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 = ;
 const int mod = 1e9+;
 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, , sizeof(c.mat));
     int i, j, k;
     for(k = ; k <= n; ++k){
         for(i = ; i <= n; ++i){
             if(a.mat[i][k] <= ) continue;
             for(j = ; j <= n; ++j){
                 if(b.mat[k][j] <= ) 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 = ; i <= n; ++i)
         for(j = ; j <= n; ++j)
             c.mat[i][j] = (i == j);//初始化为单位矩阵
     while(k){
         if(k & )
             c = c * a;
         a = a * a;
         k >>= ;
     }
     return c;
 }
 int main(){
     scanf("%d%d", &n, &m);
     int i, j;
     Mat a;
     for(i = ; i <= n; ++i)
         for(j = ; j <= n; ++j)
             scanf("%lld", &a.mat[i][j]);
     Mat c = a ^ m;
     for(i = ; i <= n; ++i){
         for(j = ; j < n; ++j)
             printf("%lld ", c.mat[i][j]);
         printf("%lld\n", c.mat[i][n]);
     }
     return ;
 }