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

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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;}

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