結果

問題 No.1097 Remainder Operation
ユーザー Coki628Coki628
提出日時 2020-11-21 00:32:22
言語 PyPy3
(7.3.15)
結果
WA  
実行時間 -
コード長 1,398 bytes
コンパイル時間 1,485 ms
コンパイル使用メモリ 82,452 KB
実行使用メモリ 153,736 KB
最終ジャッジ日時 2024-07-23 14:11:20
合計ジャッジ時間 9,430 ms
ジャッジサーバーID
(参考情報)
judge5 / judge1
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 1 WA * 1
other AC * 2 WA * 19
権限があれば一括ダウンロードができます

ソースコード

diff #
プレゼンテーションモードにする

import sys
def input(): return sys.stdin.readline().strip()
def list2d(a, b, c): return [[c] * b for i in range(a)]
def list3d(a, b, c, d): return [[[d] * c for k in range(b)] for i in range(a)]
def list4d(a, b, c, d, e): return [[[[e] * d for k in range(c)] for k in range(b)] for i in range(a)]
def ceil(x, y=1): return int(-(-x // y))
def INT(): return int(input())
def MAP(): return map(int, input().split())
def LIST(N=None): return list(MAP()) if N is None else [INT() for i in range(N)]
def Yes(): print('Yes')
def No(): print('No')
def YES(): print('YES')
def NO(): print('NO')
sys.setrecursionlimit(10**9)
INF = 10**19
MOD = 10**9 + 7
EPS = 10**-10
def doubling(MAX, A):
""" """
N = len(A)
nxt = list2d(MAX, N, -1)
sm = list2d(MAX, N, 0)
for i, a in enumerate(A):
nxt[0][i] = i + a
sm[0][i] = a
for k in range(1, MAX):
for i in range(N):
nxt[k][i] = nxt[k-1][(i+nxt[k-1][i])%N]
sm[k][i] = sm[k-1][i] + sm[k-1][(i+nxt[k-1][i])%N]
return (nxt, sm)
N = INT()
A = LIST()
MX = 40
nxt, sm = doubling(MX, A)
# x = cur = 0
# for i in range(10):
# x += sm[0][cur]
# cur = nxt[0][cur] % N
# print(x, cur)
for _ in range(INT()):
k = INT()
x = cur = 0
for i in range(MX-1, -1, -1):
if k>>i & 1:
x += sm[i][cur]
cur = nxt[i][cur] % N
print(x)
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