POJ 3070 Fibonacci

　　裸奔的矩阵乘法，当模板了。

#include <iostream>
#include <cstring>
#include <cstdio>

using namespace std;

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

struct Mat {
    long long mat[N][N];
    void init() {
        for(int i = 0; i < N; ++i) {
            for(int j = 0; j < N; ++j)
                mat[i][j] = (i == j);
        }
    }
};

Mat a;

Mat operator * (Mat a, Mat b) {
    Mat c;
    memset(c.mat, 0, sizeof(c.mat));

    int i, j, k;
    for(k = 0; k < N; ++k) {
        for(i = 0; i < N; ++i) {
            if(!a.mat[i][k])    continue;
            for(j = 0; j < N; ++j) {
                if(!b.mat[k][j])    continue;
                c.mat[i][j] += (a.mat[i][k] * b.mat[k][j] % MOD);
            }
        }
    }
    for(i = 0; i < N; ++i) {
        for(j = 0; j < N; ++j)
            c.mat[i][j] %= MOD;
    }
    return c;
}

Mat operator ^ (Mat a, int k) {
    Mat c;
    c.init();

    for(; k; k >>= 1) {
        if(k&1)    {
            c = c * a;
        }
        a = a * a;
    }
    return c;
}

int main() {
    //freopen("data.in", "r", stdin);

    int n, res;
    while(~scanf("%d", &n)) {
        if(n < 0)    break;
        if(n == 0)    {puts("0"); continue;}
        if(n == 1)    {puts("1"); continue;}

        a.mat[0][0] = 1; a.mat[0][1] = 1;
        a.mat[1][0] = 1; a.mat[1][1] = 0;
        a = a^n; res = 0;

        printf("%lld\n", a.mat[0][1]);
    }
    return 0;
}