SPOJ 2832 DETER3 - Find The Determinant III(矩阵行列式)

DETER3 - Find The Determinant III

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Given a NxN matrix A, find the Determinant of A % P.

Input

Multiple test cases (the size of input file is about 3MB, all numbers in each matrix are generated randomly).

The first line of every test case contains two integers , representing N (0 < N < 201) and P (0 < P < 1,000,000,001). The following N lines each contain N integers, the j-th number in i-th line represents A[i][j] (- 1,000,000,001 < A[i][j] < 1,000,000,001).

Output

For each test case, print a single line contains the answer.

Example
Input:
1 10
-528261590
2 2
595698392 -398355861
603279964 -232703411
3 4
-840419217 -895520213 -303215897
537496093 181887787 -957451145
-305184545 584351123 -257712188

Output:
0
0
2

求出矩阵的行列式，模p 伪代码：

 #include <map>
 #include <set>
 #include <queue>
 #include <stack>
 #include <vector>
 #include <cmath>
 #include <cstdio>
 #include <cstdlib>
 #include <cstring>
 #include <iostream>
 #include <algorithm>
 using namespace std;
 const double pi = acos(-1);
 const int inf = 0x3f3f3f3f;
 const double eps = 1e-15;
 typedef long long LL;
 typedef pair <int, int> PLL;
 const int N = 300;

 LL mat[N][N];
 LL Det (int n, int mod)        //按列化为下三角
 {
     for (int i = 0; i < n; ++i)
     {
         for (int j = 0; j < n; ++j)
         {
             mat[i][j] %= mod;
         }
     }

     LL res = 1;
     for(int i = 0; i < n; ++i)
     {
         if (!mat[i][i])
         {
             bool flag = false;
             for (int j = i + 1; j < n; ++j)
             {
                 if (mat[j][i])
                 {
                     flag = true;
                     for (int k = i; k < n; ++k)
                     {
                         swap (mat[i][k], mat[j][k]);
                     }
                     res = -res;
                     break;
                 }
             }

             if (!flag)
             {
                 return 0;
             }
         }

         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] - t * mat[j][k]) % mod;
                     swap (mat[i][k], mat[j][k]);
                 }
                 res = -res;
             }
          }
          res = (res * mat[i][i]) % mod;
      }
      return (res + mod) % mod;
}
int main()
{
    freopen("in.txt","r",stdin);
    int n;
    LL p;
    while (~scanf("%d%I64d",&n, &p))
    {
        for (int i = 0; i < n; ++i)
        {
            for (int j = 0; j < n; ++j)
            {
                scanf("%I64d",&mat[i][j]);
            }
        }
        LL ans = Det (n, p);
        printf("%I64d\n",ans);
    }
    return 0;
}