SPOJ

传送门
按理说，在求行列式时，化简成上三角新式，需要除以数字，如果要除以mod，那么肯定会产生逆元
而对于逆元的处理，这里肯定需要保证模是质数。

但对于模是任意数时

不是特别懂，好像是类似gcd的辗转相除法求的

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <climits>
#include <cmath>
#include <ctime>
#include <cassert>
#define IOS ios_base::sync_with_stdio(0); cin.tie(0);
using namespace std;
typedef long long ll;
const int MAX_N = 210;
ll mod;
ll C[MAX_N][MAX_N];
ll det(ll mat[][MAX_N], int n){
    for(int i = 1; i <= n; ++i) {
        for(int j = 1; j <= n; ++j) {
            mat[i][j] = (mat[i][j] % mod + mod) % mod;
        }
    }
    ll res = 1;
    int cnt = 0;
    for(int i = 1; i <= n; ++i) {
        for(int j = i + 1; j <= n; ++j) {
            while(mat[j][i]) {
                ll t = mat[i][i] / mat[j][i];
                for(int k = i; k <= n; ++k) {
                    mat[i][k] = (mat[i][k] - mat[j][k] * t % mod) % mod;
                    // cout << mat[i][k] << endl;
                    swap(mat[i][k], mat[j][k]);
                }
                cnt++;
            }
        }
        if(mat[i][i] == 0) return 0;
        res = res * mat[i][i] % mod;
    }
    if(cnt & 1) res = -res;
    return  (res + mod) % mod;
}
int main(){
    int n;
    while(~scanf("%d%lld", &n, &mod)){
        for(int i = 1; i <= n; ++i) {
            for(int j = 1; j <= n; ++j) {
                scanf("%lld", &C[i][j]);
            }
        }
        printf("%lld
", det(C, n));
    }
    return 0;
}