刚刚看了高斯消元，感觉主观上可以理解。= =

于是就来码了码这题，高斯消元是$O(n^3)$的，毫无压力~

于是这题需要进行一些转化

设球心$(x_1,x_2,x_3,...,x_n)$，题目给出了$n+1$个球面上的点的坐标，我们设为

$a_{1_{1}},a_{1_{2}},a_{1_{3}},...,a_{1_{n}}$

$a_{2_{1}},a_{2_{2}},a_{2_{3}},...,a_{2_{n}}$

$...$

$a_{n+1_{1}},a_{n+1_{2}},a_{n+1_{3}},...,a_{n+1_{n}}$

于是列出$n+1$个方程，设半径（雾？）为dist

$\sqrt{(a_{i_{1}}-x_1)^2+(a_{i_{2}}-x_2)^2+...+(a_{i_{n}}-x_n)^2}=dist,(1 \leq i \leq n+1)$

然后$i$与$i+1$组合一下，相减就变成高斯消元的形式了。

代码~

# include <stdio.h>
# include <algorithm>
# include <math.h>
using namespace std;
int n;
double p[13],pp[13],a[13][13],b[13];
inline void gauss() {
	for (int i=1;i<=n;++i) {
		int k=0;
		for (int j=i;j<=n;++j)
			if(fabs(a[j][i])>fabs(a[k][i])) k=j;
		for (int j=1;j<=n+1;++j) swap(a[i][j],a[k][j]);
		for (int j=i+1;j<=n;++j) {
			double temp=-a[j][i]/a[i][i];
			for (int k=i;k<=n+1;++k) a[j][k]+=a[i][k]*temp; 
		}
	}
	for (int i=n;i>=1;--i) {
		for (int j=n;j>i;--j) a[i][n+1]-=a[i][j]*b[j];
		b[i]=a[i][n+1]/a[i][i];
	}
}
int main() {
	scanf("%d",&n);
	for (int i=1;i<=n;++i) scanf("%lf",&p[i]);
	for (int i=1;i<=n;++i) {
		for (int j=1;j<=n;++j) {
			scanf("%lf",&pp[j]);
			a[i][j]=p[j]-pp[j];
			a[i][n+1]+=p[j]*p[j]-pp[j]*pp[j];
		}
		a[i][n+1]/=2;
	}
	gauss();
	for (int i=1;i<=n-1;++i) printf("%.3lf ",b[i]);
	printf("%.3lf\n",b[n]);
}