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2 years ago
#include <bits/stdc++.h>
using namespace std;
//矩阵快速幂
typedef long long LL;
const int MOD = 1e9 + 7;
const int N = 110;
LL n, k; //本题数据范围很大用int直接wa哭了
//矩阵声明
struct JZ {
LL m[N][N];
} A, res, base;
//矩阵乘法
inline JZ mul(JZ A, JZ B) {
JZ C;
memset(C.m, 0, sizeof(C.m));
for (LL i = 0; i < n; i++)
for (LL j = 0; j < n; j++)
for (LL k = 0; k < n; k++) {
C.m[i][j] += (A.m[i][k] % MOD) * (B.m[k][j] % MOD);
C.m[i][j] %= MOD;
}
return C;
}
void qmi() {
//将结果矩阵初始化为单位矩阵
memset(res.m, 0, sizeof res.m);
for (int i = 0; i < n; i++) res.m[i][i] = 1;
//其实就是把整数快速幂修改为矩阵快速幂
while (k) { //二进制快速幂
if (k & 1) res = mul(res, A); // 联想一下整数快速幂
A = mul(A, A); // base 翻倍
k >>= 1;
}
}
int main() {
cin >> n >> k;
//输入原始矩阵
for (int i = 0; i < n; i++)
for (int j = 0; j < n; j++)
cin >> A.m[i][j];
//计算矩阵快速幂
qmi();
//输出
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++)
cout << res.m[i][j] << " ";
cout << endl;
}
return 0;
}