如题，Subtask #0 全过，Subtask #1 只过了 #15，其余 $WA$。用高斯-约旦消元法写的。 代码

#include<bits/stdc++.h>
using namespace std;
const int N=1e3+10;
int n,i,j,k,p;
double x,a[N][N];
int main(){
    cin>>n;
    for(i=1;i<=n;i++)for(j=1;j<=n+1;j++)cin>>a[i][j];
    for(i=1;i<=n;i++){
        p=i;
        for(j=i;j<=n;j++)if(a[j][i]>a[p][i])p=i;
        if(!a[p][i])return cout<<"No Solution",0;
        for(j=1;j<=n+1;j++)swap(a[i][j],a[p][j]);
        x=a[i][i];
        for(j=1;j<=n+1;j++)a[i][j]=a[i][j]/x;
        for(j=1;j<=n;j++)
            if(i!=j){
                x=a[j][i];
                for(k=1;k<=n+1;k++)a[j][k]=a[j][k]-x*a[i][k];
            }
    }
    for(i=1;i<=n;i++)printf("%.2lf\n",a[i][n+1]);
}