SPOJ DETER3 (矩阵行列式)

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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-10long 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;}

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