POJ 3070 Fibonacci（矩阵乘法logN）

题意：

求第n个斐波那契数，结果mod10000.

思路：

 

矩阵的n次方可以由二分法来解决，于是把时间缩短到了logN

#include <cstdio> #include <cstdlib> #include <cstring>



const int M = 10000; typedef struct _Matrix { int v[2][2]; } Matrix; Matrix m; Matrix matrix_mul(Matrix a, Matrix b) { Matrix c; c.v[0][0] = (a.v[0][0] * b.v[0][0] + a.v[0][1] * b.v[1][0]) % M; c.v[0][1] = (a.v[0][0] * b.v[0][1] + a.v[0][1] * b.v[1][1]) % M; c.v[1][0] = (a.v[1][0] * b.v[0][0] + a.v[1][1] * b.v[1][0]) % M; c.v[1][1] = (a.v[1][0] * b.v[0][1] + a.v[1][1] * b.v[1][1]) % M; return c; } Matrix matrix_pow(Matrix b, int n) { if (n == 1) return b; Matrix c = matrix_pow(b, n >> 1); if (n % 2) return matrix_mul(matrix_mul(c, c), m); else

        return matrix_mul(c, c); } int main() { int n; m.v[0][0] = 1, m.v[0][1] = 1; m.v[1][0] = 1, m.v[1][1] = 0; while (scanf("%d", &n) && n != -1) { if (n == 0) { printf("0\n"); continue; } Matrix b; b = matrix_pow(m, n); printf("%d\n", b.v[1][0]); } return 0; }