poj3070——Fibonacci

In the Fibonacci integer sequence, F0 = 0, F1 = 1, and Fn = Fn − 1 + Fn − 2 for n ≥ 2. For example, the first ten terms of the Fibonacci sequence are:
0, 1, 1, 2, 3, 5, 8, 13, 21, 34, …
An alternative formula for the Fibonacci sequence is

Given an integer n, your goal is to compute the last 4 digits of Fn.
Input
The input test file will contain multiple test cases. Each test case consists of a single line containing n (where 0 ≤ n ≤ 1,000,000,000). The end-of-file is denoted by a single line containing the number −1.
Output
For each test case, print the last four digits of Fn. If the last four digits of Fn are all zeros, print ‘0’; otherwise, omit any leading zeros (i.e., print Fn mod 10000).
Sample Input
0
9
999999999
1000000000
-1
Sample Output
0
34
626
6875

学习矩阵快速幂的一道例题

代码：

#include <cstdio>
#include <algorithm>
using namespace std;
const int MAXN=2;
const int MOD=10000;
struct Mat{
    int m[MAXN][MAXN];
}mat;
Mat mul(Mat a,Mat b){
    Mat t;
    for(int i=0;i<MAXN;i++){
        for(int j=0;j<MAXN;j++){
            t.m[i][j]=0;
            for(int k=0;k<MAXN;k++){
                t.m[i][j]=(t.m[i][j]+(a.m[i][k]*b.m[k][j]))%MOD;
            }
        }
    }
    return t;
}
Mat quick_pow(Mat a,int n){
    Mat t;
    for(int i=0;i<MAXN;i++){
        for(int j=0;j<MAXN;j++){
            t.m[i][j]= i==j ? 1 : 0;
        }
    }
    while(n>0){
        if(n%2==1){
            t=mul(t,a);
        }
        n=n/2;
        a=mul(a,a);
    }
    return t;
}
int main(void){
    int n;
    while(~scanf("%d",&n)){
        if(n==-1){
            break;
        }
        mat.m[0][0]=1;
        mat.m[0][1]=1;
        mat.m[1][0]=1;
        mat.m[1][1]=0;
        printf("%d\n",quick_pow(mat,n).m[0][1]);
    }
    return 0;
}