結果
問題 | No.194 フィボナッチ数列の理解(1) |
ユーザー |
![]() |
提出日時 | 2020-07-25 19:34:30 |
言語 | PyPy3 (7.3.15) |
結果 |
MLE
|
実行時間 | - |
コード長 | 2,042 bytes |
コンパイル時間 | 615 ms |
コンパイル使用メモリ | 82,304 KB |
実行使用メモリ | 852,096 KB |
最終ジャッジ日時 | 2024-06-27 13:42:30 |
合計ジャッジ時間 | 6,577 ms |
ジャッジサーバーID (参考情報) |
judge3 / judge1 |
(要ログイン)
ファイルパターン | 結果 |
---|---|
sample | AC * 3 |
other | AC * 18 MLE * 1 -- * 18 |
ソースコード
from itertools import accumulateMOD = 1000000007N, K = map(int, input().split())A = list(map(int, input().split()))C = [0] + list(accumulate(A))if 1==2 and N <= 10000 and K <= 1000000:res = [0 for _ in range(K)]for i in range(N):res[i] = A[i]res[N] = sum(A)for i in range(N + 1, K):res[i] = res[i - 1] * 2 - res[i - N - 1]res[i] %= MODprint(res[-1], sum(res) % MOD)else:def inc(arr, base):n = len(arr)res = [0 for _ in range(n)]res[0] = arr[n - 1] * base[0]for i in range(1, n):res[i] = arr[i - 1] + arr[n - 1] * base[i]res[i] %= MODreturn resdef dec(arr, base):n = len(arr)res = [0 for _ in range(n)]res[n - 1] = arr[0] * pow(base[0], MOD - 2, MOD)for i in range(n - 1)[::-1]:res[i] = arr[i + 1] - res[n - 1] * base[i + 1]res[i] %= MODreturn resdef dub(arr, base):n = len(arr)res = [0 for _ in range(n)]tmp = [arr]for i in range(n - 1):tmp.append(inc(tmp[-1], base))for i in range(n):for j in range(n):res[i] += tmp[0][j] * tmp[j][i]res[i] %= MODreturn resres = [1 for _ in range(N)]exp = K - 1base = [1 for _ in range(N)]for _ in range(N):res = dec(res, base)for bit in range(exp.bit_length())[::-1]:res = dub(res, base)if (exp >> bit) & 1:res = inc(res, base)F = sum([res[i] * A[i] for i in range(N)]) % MODres = [0 for _ in range(N + 1)]res[0] = -1res[-1] = 2base = [0 for _ in range(N + 1)]base[0] = -1base[-1] = 2exp = Kfor _ in range(N + 1):res = dec(res, base)for bit in range(exp.bit_length())[::-1]:res = dub(res, base)if (exp >> bit) & 1:res = inc(res, base)S = sum([res[i] * C[i] for i in range(N + 1)]) % MODprint(F, S)