题解 | #矩阵幂#

矩阵的快速幂和整数快速幂类似，只不过初始值是单位矩阵

#include <iostream>
using namespace std;
struct matrix {
    int m[10][10];
    int row;
    matrix(int r) {
        row = r;
    };
};
matrix multiply(matrix a, matrix b) {
    matrix ans(a.row);
    for (int i = 0; i < ans.row; i++) {
        for (int j = 0; j < ans.row; j++) {
            ans.m[i][j] = 0;
            for (int k = 0; k < a.row; k++) {
                ans.m[i][j] += a.m[i][k] * b.m[k][j];
            }
        }
    }
    return ans;
}
void print(matrix a) {
    for (int i = 0; i < a.row; i++) {
        for (int j = 0; j < a.row; j++) {
            if (j == a.row - 1) {
                cout << a.m[i][j] << endl;
            } else {
                cout << a.m[i][j] << ' ';
            }
        }
    }
}
matrix fastPower(matrix a, int k) {
    matrix ans(a.row);
    for (int i = 0; i < a.row; i++) {
        for (int j = 0; j < a.row; j++) {
            if (i == j) ans.m[i][j] = 1;
            else ans.m[i][j] = 0;
        }
    }
    while (k != 0) {
        if (k % 2 == 1) {
            ans = multiply(ans, a);
        }
        k /= 2;//右移
        a = multiply(a, a); //平方
    }
    return ans;
}
int main() {
    int n, k;
    while (cin >> n >> k) { // 注意 while 处理多个 case
        // cout << a + b << endl;
        matrix a(n);
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                cin >> a.m[i][j];
            }
        }
        matrix b = fastPower(a, k);
        print(b);
    }
}
// 64 位输出请用 printf("%lld")