())\nA_list = list(map(int, input().split()))\nB_list = list(map(int, input().split()))\n\n\n#N, M, K = 3, 4, 730\n#A_list = [60, 90, 120]\n#B_list = [80, 150, 80, 150]\n\n\nA_list.insert(0, 0)\nB_list.insert(0, 0)\nB_cumsum = np.cumsum(B_list)\n\ntime = B_cumsum[-1]\nans, j = 0, M\n\nfor i in range(N+1):\n while (j > 0 and time > K):\n j -= 1\n time -= B_cumsum[j]\n if (time > K):\n break\n ans = max(ans, i+j)\n time += A_list[i]\n\nprint(ans)\n', '\nN, M, K = map(int, input().split())\nA_list = list(map(int, input().split()))\nB_list = list(map(int, input().split()))\n\n\nA_list.insert(0, 0)\nB_list.insert(0, 0)\nB_cumsum = np.cumsum(B_list)\n\ntime = B_cumsum[-1]\nans, j = 0, M\n\nfor i in range(N+1):\n if (time > K):\n break\n while (j > 0 and time > K):\n time -= B_list[j]\n j -= 1\n\n ans = max(ans, i+j)\n time += A_list[i]\n\nprint(ans)\n', 'import sys\nimport numpy as np\ninput = sys.stdin.readline\n\nN, M, K = map(int, input().split())\nA_list = list(map(int, input().split()))\nB_list = list(map(int, input().split()))\n\n\nA_list.insert(0, 0)\nB_list.insert(0, 0)\nB_cumsum = np.cumsum(B_list)\n\ntime = B_cumsum[-1]\nans, j = 0, M\n\nfor i in range(N+1):\n if (time > K):\n break\n while (j > 0 and time > K):\n time -= B_list[j]\n j -= 1\n\n ans = max(ans, i+j)\n time += A_list[i]\n\nprint(ans)', 'N, M, K = map(int, input().split())\nA_list = list(map(int, input().split()))\nB_list = list(map(int, input().split()))\n\n\nf = open(\'b20.txt\')\ndata = f.readlines()\ndata = [d.replace("\\n", "").split(" ") for d in data]\nN, M, K = map(int, data[0])\nA_list = list(map(int, data[1]))\nB_list = list(map(int, data[2]))\n\nA_list.insert(0, 0)\nB_list.insert(0, 0)\nB_cumsum = np.cumsum(B_list)\n\ntime = B_cumsum[-1]\nans, j = 0, M\n\nB_list[j]\nfor i in range(N+1):\n if (time > K):\n break\n while (j > 0 and time > K):\n time -= B_list[j]\n j -= 1\n \n ans = max(ans, i+j)\n time += A_list[i]\n\nprint(ans)', 'import sys\nimport numpy as np\ninput = sys.stdin.readline\n\nN, M, K = map(int, input().split())\nA_list = list(map(int, input().split()))\nB_list = list(map(int, input().split()))\n\n\n#N, M, K = 3, 4, 730\n#A_list = [60, 90, 120]\n#B_list = [80, 150, 80, 150]\n\n\nA_list.insert(0, 0)\nB_list.insert(0, 0)\nB_cumsum = np.cumsum(B_list)\n\ntime = B_cumsum[-1]\nans, j = 0, M\n\nfor i in range(N+1):\n while (j > 0 and time > K):\n j -= 1\n time -= B_cumsum[j]\n if (time > K):\n break\n ans = max(ans, i+j)\n time += A_list[i]\n\nprint(ans)\n', '\nimport sys\nimport numpy as np\ninput = sys.stdin.readline\n\nN, M, K = map(int, input().split())\nA_list = list(map(int, input().split()))\nB_list = list(map(int, input().split()))\n\n\nA_list.insert(0, 0)\nB_list.insert(0, 0)\nA_cumsum = np.cumsum(A_list)\nB_cumsum = np.cumsum(B_list)\n\ntime = B_cumsum[-1]\nans, j = 0, M\n\n\n\nfor i in range(N+1):\n if A_cumsum[i] > K:\n break\n while B_cumsum[j] > K - A_cumsum[i]:\n j -= 1\n ans = max(ans, i+j)\nprint(ans)\n'] | ['Wrong Answer', 'Runtime Error', 'Wrong Answer', 'Runtime Error', 'Wrong Answer', 'Accepted'] | ['s142936493', 's167678407', 's356577359', 's786390039', 's891521486', 's221512611'] | [59348.0, 40548.0, 59092.0, 41188.0, 58840.0, 59220.0] | [375.0, 110.0, 365.0, 115.0, 375.0, 515.0] | [546, 412, 468, 614, 546, 465] |
p02623 | u440613652 | 2,000 | 1,048,576 | 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. | ['n,m,k=map(int,input().split())\n\na=list(map(int,input().split()))\nb=list(map(int,input().split()))\n\n\na_pre,b_pre=[0],[0]\n\nfor i in range(n):\n\ta_pre.append(a_pre[i]+a[i])\n\nfor i in range(m):\n\tb_pre.append(b_pre[i]+b[i])\n\nans,j=0,m\nfor i in range(n+1):\n\tif a_pre[i]>k:\n\t\tbreak\n\twhile b[j]>k-a[i]:\n\t\tj-=1\n\n\tans=max(ans,i+j)\n\nprint(ans)\n', 'n,m,k=map(int,input().split())\n\na=list(map(int,input().split()))\nb=list(map(int,input().split()))\n\n\na_pre,b_pre=[0],[0]\n\nfor i in range(n):\n\ta_pre.append(a_pre[i]+a[i])\n\nfor i in range(m):\n\tb_pre.append(b_pre[i]+b[i])\n\nans,j=0,m\nfor i in range(n+1):\n\tif a_pre[i]>k:\n\t\tbreak\n\twhile b_pre[j]>k-a_pre[i]:\n\t\tj-=1\n\n\tans=max(ans,i+j)\n\nprint(ans)\n'] | ['Runtime Error', 'Accepted'] | ['s291734250', 's639709290'] | [47404.0, 47660.0] | [189.0, 276.0] | [332, 340] |
p02623 | u440975163 | 2,000 | 1,048,576 | 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. | ['n,m,k=map(int,input().split())\nA=list(map(int,input().split()))\nB=list(map(int,input().split()))\n\na,b=[0],[0]\nfor i in range(n):\n a.append(a[i]+A[i])\nfor i in range(m):\n b.append(b[i]+B[i])\nprint(a)\nprint(b)\nans,j=0,m\nfor i in range(n+1):\n if a[i]>k:\n break\n while b[j]>k-a[i]:\n j-=1\n ans=max(ans,i+j)\nprint(ans)', 'n,m,k=map(int,input().split())\nA=list(map(int,input().split()))\nB=list(map(int,input().split()))\n\na,b=[0],[0]\nfor i in range(n):\n a.append(a[i]+A[i])\nfor i in range(m):\n b.append(b[i]+B[i])\nans,j=0,m\nfor i in range(n+1):\n if a[i]>k:\n break\n while b[j]>k-a[i]:\n j-=1\n ans=max(ans,i+j)\nprint(ans)'] | ['Wrong Answer', 'Accepted'] | ['s187947173', 's188678966'] | [54460.0, 47460.0] | [361.0, 298.0] | [341, 323] |
p02623 | u444444641 | 2,000 | 1,048,576 | 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. | ['N, M, K = map(int, input().split())\nA = list(input().split())\nB = list(input().split())\n\na=0;b=0\ncnt=0\nwhile a < len(A) and b < len(B) and K > 0:\n\ta_ = A[a]\n\tb_ = B[b]\n \n if a_ > K and b_ > K:\n break\n \n if a_ < b_ and a_ <= K:\n\t\tcnt += 1\n\t\tK -= a_\n\t\tA.pop(0)\n\telif a_ >= b_ and b_ <= K:\n cnt += 1\n K -= b_\n B.pop(0)\n \nprint(cnt) \n ', 'N, M, K = map(int, input().split())\nA = list(input().split())\nB = list(input().split())\n\na=0;b=0\ncnt=0\nwhile a < len(A) and b < len(B) and K > 0:\n a_ = A[a]\n b_ = B[b]\n \n if a_ > K and b_ > K:\n break\n \n if a_ < b_ and a_ <= K:\n cnt += 1\n\tK -= a_\n a += 1\n elif a_ >= b_ and b_ <= K:\n cnt += 1\n K -= b_\n b += 1\n \nprint(cnt) \n', 'a=0;b=0\ncnt=0\nwhile a < len(A) and b < len(B) and K > 0:\n\ta_ = A[a]\n\tb_ = B[b]\n \n if a_ > K and b_ > K:\n break\n \n if a_ < b_ and a_ <= K:\n\t\tcnt += 1\n\t\tK -= a_\n a += 1\n\telif a_ >= b_ and b_ <= K:\n cnt += 1\n K -= b_\n b += 1\n \nprint(cnt) ', 'N, M, K = map(int, input().split())\nA=list(map(int, input().split()))\nB=list(map(int,input().split()))\n\na=0;b=0\ncnt=0\nwhile a < len(A) and b < len(B) and K > 0:\n a_ = A[a]\n b_ = B[b]\n print(a_, b_,K)\n \n if a_ > K and b_ > K:\n break\n \n if a_ < b_ and a_ <= K:\n cnt += 1\n K -= a_\n a += 1\n elif a_ >= b_ and b_ <= K:\n cnt += 1\n K -= b_\n b += 1\n \nprint(cnt)', 'n, m, k = map(int, input().split())\na = list(map(int, input().split()))\nb = list(map(int, input().split()))\n\nA = [0]\nB = [0]\nans = 0\n\nfor i in range(n):\n A.append(A[i] + a[i])\n\nfor i in range(m):\n B.append(B[i] + b[i])\n\nj = m\n\nfor i in range(n + 1):\n if A[i] > k:\n break\n while A[i] + B[j] > k:\n j -= 1\n ans = max(ans, i + j)\n\nprint(ans)'] | ['Runtime Error', 'Runtime Error', 'Runtime Error', 'Wrong Answer', 'Accepted'] | ['s219843448', 's244680031', 's325851974', 's337414258', 's055070452'] | [9004.0, 8988.0, 8960.0, 40616.0, 47680.0] | [22.0, 21.0, 27.0, 379.0, 294.0] | [377, 360, 282, 431, 366] |
p02623 | u445624660 | 2,000 | 1,048,576 | 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. | ['package main\nimport (\n\t"bufio"\n\t"fmt"\n\t"os"\n\t"strconv"\n)\n\nvar sc = bufio.NewScanner(os.Stdin)\nvar out = bufio.NewWriter(os.Stdout)\n\nfunc main(){\n\tbuf := make([]byte, 1024*1024)\n\tsc.Buffer(buf, bufio.MaxScanTokenSize)\n\tsc.Split(bufio.ScanWords)\n\tdefer out.Flush()\n\tN, M, K := nextInt(), nextInt(), nextInt()\n\tA := append([]int{0}, nextInts(N)...)\n\tB := append([]int{0}, nextInts(M)...)\n\tfor i:=1; i<=N; i++{\n\t\tA[i] += A[i-1]\n\t}\n\tfor i:=1; i<=M; i++{\n\t\tB[i] += B[i-1]\n\t}\n\tans := 0\n\tfor i:=0; i<=N; i++{\n\t\trest := K-A[i]\n\t\tif rest < 0{\n\t\t\tbreak\n\t\t}\n\t\t// golangでupper_boundどうやるんだ？？？？？？・\n\t\t/// bidx := max(0, sort.Search(len(B), func(j int)bool{return rest<=B[j]}))\n\t\t// bidx := sort.Search(len(B), func(j int)bool{return rest<=B[j]})-1\n\t\t// bidx := sort.SearchInts(B, rest)\n\t\t// わからんので仕方なく手書き\n\t\tleft, right := -1, len(B)\n\t\tfor right - left > 1{\n\t\t\tmid := (left+right)/2\n\t\t\tif rest-B[mid]>=0{\n\t\t\t\tleft = mid\n\t\t\t}else{\n\t\t\t\tright = mid\n\t\t\t}\n\t\t}\n\t\tbidx := left\n\t\tans = max(ans, i+bidx)\n\t}\n\tfmt.Fprintln(out, ans)\n}\n\nfunc min(a, b int) int {\n\tif a<b{\n\t\treturn a\n\t}\n\treturn b\n}\n\nfunc max(a, b int) int {\n\tif a<b{\n\t\treturn b\n\t}\n\treturn a\n}\nfunc next() string {\n\tsc.Scan()\n\treturn sc.Text()\n}\n\nfunc nextInt() int {\n\ta, _ := strconv.Atoi(next())\n\treturn a\n}\n\nfunc nextInts(n int) []int{\n\tret := make([]int, n)\n\tfor i := 0; i < n; i++ {\n\t\tret[i] = nextInt()\n\t}\n\treturn ret\n}\n\nfunc nextStrings(n int) []string {\n\tret := make([]string, n)\n\tfor i := 0; i < n; i++ {\n\t\tret[i] = next()\n\t}\n\treturn ret\n}\n\nfunc chars(s string) []string {\n\tret := make([]string, len([]rune(s)))\n\tfor i, v := range []rune(s) {\n\t\tret[i] = string(v)\n\t}\n\treturn ret\n}\n', 'n, m, k = map(int, input().split())\na = [0] + list(map(int, input().split()))\nb = [0] + list(map(int, input().split()))\nfor i in range(1, n + 1):\n a[i] += a[i - 1]\nfor i in range(1, m + 1):\n b[i] += b[i - 1]\n\n\ndef search(arr, x):\n l, r = 0, len(arr)\n while abs(r - l) > 1:\n mid = (l + r) // 2\n print(l, r, mid)\n if arr[mid] <= x:\n l = mid\n else:\n r = mid\n return l\n\n\narrr = [0, 80, 230, 310, 460]\nt = 90\nti = search(arrr, t)\n\nans = 0\nfor i in range(len(a)):\n rest = k - a[i]\n if rest < 0:\n break\n bidx = search(b, rest)\n if bidx == m:\n break\n ans = max(ans, i + bidx)\nprint(ans)', 'n, m, k = map(int, input().split())\na = [0] + list(map(int, input().split()))\nb = [0] + list(map(int, input().split()))\nfor i in range(1, n + 1):\n a[i] += a[i - 1]\nfor i in range(1, m + 1):\n b[i] += b[i - 1]\n\n\ndef search(arr, x):\n l, r = -1, len(arr)\n while abs(r - l) > 1:\n mid = (l + r) // 2\n print(l, r, mid)\n if arr[mid] <= x:\n l = mid\n else:\n r = mid\n return l\n\n\narrr = [0, 80, 230, 310, 460]\nt = 90\nti = search(arrr, t)\n\nans = 0\nfor i in range(len(a)):\n rest = k - a[i]\n if rest < 0:\n break\n bidx = search(b, rest)\n ans = max(ans, i + bidx)\nprint(ans)', 'n, m, k = map(int, input().split())\na = [0] + list(map(int, input().split()))\nb = [0] + list(map(int, input().split()))\nfor i in range(1, n + 1):\n a[i] += a[i - 1]\nfor i in range(1, m + 1):\n b[i] += b[i - 1]\n\n\ndef search(arr, x):\n l, r = -1, len(arr)\n while abs(r - l) > 1:\n mid = (l + r) // 2\n if arr[mid] <= x:\n l = mid\n else:\n r = mid\n return l\n\n\nans = 0\nfor i in range(len(a)):\n rest = k - a[i]\n if rest < 0:\n break\n bidx = search(b, rest)\n if bidx == -1 or bidx == len(b):\n continue\n ans = max(ans, i + bidx)\nprint(ans)'] | ['Runtime Error', 'Wrong Answer', 'Wrong Answer', 'Accepted'] | ['s281290356', 's893990176', 's916230580', 's561025209'] | [9004.0, 71344.0, 74568.0, 42252.0] | [29.0, 2266.0, 2279.0, 791.0] | [1679, 672, 641, 611] |
p02623 | u446711904 | 2,000 | 1,048,576 | 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. | ['n,m,k=map(int,input().split())\nA=list(map(int,input().split()))\nB=list(map(int,input().split()))\n\na=[0]\nb=[0]\n\nfor i in range(n):\n a.append(a[i]+A[i])\nfor i in range(m):\n b.append(b[i]+B[i])\n \nc=0\nidx=m\n\nfor i in range(n+1):\n while 1:\n if idx<0:\n break\n if a[i]+b[idx]>k:\n print(a[i]+b[idx])\n idx-=1\n continue\n else:\n c=max(i+idx,c)\n break\nelse:\n print(c)', 'n,m,k=map(int,input().split())\nA=list(map(int,input().split()))\nB=list(map(int,input().split()))\n\na=[0]\nb=[0]\n\nfor i in range(n):\n a.append(a[i]+A[i])\nfor i in range(m):\n b.append(b[i]+B[i])\n \nc=0\nidx=m\n\nfor i in range(n+1):\n while 1:\n if idx<0:\n break\n if a[i]+b[idx]>k:\n idx-=1\n continue\n else:\n c=max(i+idx,c)\n break\nelse:\n print(c)'] | ['Wrong Answer', 'Accepted'] | ['s199063082', 's490467136'] | [47368.0, 47376.0] | [359.0, 292.0] | [457, 426] |
p02623 | u450288159 | 2,000 | 1,048,576 | 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. | ["# -*- coding: utf-8 -*-\n\n\ndef main():\n n, m, k = map(int, input().split())\n A = list(map(int, input().split()))\n B = list(map(int, input().split()))\n a = A[0]\n A2 = [a]\n b = B[0]\n B2 = [b]\n for i in range(1, n):\n a += A[i]\n A2.append(a)\n\n for i in range(1, m):\n b += B[i]\n B2.append(b)\n max = -2\n for ia, aa in enumerate(A2):\n for ib, bb in enumerate(B2):\n if aa + bb > k:\n break\n if ia + ib > max:\n\n max = ia + ib\n print(max + 2)\n\n\ndef answer():\n N, M, K = map(int, input().split())\n\n\n A = list(map(int, input().split()))\n B = list(map(int, input().split()))\n\n a, b = [0], [0]\n for i in range(N):\n a.append(a[i] + A[i])\n for i in range(M):\n b.append(b[i] + B[i])\n\n ans, j = 0, M\n for i in range(N + 1):\n if a[i] > K:\n break\n while b[j] > K - a[i]:\n j -= 1\n ans = max(ans, i + j)\n print(ans)\nif __name__ == '__main__':\n answer()", "# -*- coding: utf-8 -*-\n\n\n\n\n# A = list(map(int, input().split()))\n# B = list(map(int, input().split()))\n# a = A[0]\n# A2 = [a]\n# b = B[0]\n# B2 = [b]\n\n# a += A[i]\n# A2.append(a)\n# \n\n# b += B[i]\n# B2.append(b)\n# max = -2\n\n# for ib, bb in enumerate(B2):\n\n# break\n# if ia + ib > max:\n# \n# max = ia + ib\n# print(max + 2)\n# \n\ndef answer():\n N, M, K = map(int, input().split())\n\n\n A = list(map(int, input().split()))\n B = list(map(int, input().split()))\n\n a, b = [0], [0]\n for i in range(N):\n a.append(a[i] + A[i])\n for i in range(M):\n b.append(b[i] + B[i])\n\n ans, j = 0, M\n for i in range(N + 1):\n if a[i] > K:\n break\n while b[j] > K - a[i]:\n j -= 1\n ans = max(ans, i + j)\n print(ans)\nif __name__ == '__main__':\n answer()"] | ['Runtime Error', 'Accepted'] | ['s392235308', 's953429087'] | [9128.0, 47332.0] | [26.0, 229.0] | [1050, 1088] |
p02623 | u454866890 | 2,000 | 1,048,576 | 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. | ['n,m,l = map(int,input().split(" "))\nNs = input().split(" ")\nN = [int (n) for n in Ns]\nMs = input().split(" ")\nM = [int(n) for n in Ms]\n#print(n,m,l)\n#print(N,M)\ni = 0\nj = 0\ntotal = 0\ncnt = 0\nwhile total <= l:\n if N[i] <= M[j]:\n total = total + N[i]\n i = i+1\n cnt = cnt +1\n else:\n total = total + M[j]\n j = j+1\n cnt = cnt +1\nprint(cnt)', 'N,M,K = list(map(int,input().split()))\nA = list(map(int,input().split()))\nB = list(map(int,input().split()))\na,b = [0],[0]\nfor i in range(N):\n a.append(a[i]+A[i])\nfor j in range(M):\n b.append(b[j]+B[j])\nans = 0\n#print(a,b,N+1)\nj = M\nfor i in range(N+1):\n if a[i] > K:\n break\n\n while K - a[i] < b[j]:\n j = j-1\n ans = max(ans,i+j)\n\nprint(ans)'] | ['Runtime Error', 'Accepted'] | ['s892285426', 's876996692'] | [55284.0, 47496.0] | [183.0, 284.0] | [355, 351] |
p02623 | u455354923 | 2,000 | 1,048,576 | 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. | ['#import sys\nN,M,K = map(int,input().split())\nA = list(map(int,input().split()))\nB = list(map(int,input().split()))\n\na=[0]\nb=[0]\nfor i in range(N):\n a.append(a[i]+A[i])\nfor j in range(M):\n b.append(b[j]+B[j])\nans = 0\nl = M\nfor k in range(N+1):\n if a[k] > K:\n break:\n while b[l] + a[k] > K:\n l -= 1\n ans = max(ans,k+l)\nprint(ans)\n', 'N,M,K = map(int,input().split())\nA = list(map(int,input().split()))\nB = list(map(int,input().split()))\nans = 0\ntime = 0\nwhile time < K:\n if not A:\n ans += 1\n if time + B[0] > K:\n break\n time += B[0]\n B.pop(0)\n elif not B:\n ans += 1\n if time + A[0]> K:\n break\n time += A[0]\n A.pop(0)\n elif A[0] > K or B[0] > K:\n break\n elif A[0] == B[0]:\n if A[1] > B[1]:\n if time + B[0] > K:\n break\n time += B[0]\n ans += 1\n B.pop(0)\n elif A[1] < B[1]:\n if time + A[0] > K:\n break\n time += B[0]\n ans += 1\n A.pop(0)\n elif A[0] > B[0] :\n if time + B[0] > K:\n break\n time += B[0]\n ans += 1\n B.pop(0)\n elif A[0] < B[0]:\n if time + A[0] > K:\n break\n time += B[0]\n ans += 1\n A.pop(0)\nprint(ans)\n', '#import sys\nN,M,K = map(int,input().split())\nA = list(map(int,input().split()))\nB = list(map(int,input().split()))\n\na=[0]\nb=[0]\nfor i in range(N):\n a.append(a[i]+A[i])\nfor j in range(M):\n b.append(b[j]+B[j])\nans = 0\nl = M\nfor k in range(N+1):\n if a[k] > K:\n break:\n while b[l] > K - a[k]:\n l -= 1\n ans = max(ans,k+l)\nprint(ans)\n', 'N,M,K = map(int,input().split())\nA = list(map(int,input().split()))\nB = list(map(int,input().split()))\nans = 0\ntime = 0\nwhile time < K:\n if not A:\n ans += 1\n time += B[0]\n B.pop(0)\n elif not B:\n ans += 1\n time += A[0]\n A.pop(0)\n elif A[0] = B[0]:\n if A[1] > B[1]:\n time += B[0]\n ans += 1\n B.pop(0)\n else A[1] < B[1]:\n time += B[0]\n ans += 1\n A.pop(0)\n elif A[0] > B[0] :\n time += B[0]\n ans += 1\n B.pop(0)\n elif A[0] < B[0]:\n time += B[0]\n ans += 1\n A.pop(0)\nprint(ans)\n', 'import sys\nN,M,K = map(int,input().split())\nA = list(map(int,input().split()))\nB = list(map(int,input().split()))\n\na=[0]\nb=[0]\nfor i in range(N):\n a.append(a[i]+A[i])\nfor j in range(M):\n b.append(b[i]+B[i])\nans = 0\nl = M\nfor k in range(N+1):\n if a[k] > K:\n break:\n while b[l] + a[k] > K:\n j -= 1\n ans = max(ans,k+j)\nprint(ans) \n', 'N,M,K = map(int,input().split())\nA = list(map(int,input().split()))\nB = list(map(int,input().split()))\nans = 0\ntime = 0\nwhile time < K:\n if not A:\n ans += 1\n time += B[0]\n B.pop(0)\n elif not B:\n ans += 1\n time += A[0]\n A.pop(0)\n elif A[0] == B[0]:\n if A[1] > B[1]:\n time += B[0]\n ans += 1\n B.pop(0)\n else A[1] < B[1]:\n time += B[0]\n ans += 1\n A.pop(0)\n elif A[0] > B[0] :\n time += B[0]\n ans += 1\n B.pop(0)\n elif A[0] < B[0]:\n time += B[0]\n ans += 1\n A.pop(0)\nprint(ans)\n', '#import sys\nN,M,K = map(int,input().split())\nA = list(map(int,input().split()))\nB = list(map(int,input().split()))\n\na=[0]\nb=[0]\nfor i in range(N):\n a.append(a[i]+A[i])\nfor j in range(M):\n b.append(b[i]+B[i])\nans = 0\nl = M\nfor k in range(N+1):\n if a[k] > K:\n break:\n while b[l] + a[k] > K:\n l -= 1\n ans = max(ans,k+l)\nprint(ans)\n', 'N,M,K = map(int,input().split())\nA = list(map(int,input().split()))\nB = list(map(int,input().split()))\nans = 0\ntime = 0\nwhile time < K:\n if not A:\n ans += 1\n if time + B[0]:\n break\n time += B[0]\n B.pop(0)\n elif not B:\n ans += 1\n if time + A[0]:\n break\n time += A[0]\n A.pop(0)\n elif A[0] > K or B[0] > K:\n break\n elif A[0] == B[0]:\n if A[1] > B[1]:\n if time + B[0]:\n break\n time += B[0]\n ans += 1\n B.pop(0)\n elif A[1] < B[1]:\n if time + A[0]:\n break\n time += B[0]\n ans += 1\n A.pop(0)\n elif A[0] > B[0] :\n if time + B[0]:\n break\n time += B[0]\n ans += 1\n B.pop(0)\n elif A[0] < B[0]:\n if time + A[0]:\n break\n time += B[0]\n ans += 1\n A.pop(0)\nprint(ans)\n', '#import sys\nN,M,K = map(int,input().split())\nA = list(map(int,input().split()))\nB = list(map(int,input().split()))\n\na=[0]\nb=[0]\nfor i in range(N):\n a.append(a[i]+A[i])\nfor j in range(M):\n b.append(b[j]+B[j])\nans = 0\nl = M\nfor k in range(N+1):\n if a[k] > K:\n break\n while b[l] > K - a[k]:\n l -= 1\n ans = max(ans,k+l)\nprint(ans)\n'] | ['Runtime Error', 'Runtime Error', 'Runtime Error', 'Runtime Error', 'Runtime Error', 'Runtime Error', 'Runtime Error', 'Runtime Error', 'Accepted'] | ['s071528214', 's166750424', 's322750788', 's526192119', 's663763471', 's751265318', 's835588767', 's968549703', 's873200760'] | [8992.0, 40052.0, 9048.0, 8988.0, 9016.0, 9012.0, 8916.0, 40056.0, 47608.0] | [24.0, 2206.0, 28.0, 24.0, 28.0, 25.0, 25.0, 2206.0, 290.0] | [357, 985, 357, 648, 357, 649, 357, 962, 356] |
p02623 | u459419927 | 2,000 | 1,048,576 | 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. | ['import bisect\nN,M,K=list(map(int,input().split()))\nA=list(map(int,input().split()))\nB=list(map(int,input().split()))\nbka=[0]*N\nbkb=[0]*M\nbka[0]=A[0]\nbkb[0]=B[0]\n\nfor i in range(1,N):\n bka[i]+=bka[i-1]+A[i]\nfor j in range(1,M):\n bkb[j]+=bkb[j-1]+B[j]\nans=bisect.bisect_left(bka,K)\nfor i in range(N):\n count=i+1\n k=K-bka[i]\n if k<0:continue\n c = bisect.bisect_left(bkb,k)\n if c!=M:\n if bkb[c]==k:\n c+=count+1\n else:c+=count\n\n # print(bkb,k,c,count)\n ans=max(ans,c)\nprint(ans)', 'import bisect\nN,M,K=list(map(int,input().split()))\nA=list(map(int,input().split()))\nB=list(map(int,input().split()))\nbka=[0]*N\nbkb=[0]*M\nbka[0]=A[0]\nbkb[0]=B[0]\n\nfor i in range(1,N):\n bka[i]+=bka[i-1]+A[i]\nfor j in range(1,M):\n bkb[j]+=bkb[j-1]+B[j]\nans=0\nc=bisect.bisect_left(bkb,K)\nif c != M:\n if bkb[c] == K:\n c +=+ 1\nans = max(ans, c)\nfor i in range(N):\n count=i+1\n k=K-bka[i]\n if k<0:continue\n c = bisect.bisect_left(bkb,k)\n b=c\n if c!=M:\n if bkb[c]==k:\n c+=count+1\n else:c+=count\n else:c+=count\n ans=max(ans,c)\nprint(ans)'] | ['Wrong Answer', 'Accepted'] | ['s650824533', 's824407206'] | [47300.0, 48976.0] | [386.0, 399.0] | [521, 592] |
p02623 | u463602788 | 2,000 | 1,048,576 | 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. | ['#n = int(input())\nn,m,k = [int(z) for z in input().split()]\na = [int(z) for z in input().split()]\nb = [int(z) for z in input().split()]\ntime = 0\ni = j = 0\nwhile (time<=k and i<n and j<m):\n if (a[i]>b[j]):\n if (time+a[i]<=k):\n time += a[i]\n i+=1\n else:\n break\n else:\n if (time+b[j]<=k):\n time += b[j]\n j+=1\n else:\n break\nwhile (i<n and time<=k):\n if (time+a[i]<=k):\n time += a[i]\n i+=1\n else:\n break\nwhile (j<n and time<=k):\n if (time+b[j]<=k):\n time += b[j]\n j+=1\n else:\n break\nprint(i+j)\n', "import sys\nimport bisect\nimport itertools\nimport collections\nimport fractions\nimport heapq\nimport math\nfrom operator import mul\nfrom functools import reduce\nfrom functools import lru_cache\n\n\ndef solve():\n readline = sys.stdin.buffer.readline\n mod = 10 ** 9 + 7\n N, M, K = map(int, readline().split())\n A = list(map(int, readline().split()))\n B = list(map(int, readline().split()))\n\n\n nowtime = 0\n\n Asum = [0]\n Bsum = [0]\n tmp = 0\n for a in A:\n tmp += a\n Asum.append(tmp)\n tmp = 0\n for b in B:\n tmp += b\n Bsum.append(tmp)\n\n maxbooks = 0\n before = M\n bcount = M\n for acount in range(N+1):\n if Asum[acount] > K:\n continue\n while K < Asum[acount] + Bsum[bcount]:\n bcount -= 1\n maxbooks = max(maxbooks, acount+bcount)\n\n print(maxbooks)\n\nif __name__ == '__main__':\n solve()\n"] | ['Runtime Error', 'Accepted'] | ['s219847187', 's227329123'] | [39744.0, 49392.0] | [252.0, 219.0] | [755, 893] |
p02623 | u469254913 | 2,000 | 1,048,576 | 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. | ['# import numpy as np\n# import math\n# import copy\n# from collections import deque\nimport sys\ninput = sys.stdin.readline\n\n\n\ndef main():\n N,M,K = map(int,input().split())\n A = list(map(int,input().split()))\n B = list(map(int,input().split()))\n\n a = 0\n b = 0\n\n time = 0\n res = 0\n\n isAexist = True\n isBexist = True\n\n for i in range(N+M+1):\n if isAexist:\n tA = A[a]\n else:\n tA = 10 ** 9 + 1\n if isBexist:\n tB = B[b]\n else:\n tB = 10 ** 9 + 1\n if (tA <= tB) & isAexist:\n if time + tA <= K:\n time += tA\n a += 1\n res += 1\n if a == N:\n isAexist = False\n elif (tA >= tB) & isBexist:\n if time + tB <= K:\n time += tB\n b += 1\n res += 1\n if b == M:\n isBexist = False\n print(res,time,a,b)\n if isAexist:\n if time + A[a] > K:\n isAexist = False\n if isBexist:\n if time + B[b] > K:\n isBexist = False\n if (not isAexist) & (not isBexist):\n break\n\n print(res)\n\n\n\nmain()\n', '# import numpy as np\n# import math\n# import copy\n# from collections import deque\nimport sys\ninput = sys.stdin.readline\n\n\n\ndef main():\n N,M,K = map(int,input().split())\n A = list(map(int,input().split()))\n B = list(map(int,input().split()))\n\n sumA = [0] * (N+1)\n sumB = [0] * (M+1)\n\n for i in range(1,N+1):\n sumA[i] = sumA[i-1] + A[i-1]\n\n for i in range(1,M+1):\n sumB[i] = sumB[i-1] + B[i-1]\n\n for i in range(M,-1,-1):\n if sumB[i] <= K:\n endB = i\n break\n\n for i in range(N,-1,-1):\n if sumA[i] <= K:\n endA = i\n break\n\n res = 0\n\n for i in range(0,endA+1):\n for j in range(endB,-1,-1):\n if sumA[i] + sumB[j] <= K:\n endB = j\n res = max(res,i+j)\n break\n\n print(res)\n\n\n\nmain()\n'] | ['Wrong Answer', 'Accepted'] | ['s017794776', 's606992505'] | [41664.0, 47760.0] | [635.0, 288.0] | [1265, 871] |
p02623 | u471503862 | 2,000 | 1,048,576 | 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. | ['n, m, k = map(int, input().split())\nA = list(map(int, input().split()))\nB = list(map(int, input().split()))\n\na_max = 0\nsum = 0\nfor i in range(n):\n sum += A[i]\n if sum > k:\n sum -= A[i]\n break\n a_max += 1\n \nans = a_max\nkotae =ans\nj = 0\n\nprint(a_max, 0)\nfor i in range(m):\n sum += B[i]\n ans += 1\n while sum > k and j < a_max:\n sum -= A[a_max-1-j]\n ans -= 1\n j += 1\n if sum > k:\n print(kotae)\n exit()\n kotae = max(kotae, ans)\n if a_max-j == 0:\n break\n \nprint(kotae)\n \n \n \n ', 'n, m, k = map(int, input().split())\nA = list(map(int, input().split()))\nB = list(map(int, input().split()))\n\na_max = 0\nsum = 0\nfor i in range(n):\n sum += A[i]\n if sum > k:\n sum -= A[i]\n break\n a_max += 1\n \nans = a_max\nkotae =ans\nj = 0\n\n\nfor i in range(m):\n sum += B[i]\n ans += 1\n while sum > k and j < a_max:\n sum -= A[a_max-1-j]\n ans -= 1\n j += 1\n if sum > k:\n print(kotae)\n exit()\n kotae = max(kotae, ans)\n\n \nprint(kotae)\n \n \n \n '] | ['Wrong Answer', 'Accepted'] | ['s819207974', 's526498446'] | [40412.0, 40624.0] | [322.0, 299.0] | [513, 470] |
p02623 | u473291366 | 2,000 | 1,048,576 | 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. | ['N, M, K = map(int,input().split())\nA = list(map(int,input().split()))\nB = list(map(int,input().split()))\nSUM_A=[0 for _ in range(N+1)]\nfor i in range(N):\n SUM_A[i+1] = A[i] + SUM_A[i]\n if SUM_A[i+1] > K:\n N=i\n break\n\nSUM_B=[0 for _ in range(M+1)]\nfor i in range(M):\n SUM_B[i+1] = B[i] + SUM_B[i]\n if SUM_B[i+1] > K:\n M=i\n break\n\nprint(N,M)\ncnt = 0\nfor i in range(N+1):\n for j in range(M+1):\n if SUM_A[i]+SUM_B[j] <= K:\n if i+j > cnt:\n cnt = i+j\n else:\n break\nprint(cnt)', 'N, M, K = map(int,input().split())\nA = list(map(int,input().split()))\nB = list(map(int,input().split()))\nSUM_A=[0 for _ in range(N+1)]\nfor i in range(N):\n SUM_A[i+1] = A[i] + SUM_A[i]\n if SUM_A[i+1] > K:\n N=i\n break\n\nSUM_B=[0 for _ in range(M+1)]\nfor i in range(M):\n SUM_B[i+1] = B[i] + SUM_B[i]\n if SUM_B[i+1] > K:\n M=i\n break\n\ncnt = N\nfor i in range(N+1)[::-1]:\n V=K-SUM_A[i]\n j = cnt-i\n while j < M+1 and SUM_B[j] <= V:\n j += 1\n j-=1\n if i+j > cnt:\n cnt = i+j\nprint(cnt)'] | ['Wrong Answer', 'Accepted'] | ['s893745440', 's297408801'] | [40812.0, 42688.0] | [2207.0, 338.0] | [563, 542] |
p02623 | u474423089 | 2,000 | 1,048,576 | 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. | ['import sys\ninput = sys.stdin.readline\nfrom itertools import accumulate\n\nN,M,K = map(int,input().split())\nA = list(map(int,input().split()))\nB = list(map(int,input().split()))\na = [0]+list(accumulate(A))\nb = [0]+list(accumulate(B))\nans,j = 0,M\nfor i in range(N+1):\n if i > K:\n break\n while b[j] > K - a[i]:\n j -= 1\n ans = max(ans,i+j)\nprint(ans)', 'import sys\nfrom itertools import accumulate\ninput = sys.stdin.readline\n\nN,M,K=map(int,input().split())\nA = list(map(int,input().split()))\nB = list(map(int,input().split()))\nA_acc = list(accumulate(A))\nB_acc = list(accumulate(B))\na = 0\nb = 0\nans = 0\ndeep = ""\nfor n,z in enumerate(zip(A_acc,B_acc)):\n i,j = z\n if i < j and i <= K:\n deep = \'A\'\n elif j <= K:\n deep = \'B\'\n else:\n break\n ans += 1\nif deep == \'A\':\n K -= A_acc[n]\n s = B\nelse:\n K -= B_acc[n]\n s = A\nfor i in s:\n if K >= i:\n ans += 1\n K -= i\n else:\n break\nprint(ans)', 'import sys\ninput = sys.stdin.readline\nfrom itertools import accumulate\n\nN,M,K = map(int,input().split())\nA = list(map(int,input().split()))\nB = list(map(int,input().split()))\na = [0]+list(accumulate(A))\nb = [0]+list(accumulate(B))\nans,j = 0,M\nfor i in range(N+1):\n if a[i] > K:\n break\n while b[j] > K - a[i]:\n j -= 1\n ans = max(ans,i+j)\nprint(ans)'] | ['Runtime Error', 'Wrong Answer', 'Accepted'] | ['s132367956', 's877901072', 's739195011'] | [49076.0, 49032.0, 49056.0] | [275.0, 217.0, 234.0] | [367, 599, 370] |
p02623 | u475402977 | 2,000 | 1,048,576 | 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. | ['N,M,K=map(int, input().split())\nA=list(map(int, input().split()))\nB=list(map(int, input().split()))\n\na, b = [0], [0]\nfor i in range(N):\n a.append(a[i] + A[i])\nfor i in range(M):\n b.append(b[i] + B[i])\n\nans, j = 0, M\nfor i in range(N + 1):\n if a[i] > K:\n break\nwhile b[j] > K - a[i]:\n j -= 1\n ans = max(ans, i + j)\nprint(ans)\n', 'N,M,K=map(int, input().split())\nA=list(map(int, input().split()))\nB=list(map(int, input().split()))\n\na, b = [0], [0]\nfor i in range(N):\na.append(a[i] + A[i])\nfor i in range(M):\nb.append(b[i] + B[i])\n\nans, j = 0, M\nfor i in range(N + 1):\n if a[i] > K:\n break\nwhile b[j] > K - a[i]:\n j -= 1\n ans = max(ans, i + j)\nprint(ans)', 'N,M,K=map(int, input().split())\nA=list(map(int, input().split()))\nB=list(map(int, input().split()))\n\na, b = [0], [0]\nfor i in range(N):\n a.append(a[i] + A[i])\nfor i in range(M):\n b.append(b[i] + B[i])\n\nans, j = 0, M\nfor i in range(N + 1):\n if a[i] > K:\n break\n while b[j] > K - a[i]:\n j -= 1\n ans = max(ans, i + j)\nprint(ans)'] | ['Runtime Error', 'Runtime Error', 'Accepted'] | ['s047536118', 's849669423', 's569818497'] | [47388.0, 8752.0, 47336.0] | [362.0, 29.0, 293.0] | [347, 338, 354] |
p02623 | u475546258 | 2,000 | 1,048,576 | 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. | ['N, M, K = map(int, input().split())\nA = list(map(int, input().split()))\nB = list(map(int, input().split()))\n\na, b = [0], [0]\nfor i in range(N):\na.append(a[i] + A[i])\nfor i in range(M):\nb.append(b[i] + B[i])\n\nans, j = 0, M\nfor i in range(N + 1):\nif a[i] > K:\nbreak\nwhile b[j] > K - a[i]:\nj -= 1\nans = max(ans, i + j)\nprint(ans)', 'N, M, K = map(int, input().split())\nA = list(map(int, input().split()))\nB = list(map(int, input().split()))\n\na, b = [0], [0]\nfor i in range(N):\n a.append(a[i] + A[i])\nfor i in range(M):\n b.append(b[i] + B[i])\n\nans, j = 0, M\nfor i in range(N + 1):\n if a[i] > K:\n break\n while b[j] > K - a[i]:\n j -= 1\n ans = max(ans, i + j)\nprint(ans)\n'] | ['Runtime Error', 'Accepted'] | ['s337723241', 's934920686'] | [8896.0, 47544.0] | [29.0, 282.0] | [326, 345] |
p02623 | u476674874 | 2,000 | 1,048,576 | 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. | ["import numpy as np\nfrom collections import deque\ndef main():\n N, M, K = map(int, input().split())\n A = deque(list(map(int, input().split())))\n B = deque(list(map(int, input().split())))\n \n cost = K\n count = 0\n while True:\n if len(A) ==0 and len(B) == 0:\n break\n elif len(A) == 0:\n cost -= B.popleft()\n elif len(B) == 0:\n cost -= A.popleft()\n\n else:\n a = A[0] + A[1]\n b = B[0]\n if a < b:\n cost -= A.popleft()\n else:\n cost -= B.popleft()\n\n if cost < 0:\n break\n count += 1\n \n print(count)\n\n\nif __name__ == '__main__':\n main()\n", "import numpy as np\ndef main():\n N, M, K = map(int, input().split())\n A = list(map(int, input().split()))\n B = list(map(int, input().split()))\n \n ia = 0\n ib = 0\n cost = 0\n while True:\n if ia < len(A) and ib < len(B):\n if A[ia] < B[ia]:\n if cost + A[ia] > K:\n break\n ia += 1\n cost += A[ia]\n else:\n if cost + B[ib] > K:\n break\n ib += 1\n cost += B[ib]\n elif ia < len(A):\n if cost + A[ia] > K:\n break\n ia += 1\n cost += A[ia]\n elif ib < len(B):\n if cost + B[ib] > K:\n break\n ib += 1\n cost += B[ib]\n else:\n break\n \n count = ia + ib\n print(count)\n\n\nif __name__ == '__main__':\n main()\n", "from bisect import bisect_right\nimport numpy as np\ndef main():\n N, M, K = map(int, input().split())\n A = np.array([0] + list(map(int, input().split())), dtype=np.int64)\n B = np.array([0] + list(map(int, input().split())), dtype=np.int64)\n\n As = np.cumsum(A)\n Bs = np.cumsum(B)\n\n count = 0\n\n for i, a in enumerate(As):\n if a > K:\n break\n j = bisect_right(Bs, K-a)\n \n count = max(count, (i+j-1))\n\n print(count)\n\nif __name__ == '__main__':\n main()"] | ['Runtime Error', 'Runtime Error', 'Accepted'] | ['s709693006', 's876767327', 's536148082'] | [59760.0, 58420.0, 53996.0] | [320.0, 254.0, 656.0] | [713, 897, 551] |
p02623 | u477837488 | 2,000 | 1,048,576 | 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. | ['n, m, k = map(int, input().split())\n\na = list(map(int, input().split()))\nb = list(map(int, input().split()))\n\na_, b_ = [0], [0]\n\nfor i in range(n):\n a_.append(a_[i] + a[i])\nfor j in range(m):\n b_.append(b_[j] + b[j])\n\nans = 0\nfor i in range(n+1):\n \n idx2 = m\n while a_[i] + b_[idx2] > k:\n idx2 -= 1\n \n ans = max(ans, i + idx2)\nprint(ans)', 'n, m, k = map(int, input().split())\n\na = list(map(int, input().split()))\nb = list(map(int, input().split()))\n\na_, b_ = [0], [0]\n\nfor i in range(n):\n a_.append(a_[i] + a[i])\nfor i in range(m):\n b_.append(b_[i] + b[i])\n\nans = 0\nj = m\n\nfor i in range(n+1):\n \n if a_[i] > k:\n break\n while b_[j] > k - a_[i]:\n j -= 1\n ans = max(ans, i + j)\nprint(ans)'] | ['Runtime Error', 'Accepted'] | ['s163502021', 's535705159'] | [47580.0, 47516.0] | [2207.0, 280.0] | [365, 377] |
p02623 | u480168496 | 2,000 | 1,048,576 | 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. | ['n, m, k = map(int, input().split())\na = list(map(int, input().split()))\nb = list(map(int, input().split()))\n\nA = [0], B = [0]\nfor i in range(n):\n A.append(A[i] + a[i])\nfor i in range(m):\n B.append(B[i] + b[i])\n\nans, j = 0, m\nfor i in range(n+1):\n if A[i] > k:\n break\n while B[j] > k - A[i]:\n j -= 1\n ans = max(ans, i+j)\nprint(ans)', 'n, m, k = map(int, input().split())\na = list(map(int, input().split()))\nb = list(map(int, input().split()))\n\nA, B = [0], [0]\nfor i in range(n):\n A.append(A[i] + a[i])\nfor i in range(m):\n B.append(B[i] + b[i])\n\nans, j = 0, m\nfor i in range(n+1):\n if A[i] > k:\n break\n while B[j] > k - A[i]:\n j -= 1\n ans = max(ans, i+j)\nprint(ans)\n'] | ['Runtime Error', 'Accepted'] | ['s575853523', 's695368672'] | [9140.0, 47504.0] | [28.0, 295.0] | [359, 359] |
p02623 | u484052148 | 2,000 | 1,048,576 | 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. | ['from itertools import accumulate\nfrom bisect import bisect_right\nN, M, K = map(int, input().split())\nA = [0] + list(map(int, input().split()))\nB = list(map(int, input().split()))\nSA = list(accumulate(A))\nSB = list(accumulate(B))\n\nans = 0\nfor i in range(len(SA)):\n if tmp := K-SA[i] < 0:\n break\n ans = max(ans, bisect_right(SB, tmp))\nprint(ans)', 'from itertools import accumulate\nfrom bisect import bisect_right\nN, M, K = map(int, input().split())\nA = [0] + list(map(int, input().split()))\nB = list(map(int, input().split()))\nSA = list(accumulate(A))\nSB = list(accumulate(B))\n\nans = 0\nfor i in range(len(SA)):\n ans = max(ans, bisect_right(SB, K-SA[i]))\nprint(ans)', 'from itertools import accumulate\nfrom bisect import bisect_right\nN, M, K = map(int, input().split())\nA = [0] + list(map(int, input().split()))\nB = list(map(int, input().split()))\nSA = list(accumulate(A))\nSB = list(accumulate(B))\n\nans = 0\nfor i in range(len(SA)):\n if tmp := K-SA[i] < 0:\n break\n ans = max(ans, bisect_right(SB, tmp)+i)\nprint(ans)', 'from itertools import accumulate\nfrom bisect import bisect_right\nN, M, K = map(int, input().split())\nA = [0] + list(map(int, input().split()))\nB = list(map(int, input().split()))\nSA = list(accumulate(A))\nSB = list(accumulate(B))\n\nans = 0\nfor n in range(len(SA)):\n tmp = K-SA[n]\n if tmp < 0:\n break\n m = bisect_right(SB, tmp)\n ans = max(ans, n+m)\nprint(ans)'] | ['Wrong Answer', 'Wrong Answer', 'Wrong Answer', 'Accepted'] | ['s033956550', 's086556840', 's227327442', 's752543448'] | [49344.0, 49308.0, 49168.0, 49140.0] | [251.0, 244.0, 256.0, 265.0] | [356, 319, 358, 375] |
p02623 | u485319545 | 2,000 | 1,048,576 | 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. | ['n,m,k=map(int,input().split())\na=list(map(int,input().split()))\nb=list(map(int,input().split()))\n\nimport numpy as np\nimport bisect\n\nra=[0]+list(np.cumsum(a))\nrb=[0]+list(np.cumsum(b))\n\nans=0\nfor i in range(n+1): \n capa = k - ra[i]\n if ra[i]>0:\n break\n j=bisect.bisect_right(rb,capa)\n\n\n ans=max(ans,i-1+j)\n \nprint(ans) ', 'n,m,k=map(int,input().split())\na=list(map(int,input().split()))\nb=list(map(int,input().split()))\n\nimport numpy as np\nimport bisect\n\nra=[0]+list(np.cumsum(a))\nrb=[0]+list(np.cumsum(b))\n\nans=0\nfor i in range(n+1): \n capa = k - ra[i]\n if ra[i]>0:\n break\n j=bisect.bisect_right(rb,capa)\n\n\n ans=max(ans,i-1+j)\n \nprint(ans) \n', 'n,m,k=map(int,input().split())\na=list(map(int,input().split()))\nb=list(map(int,input().split()))\n\nimport numpy as np\nimport bisect\n\nra=[0]+list(np.cumsum(a))\nrb=[0]+list(np.cumsum(b))\n\nans=0\nfor i in range(n+1): \n capa = k - ra[i]\n if ra[i]>k:\n break\n j=bisect.bisect_right(rb,capa)\n\n\n ans=max(ans,i-1+j)\n \nprint(ans) \n'] | ['Wrong Answer', 'Wrong Answer', 'Accepted'] | ['s386858979', 's524666932', 's460990952'] | [59972.0, 59824.0, 60200.0] | [266.0, 256.0, 504.0] | [354, 355, 355] |
p02623 | u485819963 | 2,000 | 1,048,576 | 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. | ['\nN, M, K = map(int,input().split())\n\nA = list(map(int, input().split()))\nB = list(map(int, input().split()))\n\n\na, b = [0], [0]\n\nfor i in range(N):\n a.append(a[i] + A[i])\nfor i in range(M):\n b.append(b[i] + B[i])\n\n\nans, j = 0, M\n\n\n\n\n\nfor i in range(N+1):\n if a[i] > K:\n break\n # print(a[i])\n while b[j] > K -a[i]:\n \n j -= 1\n print(b[j])\n \n ans = max(ans, i + j)\nprint(ans)\n\n\n', '\nN, M, K = map(int,input().split())\n\nA = list(map(int, input().split()))\nB = list(map(int, input().split()))\n\n\na, b = [0], [0]\n\nfor i in range(N):\n a.append(a[i] + A[i])\nfor i in range(M):\n b.append(b[i] + B[i])\n\n\nans, j = 0, M\n\n\n\n\n\nfor i in range(N+1):\n if a[i] > K:\n break\n # print(a[i])\n while b[j] > K -a[i]:\n \n j -= 1\n # print(b[j])\n \n ans = max(ans, i + j)\nprint(ans)\n'] | ['Wrong Answer', 'Accepted'] | ['s584410778', 's531224612'] | [49208.0, 47308.0] | [349.0, 303.0] | [1932, 1932] |
p02623 | u490489966 | 2,000 | 1,048,576 | 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. | ['#C\nn, m, k = map(int, input().split())\na = list(map(int, input().split()))\nb = list(map(int, input().split()))\nans = 0\ni = 0\nj = 0\nfor _ in range(n+m)\n # print(i,j,k)\n if n <= i:#\n k -= b[j]\n j += 1\n if k < 0:\n break\n ans += 1\n if k == 0:\n break\n continue\n elif n <= j:\n k -= a[i]\n i += 1\n if k < 0:\n break\n ans += 1\n if k == 0:\n break\n continue\n if a[i] < b[j]:\n k -= a[i]\n i += 1\n else:\n k -= b[j]\n j += 1\n if k < 0:\n break\n ans += 1\n if k == 0:\n break\nprint(ans)', '#C\nn, m, k = map(int, input().split())\na = list(map(int, input().split()))\nb = list(map(int, input().split()))\nA = [0]\nB = [0]\n\nfor i in range(n):\n A.append(A[i] + a[i])\nfor i in range(m):\n B.append(B[i] + b[i])\nans, j = 0, m\n\nfor i in range(n + 1):\n if a[i] > k:\n break\n while b[j] > k - a[i]:\n j -= 1\n ans = max(ans, i + j)\nprint(ans)', '#C\nn, m, k = map(int, input().split())\na = list(map(int, input().split()))\nb = list(map(int, input().split()))\nA = [0]\nB = [0]\n\nfor i in range(n):\n A.append(A[i] + a[i])\nfor i in range(m):\n B.append(B[i] + b[i])\nans = 0\nj = m\n\nfor i in range(n + 1):\n if A[i] > k:\n break\n \n while A[i] + B[j] > k:\n j -= 1\n ans = max(ans, i + j)\nprint(ans)'] | ['Runtime Error', 'Runtime Error', 'Accepted'] | ['s063233599', 's363588982', 's878235582'] | [9036.0, 47684.0, 47324.0] | [24.0, 184.0, 284.0] | [659, 480, 531] |
p02623 | u497805118 | 2,000 | 1,048,576 | 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. | ['\nimport bisect\nimport numpy\n\nn, m, k = map(int, input().split(" "))\na = list(map(int, input().split(" ")))\nb = list(map(int, input().split(" ")))\n\ncost_a = numpy.insert(numpy.cumsum(a), 0, 0)\ncost_b = numpy.insert(numpy.cumsum(b), 0, 0)\n\nans = 0\nj = len(cost_b)\nfor i in range(len(cost_a)):\n for j in range( j, 0, -1):\n if cost_a[i] + cost_b[j] <= k:\n ans = i + j if ans < i + j else ans\n break\nprint(ans)', '\nimport bisect\nimport numpy\n\nn, m, k = map(int, input().split(" "))\na = list(map(int, input().split(" ")))\nb = list(map(int, input().split(" ")))\n\ncost_a = numpy.insert(numpy.cumsum(a), 0, 0)\ncost_b = numpy.insert(numpy.cumsum(b), 0, 0)\n\nans = 0\nj = len(cost_b) -1\nfor i in range(len(cost_a)):\n for j in range( j, 0, -1):\n if cost_a[i] + cost_b[j] <= k:\n ans = i + j if ans < i + j else ans\n break\nprint(ans)\n[root@local', '\nimport bisect\nimport numpy\n\nn, m, k = map(int, input().split(" "))\na = list(map(int, input().split(" ")))\nb = list(map(int, input().split(" ")))\n\ncost_a = numpy.insert(numpy.cumsum(a), 0, 0)\ncost_b = numpy.insert(numpy.cumsum(b), 0, 0)\n\nans = 0\nj = len(cost_b) -1\nfor i in range(len(cost_a)):\n for j in range(j, -1, -1):\n if cost_a[i] + cost_b[j] <= k:\n ans = i + j if ans < i + j else ans\n break\nprint(ans)'] | ['Runtime Error', 'Runtime Error', 'Accepted'] | ['s456403324', 's508141027', 's204950520'] | [58476.0, 8884.0, 58388.0] | [229.0, 26.0, 539.0] | [439, 454, 442] |
p02623 | u497883442 | 2,000 | 1,048,576 | 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. | ['N, M, K = map(int, input().split())\nA = list(map(int, input().split()))\nB = list(map(int, input().split()))\na, b = [0], [0]\nfor i in range(N):\n\u3000a.append(a[i] + A[i])\nfor i in range(M):\n\u3000b.append(b[i] + B[i])\n\nans, j = 0, M\nfor i in range(N + 1):\n\u3000if a[i] > K:\n\u3000\u3000break\n\u3000while b[j] > K - a[i]:\n\u3000\u3000j -= 1\n\u3000ans = max(ans, i + j)\nprint(ans)\n', 'N, M, K = map(int, input().split())\nA = list(map(int, input().split()))\nB = list(map(int, input().split()))\na, b = [0], [0]\nfor i in range(N):\n a.append(a[i] + A[i])\nfor i in range(M):\n b.append(b[i] + B[i])\n\nans, j = 0, M\nfor i in range(N + 1):\n if a[i] > K:\n break\n while b[j] > K - a[i]:\n j -= 1\n ans = max(ans, i + j)\nprint(ans)\n'] | ['Runtime Error', 'Accepted'] | ['s918519865', 's564032092'] | [8908.0, 47444.0] | [26.0, 305.0] | [353, 344] |
p02623 | u498658826 | 2,000 | 1,048,576 | 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. | ['/usr/bin/env python3\nimport sys\n\n\nN, M, K = map(int, input().split())\nA = list(map(int, input().split()))\nB = list(map(int, input().split()))\na, b = [0], [0]\n\nfor i in range(N):\n a.append(a[i] + A[i])\nfor i in range(M):\n b.append(b[i] + B[i])\nans, j = 0, M\nfor i in range(N+1):\n if a[i] > K:\n break\n while b[j] > K - a[i]:\n j -= 1\n ans = max(ans, i + j)\nprint(ans)', '#!/usr/bin/env python3\nimport sys\n\n\nN, M, K = map(int, input().split())\nA = list(map(int, input().split()))\nB = list(map(int, input().split()))\na, b = [0], [0]\n\nfor i in range(N):\n a.append(a[i] + A[i])\nfor i in range(M):\n b.append(b[i] + B[i])\nans, j = 0, M\nfor i in range(N + 1):\n if a[i] > K:\n break\nwhile b[j] > K - a[i]:\n j -= 1\nans = max(ans, i + j)\nprint(ans)\n', '#!/usr/bin/env python3\nimport sys\n\ndef solve(N: int, M: int, K: int, A: "List[int]", B: "List[int]"):\n weightA = 0\n tmp_A = N\n for idx, a in enumerate(A):\n if weightA + a > K:\n tmp_A = idx\n break\n else:\n weightA += a\n max_count = current_count = tmp_A\n len_A = tmp_A\n\n weightB = 0\n terminated = False\n for b in B:\n if terminated == True and weightA + weightB > K:\n break\n\n if len_A==0 and weightA + weightB + b > K:\n break\n\n if weightA + weightB + b <= K:\n weightB += b\n current_count += 1\n\n elif terminated==False:\n while weightA + weightB + b > K:\n if len_A>0:\n weightA -= A[len_A-1]\n len_A -= 1\n current_count -= 1\n else:\n terminated=True\n\n if weightA + weightB + b <= K:\n weightB += b\n current_count += 1\n if current_count < max_count:\n break\n max_count = max(max_count, current_count)\n\n print(max_count)\n return\n\n\ndef solve_old(N: int, M: int, K: int, A: "List[int]", B: "List[int]"):\n weightA = 0\n tmp_A = N\n for idx, a in enumerate(A):\n if weightA + a > K:\n tmp_A = idx\n break\n else:\n weightA += a\n max_count = current_count = tmp_A\n len_A = tmp_A\n\n weightB = 0\n terminated = False\n for b in B:\n if terminated == True and weightA + weightB > K:\n break\n\n if len_A==0 and weightA + weightB + b > K:\n break\n\n if weightA + weightB + b <= K:\n weightB += b\n current_count += 1\n\n elif terminated==False:\n while weightA + weightB + b > K:\n if len_A>0:\n weightA -= A[len_A-1]\n len_A -= 1\n current_count -= 1\n else:\n terminated=True\n\n if weightA + weightB + b <= K:\n weightB += b\n current_count += 1\n max_count = max(max_count, current_count)\n\n print(max_count)\n return\n', '#!/usr/bin/env python3\nimport sys\n\n\nN, M, K = map(int, input().split())\nA = list(map(int, input().split()))\nB = list(map(int, input().split()))\na, b = [0], [0]\n\nfor i in range(N):\n a.append(a[i] + A[i])\nfor i in range(M):\n b.append(b[i] + B[i])\n\nans = 0\nj = N\n\nfor i in range(M+1):\n if b[i] > K:\n break\n while a[j] > K - b[i]:\n j -= 1\n ans = max(ans, i + j)\nprint(ans)\n'] | ['Runtime Error', 'Runtime Error', 'Wrong Answer', 'Accepted'] | ['s739433933', 's958070816', 's971171070', 's887558541'] | [9004.0, 47176.0, 9236.0, 47296.0] | [27.0, 269.0, 25.0, 307.0] | [393, 386, 2213, 398] |
p02623 | u500673386 | 2,000 | 1,048,576 | 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. | ['n,m,k=map(int,input().split())\na=list(map(int,input().split()))\nb=list(map(int,input().split()))\ni=0;j=0;ans=0\nwhile k>0:\n if i<n and j<m:\n if a[i]<=b[j]:\n k-=a[i]\n ans+=1\n i+=1\n continue\n else:\n k-=b[j]\n ans+=1\n j+=1\n continue\n elif i==n and j<m:\n k-=b[j]\n ans+=1\n j+=1\n continue\n elif i<n and j==m:\n k-=a[i]\n ans+=1\n i+=1\n continue\nprint(ans-1 if (i==0 and j==1)or(i==1 and j==0)else ans)', 'n,m,k=map(int,input().split())\nA=list(map(int,input().split()))\nB=list(map(int,input().split()))\na=[0];b=[0];ans=0;j=m\nfor i in range(n):\n a.append(a[i]+A[i])\nfor i in range(m):\n b.append(b[i]+B[i])\nfor i in range(n+1):\n if a[i]>k:\n break\n while b[j]<k-a[i]:\n j-=1\nans=max(ans,i+j)\nprint(ans)', 'n,m,k=map(int,input().split())\nA=list(map(int,input().split()))\nB=list(map(int,input().split()))\na=[0];b=[0];ans=0;j=m\nfor i in range(n):\n a.append(a[i]+A[i])\nfor i in range(m):\n b.append(b[i]+B[i])\nfor i in range(n+1):\n if a[i]>k:\n break\nwhile b[j]<k-a[i]:\n j-=1\nans=max(ans,i+j)\nprint(ans)', 'n,m,k=map(int,input().split())\nA=list(map(int,input().split()))\nB=list(map(int,input().split()))\na=[0];b=[0];ans=0;j=m\nfor i in range(n):\n a.append(a[i]+A[i])\nfor i in range(m):\n b.append(b[i]+B[i])\nfor i in range(n+1):\n if a[i]>k:\n break\n while b[j]>k-a[i]:\n j-=1\n ans=max(ans,i+j)\nprint(ans)'] | ['Wrong Answer', 'Runtime Error', 'Wrong Answer', 'Accepted'] | ['s090467850', 's428666193', 's961723782', 's620046845'] | [40408.0, 47480.0, 47328.0, 47496.0] | [233.0, 235.0, 209.0, 288.0] | [641, 318, 310, 322] |
p02623 | u505391462 | 2,000 | 1,048,576 | 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. | ['N, M, K = map(int, input().split())\nA = list(map(int, input().split()))\nB = list(map(int, input().split()))\na,b=[0],[0]\nfor i in range(N):\n\ta.append(a[i] + A[i])\nfor i in range(M):\n\tb.append(b[i] + B[i])\nans,j=0,M\nfor i in range(N + 1): \n\tif a[i]>K:\n \tbreak\n\twhile b[j] > K - a[i]:\n \t\tj -= 1\n\tans = max(ans, i + j)\nprint(ans)\n\n', 'N, M, K = map(int, input().split())\nA = list(map(int, input().split()))\nB = list(map(int, input().split()))\na,b=[0],[0]\nfor i in range(N):\n\ta.append(a[i] + A[i])\nfor i in range(M):\n\tb.append(b[i] + B[i])\nans=0\nj=M\nfor i in range(N + 1): \n\tif a[i]>K:\n \tbreak\n\twhile b[j] > K - a[i]:\n \tj -= 1\n\tans = max(ans, i + j)\nprint(ans)\n', 'N, M, K = map(int, input().split())\nA = list(map(int, input().split()))\nB = list(map(int, input().split()))\na,b=[0],[0]\nfor i in range(N):\n\ta.append(a[i] + A[i])\nfor i in range(M):\n\tb.append(b[i] + B[i])\nans,j=0,M\nfor i in range(N + 1): \n\tif a[i]>K:\n \tbreak\nwhile b[j] > K - a[i]:\n j -= 1\nans = max(ans, i + j)\nprint(ans)', 'N, M, K = map(int, input().split())\nA = list(map(int, input().split()))\nB = list(map(int, input().split()))\na,b=[0],[0]\nfor i in range(N):\n\ta.append(a[i] + A[i])\nfor i in range(M):\n\tb.append(b[i] + B[i])\nans,j=0,M\nfor i in range(N + 1): \n\tif a[i]>K:\n \t\tbreak\n\twhile b[j] > K - a[i]:\n \t\tj -= 1\n\tans = max(ans, i + j)\nprint(ans)\n\n'] | ['Runtime Error', 'Runtime Error', 'Runtime Error', 'Accepted'] | ['s342106787', 's406821282', 's843103144', 's567578654'] | [8900.0, 9060.0, 8968.0, 47312.0] | [26.0, 25.0, 24.0, 309.0] | [333, 331, 327, 334] |
p02623 | u511449169 | 2,000 | 1,048,576 | 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. | ['n, m, k = map(int, input().split())\nA = list(map(int, input().split()))\nB = list(map(int, input().split()))\n\na, b = [0], [0]\nfor i in range(n):\n a.append(a[i] + A[i])\nfor i in range(m):\n b.append(b[i] + B[i])\n\nans, j = 0, m\nprint(b[j])\nfor i in range(n + 1):\n if a[i] > k:\n break\n while b[j] > k - a[i]:\n j -= 1\n ans = max(ans, i + j)\nprint(ans)', 'n, m, k = map(int, input().split())\nA = list(map(int, input().split()))\nB = list(map(int, input().split()))\n\na, b = [0], [0]\nfor i in range(n):\n a.append(a[i] + A[i])\nfor i in range(m):\n b.append(b[i] + B[i])\n\nans, j = 0, m\nfor i in range(n + 1):\n if a[i] > k:\n break\n while b[j] > k - a[i]:\n j -= 1\n ans = max(ans, i + j)\nprint(ans)'] | ['Wrong Answer', 'Accepted'] | ['s161769167', 's841699513'] | [47448.0, 47368.0] | [297.0, 315.0] | [374, 362] |
p02623 | u512099209 | 2,000 | 1,048,576 | 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. | ['N, M, K = list(map(int, input().split()))\nA = list(map(int, input().split()))\nB = list(map(int, input().split()))\n\ndp = [[0] * (M + 1) for _ in range(N + 1)]\n \nres = 0\n\nfor i in range(1, N + 1):\n dp[i][0] = dp[i - 1][0] + A[-i]\n if dp[i][0] <= K:\n res = max(res, i)\n else:\n break\n\nfor j in range(1, M + 1):\n dp[0][j] = dp[0][j - 1] + B[-j]\n if dp[0][j] <= K:\n res = max(res, j)\n else:\n break\n\nfor i in range(1, N + 1):\n for j in range(1, M + 1):\n dp[i][j] = min(dp[i - 1][j] + A[-i], dp[i][j - 1] + B[-j])\n if dp[i][j] <= K:\n res = max(res, i + j)\n else:\n break\n\nprint(res)', 'N, M, K = list(map(int, input().split()))\nA = list(map(int, input().split()))\nB = list(map(int, input().split()))\n\ndp = [[0] * (M + 1) for _ in range(N + 1)]\n\nfor i in range(1, N + 1):\n dp[i][0] = dp[i - 1][0] + A[-i]\n\nfor j in range(1, M + 1):\n dp[0][j] = dp[0][j - 1] + B[-j]\n \nres = 0\n\nfor i in range(1, N + 1):\n for j in range(1, M + 1):\n dp[i][j] = min(dp[i - 1][j] + A[-i], dp[i][j - 1] + B[-j])\n if dp[i][j] <= K:\n res = max(res, i + j)\n\nprint(res)', 'N, M, K = list(map(int, input().split()))\nA = list(map(int, input().split()))\nB = list(map(int, input().split()))\n\nAs = [0] * (N + 1)\nBs = [0] * (M + 1)\n\nfor i in range(1, N + 1):\n As[i] = As[i - 1] + A[i - 1]\n if As[i] > K:\n i -= 1\n break\n\nfor j in range(1, M + 1):\n Bs[j] = Bs[j - 1] + B[j - 1]\n if Bs[j] > K:\n j -= 1\n break\n\nif i < j:\n As, Bs = Bs, As\n i, j = j, i\n\nj = 0\nn = i\nwhile i:\n while j + 1 < len(Bs) and As[i] + Bs[j + 1] <= K:\n j += 1\n n = max(n, i + j)\n #print(As, Bs, i, j, n)\n i -= 1\n\nprint(n)'] | ['Wrong Answer', 'Wrong Answer', 'Accepted'] | ['s067742384', 's156695990', 's711754143'] | [2642076.0, 2707984.0, 40776.0] | [2285.0, 2298.0, 399.0] | [612, 470, 535] |
p02623 | u514395957 | 2,000 | 1,048,576 | 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. | ['N, M, K = map(int, input().split())\nA = list(map(int, input().split()))\nB = list(map(int, input().split()))\n \na, b = [0], [0]\nfor i in range(N):\n\ta.append(a[i] + A[i])\nfor i in range(M):\n\tb.append(b[i] + B[i])\n \nans, j = 0, M\nfor i in range(N + 1):\n\tif a[i] > K:\n \t\tbreak\n \twhile b[j] > K - a[i]:\n \t\tj -= 1\n \t\tans = max(ans, i + j)\nprint(ans)', 'N, M, K = map(int, input().split())\nA = list(map(int, input().split()))\nB = list(map(int, input().split()))\n\na, b = [0], [0]\nfor i in range(N):\n a.append(a[i] + A[i])\nfor i in range(M):\n b.append(b[i] + B[i])\n\nans, j = 0, M\nfor i in range(N + 1):\n if a[i] > K:\n break\n while b[j] > K - a[i]:\n j -= 1\n ans = max(ans, i + j)\nprint(ans)'] | ['Runtime Error', 'Accepted'] | ['s699503898', 's187595556'] | [8828.0, 47512.0] | [23.0, 277.0] | [342, 344] |
p02623 | u516554284 | 2,000 | 1,048,576 | 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. | ['n,m,k=map(int,input().split())\na=list(map(int,input().split()))\nb=list(map(int,input().split()))\n\nc=[0]\nd=[0]\n\nfor i in range(n):\n c.append(c[-1]+a[i])\n \nfor j in range(m):\n d.append(d[-1]+b[j])\n\nans=0\nfor x in range(n):\n l=c[x+1]-c[x]\n B=0\n if l<=k:\n h=k-l\n mi=0\n ma=m+1\n while ma-mi>1:\n mid=(mi+ma)//2\n if d[mid]<=h:\n mi=mid\n elif d[mid]<h:\n mi=mid\n B=mi\n \n ans=max(ans,x+1+B)\n \nprint(ans)\n', 'n,m,k=map(int,input().split())\na=list(map(int,input().split()))\nb=list(map(int,input().split()))\n\nc=[0]\nd=[0]\n\nfor i in range(n):\n c.append(c[-1]+a[i])\n \nfor j in range(m):\n d.append(d[-1]+b[j])\n\nans=0\nfor x in range(n):\n l=c[x+1]-c[x]\n B=0\n if l<=k:\n h=k-l\n mi=0\n ma=m+1\n while ma-mi>1:\n mid=(mi+ma)//2\n if d[mid]<=h:\n mi=mid\n elif d[mid]>h:\n ma=mid\n B=mi\n \n ans=max(ans,x+1+B)\n \nprint(ans)', 'n,m,k=map(int,input().split())\na=list(map(int,input().split()))\nb=list(map(int,input().split()))\n\nc=[0]\nd=[0]\n\nfor i in range(n):\n c.append(c[-1]+a[i])\n \nfor j in range(m):\n d.append(d[-1]+b[j])\n\nans=0\nfor x in range(n):\n l=c[x+1]-c[x]\n B=0\n if l<=k:\n h=k-l\n mi=0\n ma=m+1\n while ma-mi>1:\n mid=(mi+ma)//2\n if d[mid]<=h:\n mi=mid\n elif d[mid]<h:\n ma=mid\n B=mi\n \n ans=max(ans,x+1+B)\n \nprint(ans)', 'n,m,k=map(int,input().split())\na=list(map(int,input().split()))\nb=list(map(int,input().split()))\nans=0\n\nfor x in range(n):\n sum1=0\n ans1=0\n \n sum1=sum(a[0:x])\n if sum1>k:\n break\n else:\n \tans1+=x+1\n for y in range(m):\n \n \n sum1+=b[y]\n if sum1>k:\n break\n else:\n ans1+=1\n if ans1>ans:\n ans=ans1\n \nprint(ans)\n ', 'n,m,k=map(int,input().split())\na=list(map(int,input().split()))\nb=list(map(int,input().split()))\n\nc=[0]\nd=[0]\n\nfor i in range(n):\n c.append(c[-1]+a[i])\n \nfor j in range(m):\n d.append(d[-1]+b[j])\n\nans=0\nfor x in range(n+1):\n l=c[x]\n B=0\n if l<=k:\n h=k-l\n mi=0\n ma=m+1\n while ma-mi>1:\n mid=(mi+ma)//2\n if d[mid]<=h:\n mi=mid\n elif d[mid]>h:\n ma=mid\n B=mi\n \n ans=max(ans,x+B)\n \n \n \nprint(ans)\n'] | ['Wrong Answer', 'Wrong Answer', 'Wrong Answer', 'Wrong Answer', 'Accepted'] | ['s033232513', 's134064026', 's508381723', 's825900808', 's179176294'] | [48832.0, 48660.0, 48700.0, 40488.0, 48700.0] | [2207.0, 1194.0, 2207.0, 2206.0, 1115.0] | [453, 452, 452, 366, 456] |
p02623 | u516927307 | 2,000 | 1,048,576 | 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. | ['N, M, K = map(int, input().split())\nA = list(map(int, input().split()))\nB = list(map(int, input().split()))\n\na, b = [0], [0]\nfor i in range(N):\n a.append(a[i]+A[1])\nfor i in range(M):\n b.append(b[i]+B[i])\n\nans, j = 0, M\nfor i in range(N + 1):\n if a[i] > K:\n break\n while b[j] > K-a[i]:\n j -= 1\n \n ans = max(ans, i+j)\n \nprint(ans)', 'N, M, K = map(int, input().split())\nA = list(map(int, input().split()))\nB = list(map(int, input().split()))\n\na, b = [0], [0]\nfor i in range(N):\n a.append(a[i]+A[i])\nfor i in range(M):\n b.append(b[i]+B[i])\n\nans, j = 0, M\nfor i in range(N + 1):\n if a[i] > K:\n break\n while b[j] > K-a[i]:\n j -= 1\n \n ans = max(ans, i+j)\n \nprint(ans)'] | ['Runtime Error', 'Accepted'] | ['s868809880', 's931894960'] | [47504.0, 47548.0] | [290.0, 296.0] | [364, 364] |
p02623 | u517388115 | 2,000 | 1,048,576 | 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. | ['n, m, k =map(int, input().split())\na=list(map(int, input().split()))\nb=list(map(int, input().split()))\nta=0\ntb=0\nans=0\nfor i in range(n+1):\n ta += a[i]\n if i=0:\n ta=0\n if ta>k:\n break\n for j in range(m):\n tb=sum(b[:j+1])\n ans=max(ans,i+j)\n if tb + ta >k:\n break\n ans=max(ans,i+j+1)\nprint(ans)\n', 'n, m, k =map(int, input().split())\na=list(map(int, input().split()))\nb=list(map(int, input().split()))\nta=sum(a)\na.append(0)\ntb=0\nans=0\nj=0\nfor i in range(n+1):\n ta -= a[n-i]\n if ta>k:\n continue\n while tb + ta<=k:\n\n if j ==m:\n ans=max(ans,n-i+j)\n break\n ans=max(ans,n-i+j)\n tb += b[j]\n j +=1\n\nprint(ans)\n'] | ['Runtime Error', 'Accepted'] | ['s904434528', 's047885182'] | [8980.0, 40384.0] | [26.0, 285.0] | [358, 370] |
p02623 | u518958552 | 2,000 | 1,048,576 | 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. | ['def aaa():\n n,m,k = map(int,input().split())\n a = list(map(int,input().split()))\n b = list(map(int,input().split()))\n c = []\n for i,j in zip(a,b):\n if i <= j:\n c.append(i)\n c.append(j)\n else:\n c.append(j)\n c.append(i)\n d = 0\n e = 0\n for i in c:\n if i+e < k:\n e += i\n d += 1\n else:\n return d\n return n + m\naaa()', 'def aaa():\n n,m,k = map(int,input().split())\n a = list(map(int,input().split()))\n b = list(map(int,input().split()))\n c = []\n for i,j in zip(a,b):\n if i <= j:\n c.append(i)\n c.append(j)\n else:\n c.append(j)\n c.append(i)\n \n a1 = a\n b1 = b\n a1[0:n+1] = []\n b1[0:m+1] = []\n c = c + a1 + b1\n d = 0\n e = 0\n for i in c:\n if i+e < k:\n e += i\n d += 1\n else:\n return d\n return n + m\naaa()', 'def aaa():\n n,m,k = map(int,input().split())\n a = list(map(int,input().split()))\n b = list(map(int,input().split()))\n if n >= m:\n for _ in range(m-n):\n a.append(10**9+1)\n else:\n for _ in range(n-m):\n b.append(10**9+1)\n c = []\n for i,j in zip(a,b):\n if i <= j:\n c.append(i)\n c.append(j)\n else:\n c.append(j)\n c.append(i)\n d = 0\n e = 0\n for i in c:\n if i+e < k:\n e += i\n d += 1\n else:\n return d\n return n + m\naaa()', 'def aaa():\n n,m,k = map(int,input().split())\n a = list(map(int,input().split()))\n b = list(map(int,input().split()))\n c = []\n for i,j in zip(a,b):\n c.append(i)\n c.append(j)\n d = 0\n e = 0\n for i in c:\n if i+e < k:\n e += i\n d += 1\n else:\n return d\n return n + m\naaa()', 'def aaa():\n n,m,k = map(int,input().split())\n a = list(map(int,input().split()))\n b = list(map(int,input().split()))\n c = []\n for i,j in zip(a,b):\n c.append(i)\n c.append(j)\n c = sorted(c)\n d = 0\n e = 0\n for i in c:\n if i+e < k:\n e += i\n d += 1\n else:\n return d\n return n + m\naaa()', 'N, M, K = map(int, input().split())\nA = list(map(int, input().split()))\nB = list(map(int, input().split()))\n \na, b = [0], [0]\nfor i in range(N):\n a.append(a[i] + A[i])\nfor i in range(M):\n b.append(b[i] + B[i])\n \nans, j = 0, M\nfor i in range(N + 1):\n if a[i] > K:\n break\n while b[j] > K - a[i]:\n j -= 1\n ans = max(ans, i + j)\nprint(ans)'] | ['Wrong Answer', 'Wrong Answer', 'Wrong Answer', 'Wrong Answer', 'Wrong Answer', 'Accepted'] | ['s061619384', 's184206454', 's407819281', 's602240273', 's958103788', 's311335032'] | [40604.0, 40108.0, 40488.0, 40452.0, 40452.0, 47548.0] | [173.0, 180.0, 171.0, 163.0, 264.0, 292.0] | [442, 531, 585, 353, 371, 364] |
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