python/TangDou/AcWingSuanFaJinJieZhiNan/P3390.cpp

#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
LL C[N][N], A[N][N], B[N][N];

//矩阵乘法
void mul(LL C[][N], LL A[][N], LL B[][N]) {
    //为什么非得要开一个临时的tmp来存结果呢，直接用C确实不行，原理不清楚
    static LL tmp[N][N];
    memset(tmp, 0, sizeof tmp);
    for (LL i = 0; i < n; i++)
        for (LL j = 0; j < n; j++)
            for (LL k = 0; k < n; k++) {
                tmp[i][j] += A[i][k] * B[k][j] % MOD; //矩阵乘法
                tmp[i][j] %= MOD;
            }
    memcpy(C, tmp, sizeof tmp);
}

void qmi() {
    //将结果矩阵初始化为单位矩阵
    memset(B, 0, sizeof B);
    for (int i = 0; i < n; i++) B[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) mul(B, B, A); // 联想一下整数快速幂
        mul(A, 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[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 << B[i][j] << " ";
        cout << endl;
    }
    return 0;
}