bzoj 2107: Spoj2832 Find The Determinant III 辗转相除法

2022-09-07 12:35:41

2107: Spoj2832 Find The Determinant III

Time Limit: 1 Sec Memory Limit: 259 MB
Submit: 154 Solved: 46
[Submit][Status][Discuss]

Description

Problem code: DETER3

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

给出一个尺寸为N×N的整数方阵A（N≤200），要求求出|A|%P的值（即A的行列式的值除以P的余数）。方阵中的数与P均为32位有符号类型可容纳的整数

Input

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.

Sample Input

3 4
-840419217 -895520213 -303215897
537496093 181887787 -957451145
-305184545 584351123 -257712188

Sample Output

　　其实一说算法名称大概都会做了吧。高斯消元的除法本质上等效与辗转相处法，而辗转相处不存在精度误差。我们为了把两行之一消掉，通过辗转相除大行减小行变成类似子问题。

　　最开始尝试用java的BigDecimal，结果发现精度和时间是不可同时满足的。

　　矩阵行列式求法可几何理解，向量（基底）可以互相加减，而不影响体积，然而基底互换，有向体积取反。

#include<iostream>

#include<cstdio>

#include<cstring>

#include<cmath>

#include<algorithm>

using namespace std;

#define MAXN 210

typedef long long qword;

qword mat[MAXN][MAXN];

int main()

{

        //freopen("input.txt","r",stdin);

        int n,m,x,y,z;

        int mod;

        while (~scanf("%d%d",&n,&mod))

        {

                for (int i=;i<=n;i++)

                {

                        for (int j=;j<=n;j++)

                        {

                                scanf("%lld",&mat[i][j]);

                                mat[i][j]%=mod;

                        }

                }

                int rev=;

                for (int i=;i<=n;i++)

                {

                        x=-;

                        for (int j=i;j<=n;j++)

                        {

                                if (mat[j][i])

                                {

                                        x=j;

                                        break;

                                }

                        }

                        if (x==-)break;

                        if (x!=i)

                        {

                        for (int j=;j<=n;j++)

                                swap(mat[x][j],mat[i][j]);

                        rev=-rev;

                        }

                        if (!mat[i][i])break;

                        for (int j=i+;j<=n;j++)

                        {

                                while (mat[i][i])

                                {

                                        qword t=mat[j][i]/mat[i][i];

                                        for (int k=;k<=n;k++)

                                                mat[j][k]=(mat[j][k]-mat[i][k]*t)%mod;

                                        for (int k=;k<=n;k++)

                                                swap(mat[j][k],mat[i][k]);

                                        rev=-rev;

                                }

                                for (int k=;k<=n;k++)

                                        swap(mat[j][k],mat[i][k]);

                                rev=-rev;

                        }

                }

                qword ans=;

                for (int i=;i<=n;i++)

                        ans=ans*mat[i][i]%mod;

                ans=(ans*rev+mod)%mod;

                printf("%lld\n",ans);

        }

}