题解 | 矩阵幂

#include <iostream>
#include<cstring>
#include<string>
using namespace std;

const int N = 12;
int n;
int mode[N][N], res[N][N];
//这里拷贝不能用a，因为这里a是参数
void multiply(int a[][N], int b[][N]) {
    int c[N][N];
    memcpy(c, a, sizeof c);
    for (int i = 0; i < n; i++)
        for (int j = 0; j < n; j++) {
            a[i][j] = 0;
            for (int k = 0; k < n; k++)
                a[i][j] += c[i][k] * b[k][j];
        }

}
int main() {
    int k;
    while (cin >> n >> k) { // 注意 while 处理多个 case
        for (int i = 0; i < n; i++)
            for (int j = 0; j < n; j++)
                cin >> mode[i][j], res[i][j] = mode[i][j];
        for (int i = 0; i < k-1; i++) {
            multiply(res, mode);
        }
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                cout << res[i][j];
                if (j != n - 1)cout << ' ';
            }
            cout << endl;
        }

    }
}
// 64 位输出请用 printf("%lld")