51nod 1113 矩阵快速幂

题目链接：51nod 1113 矩阵快速幂

模板题，学习下。

#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
const int N = 101;
const int mod = 1e9+7;
int n, m;
struct Mat{//矩阵
    ll mat[N][N];
};
Mat operator * (Mat a, Mat b){//一次矩阵乘法
    Mat c;
    memset(c.mat, 0, sizeof(c.mat));
    int i, j, k;
    for(k = 1; k <= n; ++k){
        for(i = 1; i <= n; ++i){
            if(a.mat[i][k] <= 0) continue;
            for(j = 1; j <= n; ++j){
                if(b.mat[k][j] <= 0) continue;
                c.mat[i][j] += a.mat[i][k] * b.mat[k][j];
                c.mat[i][j] %= mod;
            }
        }
    }
    return c;
}
Mat operator ^ (Mat a, int k){//矩阵快速幂，a^k
    Mat c;
    int i, j;
    for(i = 1; i <= n; ++i)
        for(j = 1; j <= n; ++j)
            c.mat[i][j] = (i == j);//初始化为单位矩阵
    while(k){
        if(k & 1)
            c = c * a;
        a = a * a;
        k >>= 1;
    }
    return c;
}
int main(){
    scanf("%d%d", &n, &m);
    int i, j;
    Mat a;
    for(i = 1; i <= n; ++i)
        for(j = 1; j <= n; ++j)
            scanf("%lld", &a.mat[i][j]);
    Mat c = a ^ m;
    for(i = 1; i <= n; ++i){
        for(j = 1; j < n; ++j)
            printf("%lld ", c.mat[i][j]);
        printf("%lld\n", c.mat[i][n]);
    }
    return 0;
}