初等变换求 |A| % Mod & A- % Mod & A* % Mod(模板)

// |A| * A- = A* (伴随矩阵) = 逆矩阵 * 矩阵的值

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<ctime>
#include<iostream>
#include<algorithm>
using namespace std;

const int MAXN = 205;
const int Mod = 1000000007;
int a[MAXN][MAXN], b[MAXN][MAXN];

int fast_pow(int a, int k){
    int res = 1;
    while(k){
        if(k & 1) res = 1LL * res * a % Mod;
        a = 1LL * a * a % Mod;
        k >>= 1;
    }
    return res;
}

void solve(int n){
    for(int i = 1; i <= n; i++){
        for(int j = 1; j <= n; j++){
            b[i][j] = (i==j);//初始 b 为单位矩阵
        }
    }

    int det = 1;
    for(int i = 1; i <= n; i++){
        int t = i;
        for(int k = i; k <= n; k++){
            if(a[k][i]) t = k;
        }

        if(t != i) det *= -1;
        for(int j = 1; j <= n; j++){
            swap(a[i][j], a[t][j]);
            swap(b[i][j], b[t][j]);
        }

        det = 1LL * a[i][i] * det % Mod;
        int inv = fast_pow(a[i][i], Mod-2); // a[i][i] 的逆元

        for(int j = 1; j <= n; j++){
            a[i][j] = 1LL * inv * a[i][j] % Mod;
            b[i][j] = 1LL * inv * b[i][j] % Mod;
        }
        for(int k = 1; k <= n; k++){
            if(k == i) continue;
            int tmp = a[k][i];
            for(int j = 1; j <= n; j++){
                a[k][j] = (a[k][j] - 1LL * a[i][j] * tmp % Mod + Mod) % Mod;
                b[k][j] = (b[k][j] - 1LL * b[i][j] * tmp % Mod + Mod) % Mod;
            }
        }
    }
    //经过增广矩阵初等变换,此时 b 为 a 的逆矩阵
    det = (det + Mod) % Mod;  //    |a| % Mod
    for(int i = 1; i <= n; i++){
        for(int j = 1; j<= n; j++){
            b[i][j] = 1LL * det * b[i][j] % Mod;  //    将 b 由逆矩阵变成伴随矩阵
        }
    }
}

int main(void){
    int n;
    while(scanf("%d",&n)!=EOF){
        for(int i = 1; i <= n; i++)
            for(int j = 1; j <= n; j++)
                scanf("%d", &a[i][j]);
        solve(n);
        for(int i = 1; i <= n; i++)
            printf("%d%c",(i & 1 ? b[i][1] : (Mod - b[i][1]) % Mod), " 
" [i == n]);
    }
    return 0;
}

时间复杂度为 O(n^3)