如果单按照距离相等的话既是高次也没有半径，所以因为给了 \(n+1\) 组点就想到两两做差。
假如一组点是 \(\{a_i\}\) 一组是 \(\{b_i\}\)，我们能轻易地得出
\[\sum_{i=1}^n(x_i-a_i)^2=\sum_{i=1}^n(x_i-b_i)^2 \Rightarrow \sum_{i=1}^n 2(b_i-a_i)x_i=\sum_{i=1}^n(b_i^2-a_i^2)\]
这是一个 \(n\) 元一次线性方程。我们能搞出 \(n\) 组这样的方程，高斯消元解之。
#include <iostream>#include <cstdio>#include <cmath>using namespace std;int n;double tmp1[15], tmp2[15], a[15][15];void gauss(){ ???for(int i=1; i<=n; i++){ ???????int maxi=i; ???????for(int j=i+1; j<=n; j++) ???????????if(fabs(a[j][i])>fabs(a[i][i])) ???????????????maxi = j; ???????double now=a[maxi][i]; ???????swap(a[maxi], a[i]); ???????for(int j=i; j<=n+1; j++) ???????????a[i][j] /= now; ???????for(int j=i+1; j<=n; j++){ ???????????now = a[j][i]; ???????????for(int k=i; k<=n+1; k++) ???????????????a[j][k] -= a[i][k] * now; ???????} ???} ???for(int i=n; i>=1; i--) ???????for(int j=1; j<=i-1; j++){ ???????????a[j][n+1] -= a[j][i] * a[i][n+1]; ???????????a[j][i] = 0; ???????}}int main(){ ???cin>>n; ???for(int i=1; i<=n; i++) scanf("%lf", &tmp1[i]); ???for(int i=1; i<=n; i++){ ???????for(int j=1; j<=n; j++) ???????????scanf("%lf", &tmp2[j]); ???????for(int j=1; j<=n; j++){ ???????????a[i][j] = 2 * (tmp2[j] - tmp1[j]); ???????????a[i][n+1] += tmp2[j] * tmp2[j] - tmp1[j] * tmp1[j]; ???????} ???????swap(tmp1, tmp2); ???} ???gauss(); ???for(int i=1; i<=n; i++) ???????printf("%.3f ", a[i][n+1]); ???printf("\n"); ???return 0;}