2017CCPC湘潭A题Determinant-优快云博客

本文介绍了一种计算矩阵去除第j列后的行列式的高效方法，并提供了一个C++实现示例，利用高斯消元法求解伴随矩阵的第一行，即逆矩阵的第一列。

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传送门：http://202.197.224.59/OnlineJudge2/index.php/Contest/read_problem/cid/43/pid/1260

Determinant

Bobo learned the definition of determinant det(A) of matrix A in ICPCCamp. He also knew determinant can be computed in O(n3) using Gaussian Elimination.

Bobo has an (n−1)×n matrix B he would like to find det(Bj) modulo (109+7) for all j∈{1,2,…,n} where Bj is the matrix after removing the j -th column from B .

Input

The input contains zero or more test cases and is terminated by end-of-file. For each test case:

The first line contains an integer n . The i -th of following n lines contains n integers Bi,1,Bi,2,…,Bi,n .

2≤n≤200
0≤Bi,j<109+7
The sum of n does not exceed 2000 .

Output

For each case, output n integers which denote the result.

Sample Input

Sample Output

0 2
2 1 999999998

Note

For the second sample,

官方题解：随机n个数把矩阵补全成n×n的。

那么就是要算伴随矩阵的第一行，也就是逆矩阵的第一列，高斯消元即可。

然后以下是Q巨的写法：

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<ctime>
#include<iostream>
#include<algorithm>
using namespace std;
const int MAXN=205;
const int Mod=1000000007;
int a[MAXN][MAXN],b[MAXN][MAXN];
int get_rand(int x)//[0,x)
{
    int t=1;
    while((1<<t)<x)t++;
    int res=x;
    while(res>=x)
    {
        res=0;
        for(int i=0;i<t;i++)
            res|=(rand()%2)<<i;
    }
    return res;
}
int fp(int a,int k)
{
    int res=1;
    while(k)
    {
        if(k&1)res=1LL*res*a%Mod;
        a=1LL*a*a%Mod;
        k>>=1;
    }
    return res;
}
void solve(int n)
{
    for(int i=1;i<=n;i++)
        for(int j=1;j<=n;j++)
            b[i][j]=(i==j);
    int det=1;
    for(int i=1;i<=n;i++)
    {
        int t=i;
        for(int k=i;k<=n;k++)
            if(a[k][i])t=k;
        if(t!=i)det*=-1;
        for(int j=1;j<=n;j++)
        {
            swap(a[i][j],a[t][j]);
            swap(b[i][j],b[t][j]);
        }
        det=1LL*a[i][i]*det%Mod;
        int inv=fp(a[i][i],Mod-2);
        for(int j=1;j<=n;j++)
        {
            a[i][j]=1LL*inv*a[i][j]%Mod;
            b[i][j]=1LL*inv*b[i][j]%Mod;
        }
        for(int k=1;k<=n;k++)
        {
            if(k==i)continue;
            int tmp=a[k][i];
            for(int j=1;j<=n;j++)
            {
                a[k][j]=(a[k][j]-1LL*a[i][j]*tmp%Mod+Mod)%Mod;
                b[k][j]=(b[k][j]-1LL*b[i][j]*tmp%Mod+Mod)%Mod;
            }
        }
    }
    det=(det+Mod)%Mod;
    for(int i=1;i<=n;i++)
        for(int j=1;j<=n;j++)
            b[i][j]=1LL*det*b[i][j]%Mod;
}
int main()
{
    srand(time(NULL));
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        for(int j=1;j<=n;j++)
            a[1][j]=2;
        for(int i=2;i<=n;i++)
            for(int j=1;j<=n;j++)
                scanf("%d",&a[i][j]);
        solve(n);
        for(int i=1;i<=n;i++)
            printf("%d%c",(i&1 ? b[i][1] : (Mod-b[i][1])%Mod)," \n"[i==n]);
    }
    return 0;
}