51nod 1242 斐波那契数列的第N项【矩阵快速幂】

题目链接：http://www.51nod.com/onlineJudge/questionCode.html#!problemId=1242

代码：

#include <stdio.h>#include <iostream>#include <string.h>#include <algorithm>#include <math.h>#include <ctype.h>#include <time.h>#include <queue>using namespace std;const int MOD = 1000000009;struct node{    long long m[2][2];}ans, base;long long n;node multi(node a, node b){    node 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] += (a.m[i][k] * b.m[k][j]);                tmp.m[i][j] %= MOD;            }        }    return tmp;}long long fast_mod(long long 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(){    while (scanf("%lld", &n) != EOF)    {        int ans = fast_mod(n);        printf("%lld\n", ans);    }    return 0;}