矩阵快速幂

矩阵快速幂#include <cstdio>#include <iostream>using namespace std;const int MOD = 10000;struct matrix{    int m[2][2];}ans, base;matrix multi(matrix a, matrix b){    matrix tmp;    for(int i = 0; i < 2; ++i)    {        for(int j = 0; j < 2; ++j)        {            tmp.m[i][j] = 0;            for(int k = 0; k < 2; ++k)                tmp.m[i][j] = (tmp.m[i][j] + a.m[i][k] * b.m[k][j]) % MOD;        }    }    return tmp;}int fast_mod(int n)  // 求矩阵 base 的  n 次幂 {    base.m[0][0] = base.m[0][1] = base.m[1][0] = 1;    base.m[1][1] = 0;    ans.m[0][0] = ans.m[1][1] = 1;  // ans 初始化为单位矩阵     ans.m[0][1] = ans.m[1][0] = 0;    while(n)    {        if(n & 1)  //实现 ans *= t; 其中要先把 ans赋值给 tmp，然后用 ans = tmp * t         {            ans = multi(ans, base);        }        base = multi(base, base);        n >>= 1;    }    return ans.m[0][1];}int main(){    int n;    while(scanf("%d", &n) && n != -1)    {           printf("%d\n", fast_mod(n));    }    return 0;}