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

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

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

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

         if(t != i) det *= -;
         for(int j = ; 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-); // a[i][i] 的逆元

         for(int j = ; j <= n; j++){
             a[i][j] = 1LL * inv * a[i][j] % Mod;
             b[i][j] = 1LL * inv * b[i][j] % Mod;
         }
         for(int k = ; k <= n; k++){
             if(k == i) continue;
             int tmp = a[k][i];
             for(int j = ; 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 = ; i <= n; i++){
         for(int j = ; 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 = ; i <= n; i++)
             for(int j = ; j <= n; j++)
                 scanf("%d", &a[i][j]);
         solve(n);
         for(int i = ; i <= n; i++)
             printf("%d%c",(i &  ? b[i][] : (Mod - b[i][]) % Mod), " \n" [i == n]);
     }
     return ;
 }

时间复杂度为 O(n^3)