Sol
设一个dis，就有n+1个方程，消掉dis，就只有n个方程，组成一个方程组，高斯消元就好(建议建立方程时推一下，很简单)
# include <bits/stdc++.h># define RG register# define IL inline# define Fill(a, b) memset(a, b, sizeof(a))using namespace std;typedef long long ll;int n;double a[20][20], b[20][20], ans[20], sqr[20];int main(RG int argc, RG char *argv[]){ ???scanf("%d", &n); ???for(RG int i = 0; i <= n; ++i) ???????for(RG int j = 1; j <= n; ++j) ???????????scanf("%lf", &b[i][j]), sqr[i] += b[i][j] * b[i][j]; ???for(RG int i = 1; i <= n; ++i){ ???????for(RG int j = 1; j <= n; ++j) ???????????a[i][j] = 2 * (b[i][j] - b[0][j]); ???????a[i][n + 1] = sqr[i] - sqr[0]; ???} ???for(RG int i = 1; i < n; ++i) ???????for(RG int j = i + 1; j <= n; ++j){ ???????????RG double div = a[j][i] / a[i][i]; ???????????for(RG int k = 1; k <= n + 1; ++k) a[j][k] -= a[i][k] * div; ???????} ???for(RG int i = n; i; --i){ ???????ans[i] = a[i][n + 1] / a[i][i]; ???????for(RG int j = 1; j < i; ++j) a[j][n + 1] -= ans[i] * a[j][i]; ???} ???for(RG int i = 1; i <= n; ++i){ ???????printf("%.3lf", ans[i]); ???????if(i != n) putchar(' '); ???} ???return 0;}