【高斯消元】BZOJ1013 [JSOI2008]球形空間産生器sphere

題面在這裡

高斯消元裸題，把球心的坐标設出來，再搞一搞把二次項消掉就好了

示例程式：

#include<cstdio>
#include<algorithm>
#define _abs(x) ((x)>0?(x):-(x))
using namespace std;

const int maxn=;
const double eps=;
int n;
double x[maxn][maxn],a[maxn][maxn],ans[maxn];
int fcmp(double x){
    if (_abs(x)<=eps) return ;
    return x>?:-;
}
void Gauss(int n){
    for (int i=;i<=n;i++){
        int where=-;
        for (int j=i;j<=n;j++)
         if (fcmp(a[j][i]-)!=) {where=j;break;}
        swap(a[where],a[i]);
        //if where<0 無解
        for (int j=i+;j<=n;j++){
            double w=a[j][i]/a[i][i];
            for (int k=i;k<=n+;k++) a[j][k]-=a[i][k]*w;
        }
    }
    for (int i=n;i>=;i--){
        ans[i]=a[i][n+]/a[i][i];
        for (int j=;j<i;j++) a[j][n+]-=ans[i]*a[j][i];
    }
}
int main(){
    scanf("%d",&n);
    for (int i=;i<=n+;i++)
     for (int j=;j<=n;j++) scanf("%lf",&x[i][j]);
    for (int i=;i<=n+;i++)
     for (int j=;j<=n;j++) a[i-][j]=*(x[][j]-x[i][j]),a[i-][n+]+=x[][j]*x[][j]-x[i][j]*x[i][j];
    Gauss(n);
    for (int i=;i<n;i++) printf("%.3lf ",ans[i]);printf("%.3lf",ans[n]);
    return ;
}