2017CCPC湘潭A题Determinant

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

22 031 2 06 3 1

Sample Output

0 22 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;}