Article From:https://www.cnblogs.com/cn-suqingnian/p/9248778.html

Look at the school sister’s code $… A similar$qwq$that feels that his own code wind is metaphysical. The center coordinates are$O (x_{1}, x_{2},… X_{3}) $Then, according to the definition of$n$, the distance between the points on the ball and the center is equal.$n$The formula of distance between two points in a dimensional space $$\sqrt{(a_{1}-a_{2})^{2}+(b_{1}-b_{2})^{2}+……}$$ So, according to$|OX_{1}|=|OX_{2}|$,$|OX_{1}|=|OX_{3}|$,…$|OX_{1}|=|OX_{n+1}|$A total of$n$equations. List the$i$equation in it: Set up$X_{i} \ (X_{i1}, X_{i2},… X_{in}) $$$\sqrt{(x_{11}-x_{1})^{2}+(x_{12}-x_{2})^{2}+……+(x_{1n}-x_{n})^{2}}=\sqrt{(x_{i1}-x_{1})^{2}+(x_{i2}-x_{2})^{2}+……+(x_{in}-x_{n})^{2}}$$ The square is square at the same time, then the square is taken apart, and then the$x_{j}^{2}$on the left and the$x_{j}^{2}$on the right are eliminated. Then it will be found that this becomes an equation once. The I equation is: $$-2*(x_{i1}-x_{11})x_{1}-2*(x_{i2}-x_{12})x_{2}-……-2*(x_{in}-x_{1n})x_{n}=x_{11}^{2}-x_{i1}^{2}+x_{12}^{2}-x_{i2}^{2}+……+x_{1n}^{2}-x_{in}^{2}$$ So,$Gauss\_ \ elimination\$.

Just fine.

The code is enlisted:

#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
int n;
const int N=20+10;
double a[N][N];
double f[N];
double p;

void Gauss_jordan()
{
for(int i=1;i<=n;i++)
{
int now=i;
for(int j=i+1;j<=n;j++)
if(a[j][i]>a[now][i])
now=j;
for(int j=i;j<=n+1;j++)
swap(a[now][j],a[i][j]);
for(int j=i+1;j<=n+1;j++)
a[i][j]/=a[i][i];
a[i][i]=1;
for(int j=i+1;j<=n;j++)
{
for(int k=i+1;k<=n+1;k++)
a[j][k]-=a[j][i]*a[i][k];
a[j][i]=0;
}
}
for(int i=n;i>=1;i--)
for(int j=i+1;j<=n;j++)
{
a[i][n+1]-=a[i][j]*a[j][n+1];
a[i][j]=0;
}
}
int main()
{
scanf("%d",&n);
for(int i=1;i<=n;i++)
scanf("%lf",&f[i]);
for(int i=1;i<=n;i++)
for(int j=1;j<=n;j++)
{
scanf("%lf",&p);
a[i][j]=2.0*(p-f[j]);
a[i][n+1]=a[i][n+1]+p*p-f[j]*f[j];
}
Gauss_jordan();
for(int i=1;i<=n;i++)
printf("%.3lf ",a[i][n+1]);
return 0;
}

View Code