結果
| 問題 |
No.194 フィボナッチ数列の理解(1)
|
| コンテスト | |
| ユーザー |
|
| 提出日時 | 2022-10-11 04:58:49 |
| 言語 | PyPy3 (7.3.15) |
| 結果 |
AC
|
| 実行時間 | 142 ms / 5,000 ms |
| コード長 | 1,854 bytes |
| コンパイル時間 | 147 ms |
| コンパイル使用メモリ | 82,508 KB |
| 実行使用メモリ | 77,504 KB |
| 最終ジャッジ日時 | 2024-06-25 01:29:47 |
| 合計ジャッジ時間 | 5,141 ms |
|
ジャッジサーバーID (参考情報) |
judge4 / judge2 |
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| ファイルパターン | 結果 |
|---|---|
| sample | AC * 3 |
| other | AC * 37 |
ソースコード
from collections import *
from itertools import *
from functools import *
from heapq import *
import sys,math
input = sys.stdin.readline
N,K = map(int,input().split())
A = list(map(int,input().split()))
mod = 10**9 + 7
def answer1():
dp = [0]*(K)
for i in range(N):
dp[i] = A[i]
S = sum(dp)%mod
for i in range(N,K):
dp[i] = S
S += dp[i] - dp[i-N]
S %= mod
print(dp[-1],sum(dp)%mod)
def matrix_mul(A,B,mod = None):
nA = len(A)
mA = len(A[0])
mB = len(B[0])
tmp = [[0]*mB for _ in range(nA)]
if mod is None:
for i in range(nA):
for j in range(mB):
tmp[i][j] = sum(A[i][k]*B[k][j] for k in range(mA))
return tmp
for i in range(nA):
for j in range(mB):
tmp[i][j] = sum(A[i][k]*B[k][j]%mod for k in range(mA))%mod
return tmp
def matrix_pow(A,n,mod = None):
nbit = list(str(bin(n))[2:])
nbit = [int(i) for i in nbit]
N = len(A)
C = [[0]*N for _ in range(N)]
B = A
for i in range(N):
C[i][i] = 1
if mod is None:
for i in range(len(nbit)):
if nbit[-1-i] == 1:
C = matrix_mul(C,B)
B = matrix_mul(B,B)
return C
for i in range(len(nbit)):
if nbit[-1-i] == 1:
C = matrix_mul(C,B,mod)
B = matrix_mul(B,B,mod)
return C
def answer2():
I = [[A[i]] for i in range(N)] + [[A[0]]]
X = [[0]*(N+1) for _ in range(N+1)]
for i in range(N-1):
X[i][i+1] = 1
for i in range(N):
X[-2][i]=1
X[-1][1]=1
X[-1][-1]=1
Y = matrix_pow(X,K-1,mod=mod)
Z = matrix_mul(Y,I,mod=mod)
print(Z[0][0],Z[-1][0])
if N<50:
answer2()
else:
answer1()