点击打开uva12470 

思路: 矩阵快速幂

分析:

1 裸题

代码:

/************************************************
 * By: chenguolin                               * 
 * Date: 2013-08-30                             *
 * Address: http://blog.csdn.net/chenguolinblog *
 ************************************************/
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;

typedef long long int64;
const int MOD = 1e9+9;
const int N = 3;

int64 n;
struct Matrix{
    int64 mat[N][N];
    Matrix operator*(const Matrix& m)const{
        Matrix tmp;
        for(int i = 0 ; i < N ; i++){
            for(int j = 0 ; j < N ; j++){
                tmp.mat[i][j] = 0;
                for(int k = 0 ; k < N ; k++)
                    tmp.mat[i][j] += mat[i][k]*m.mat[k][j]%MOD;
                tmp.mat[i][j] %= MOD;
            }
        }
        return tmp;
    }
};

int Pow(Matrix m){
    if(n <= 3) return n-1;
    Matrix ans;
    memset(ans.mat , 0 , sizeof(ans.mat));
    for(int i = 0 ; i < N ; i++)
        ans.mat[i][i] = 1;
    n -= 3;
    while(n){
        if(n%2)
            ans = ans*m;
        n /= 2;
        m = m*m;
    }
    int64 sum = 0;
    sum += ans.mat[0][0]*2%MOD;
    sum += ans.mat[0][1]*1%MOD;
    return sum%MOD;
}

int main(){
    Matrix m;
    memset(m.mat , 0 , sizeof(m.mat));
    for(int i = 0 ; i < N ; i++)
        m.mat[0][i] = 1;
    m.mat[1][0] = 1 ; m.mat[2][1] = 1;
    while(scanf("%lld" , &n) && n)
        printf("%d\n" , Pow(m));
    return 0;
}