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

;
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;
}