UESTC 1335 Fibonacci

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思路: 矩阵快速幂

分析:

1 最简单的矩阵快速幂

代码:

/************************************************
 * By: chenguolin                               * 
 * Date: 2013-08-28                             *
 * Address: http://blog.csdn.net/chenguolinblog *
 ************************************************/
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;

const int MOD = 10000;
const int N = 2;

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

int main(){
    Matrix m;
    m.mat[0][0] = m.mat[0][1] = 1;
    m.mat[1][0] = 1 ; m.mat[1][1] = 0;
    while(scanf("%d" , &n) && n != -1)
         printf("%d\n" , Pow(m));
    return 0;
}