版权声明：转载请注明出处。 https://blog.csdn.net/u014427196/article/details/48871469

题目链接： [

http://www.51nod.com/onlineJudge/questionCode.html#!problemId=1242
](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;
}