51Nod - 1242 斐波那契（快速冪）
阿新 • • 發佈：2018-11-18
斐波那契數列的定義如下：
F(0) = 0
F(1) = 1
F(n) = F(n - 1) + F(n - 2) (n >= 2)
(1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, ...)
給出n，求F(n)，由於結果很大，輸出F(n) % 1000000009的結果即可。
Input輸入1個數n(1 <= n <= 10^18)。Output輸出F(n) % 1000000009的結果。Sample Input
11
Sample Output
89
#include <iostream> #include<string.h> #include<stdio.h> #define ll long long using namespace std; const ll mod = 1000000009; struct mat { ll m[2][2]; mat() { memset(m, 0, sizeof(m)); } }; mat mul(mat &A, mat &B) { mat C; for (int i = 0; i < 2; i++) { for (int j = 0; j < 2; j++) { for (int k = 0; k < 2; k++) { C.m[i][j] = (C.m[i][j] + A.m[i][k] * B.m[k][j]) % mod; } } }return C; } mat pow(mat A, ll n) { mat B; B.m[0][0] = B.m[0][1] = 1; while (n) { if (n & 1) B = mul(A, B); A = mul(A, A); n >>= 1; } return B; } int main() { ll n; while (cin >> n) { mat A; A.m[0][0] = A.m[0][1] = A.m[1][0] = 1; mat B = pow(A, n); printf("%lld\n", B.m[1][0]); } return 0; }