SPOJ DETER3 (矩阵行列式)

题意:求矩阵的行列式模p.

这样的情况下就不能用高精度消元搞了,可以在消去的时候用辗转相除避免精度误

差.然后根据行列式的性质,矩阵的元素可以直接模p.

板子题,最终的结果要是正数.

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <vector>
#include <queue>
using namespace std;
#define maxn 333

#define eps 1e-10
long long a[maxn][maxn];
long long n, mod;
long long det ()        //按列化为下三角
{
     long long res = 1;
     for(int i = 0; i < n; ++i)
     {
         if (!a[i][i])
         {
             bool flag = false;
             for (int j = i + 1; j < n; ++j)
             {
                 if (a[j][i])
                 {
                     flag = true;
                     for (int k = i; k < n; ++k)
                     {
                         swap (a[i][k], a[j][k]);
                     }
                     res = -res;
                     break;
                 }
             }

             if (!flag)
             {
                 return 0;
             }
         }

         for (int j = i + 1; j < n; ++j)
         {
             while (a[j][i])
             {
                 long long t = a[i][i] / a[j][i];
                 for (int k = i; k < n; ++k)
                 {
                     a[i][k] = (a[i][k] - t * a[j][k]) % mod;
                     swap (a[i][k], a[j][k]);
                 }
                 res = -res;
             }
          }
          res *= a[i][i];
          res %= mod;
      }
      return (res+mod)%mod;
}

int main()
{
    //freopen("in.txt", "r", stdin);
    while (cin >> n >> mod) {
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++)
                cin >> a[i][j], a[i][j] %= mod;
        }
        cout << det () << "\n";
    }
    return 0;
}