Application of Gauss elimination

Gauss elimination is used to solve System of linear equations Of , That is, it can solve the problem that can be converted into a system of linear equations ：

n Solve the problem of dimension Center

   We know that n On the Uygur ball n+1 A little bit , It can be solved n Coordinates of the center of the sphere .
hypothesis , Known on the circle （2 Dimension ball ） Three points ：
$(a_1,b_1),(a_2,b_2),(a_3,b_3)$
Then we can get the equations ：
$$\{\begin{array}{l}(x-a_1{)}^2+(y-b_1{)}^2=r^2\\ (x-a_2{)}^2+(y-b_2{)}^2=r^2\\ (x-a_3{)}^2+(y-b_3{)}^2=r^2\end{array}$$
Do a subtraction conversion ：
$$\{\begin{array}{l}2(a_1-a_2)x+2(b_1-b_2)y=(a_1^2-a_2^2)+(b_1^2-b_2^2)\\ 2(a_2-a_3)x+2(b_2-b_3)y=(a_2^2-a_3^2)+(b_2^2-b_3^2)\end{array}$$ This is the well-known system of linear equations , And then gauss That's it .
Expand to n dimension ：
$$\{\begin{array}{l}2(a_1-a_2)x+2(b_1-b_2)y+\dots +2(z_1-z_2)p=(a_1^2-a_2^2)+(b_1^2-b_2^2)+\dots +(z_1^2-z_2^2)\\ 2(a_2-a_3)x+2(b_2-b_3)y+\dots +2(z_2-z_3)p=(a_2^2-a_3^2)+(b_2^2-b_3^2)+\dots +(z_2^2-z_3^2)\\ \dots \dots \\ 2(a_{n-1}-a_n)x+2(b_{n-1}-b_n)y+\dots +2(z_{n-1}-z_n)p=(a_{n-1}^2-a_n^2)+\dots +(z_{n-1}^2-z_n^2)\end{array}$$
Example ： Spherical space generator
There's a spherical space generator that can be used in n A solid sphere is created in dimensional space .

Now? , You're trapped in this n In the sphere , You only know that on the sphere n+1 Coordinates of points , You need to determine this as quickly as possible n The coordinates of the center of the sphere , In order to destroy this spherical space generator .

Input format
The first line is an integer n.

Next n+1 That's ok , Each row has n A real number , Representing a point on a sphere n Dimensional coordinates .

Every real number is accurate to the decimal point 6 position , And its absolute value does not exceed 20000.

Output format
There is only one line , Give the center of the ball in turn n Dimensional coordinates （n A real number ）, Two real numbers are separated by a space .

Every real number is accurate to the decimal point 3 position .

The data is guaranteed to have a solution .

Data range
1≤n≤10
sample input ：
2
0.0 0.0
-1.0 1.0
1.0 0.0
sample output ：
0.500 1.500

#include<cstdio>
#include<iostream>
#include<string>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<vector>
#include<map>
#include<queue>
#include<utility>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
const int maxn = 110;
const int inf = 0x3f3f3f3f3f;
const double eps = 1e-7;
int n;
double a[maxn][maxn];
double b[maxn][maxn];
void gauss(){
    
    double del;
    for(int i=1;i<=n;i++){
    
        int k=i;
        for(int j=i+1;j<=n;j++){
    
            if(fabs(a[j][i])>fabs(a[k][i]))
                k=j;
        }
        if(fabs(del=a[k][i])<eps)   continue;
        if(i!=k){
    
            for(int j=i;j<=n+1;j++)
                swap(a[i][j],a[k][j]);
        }
        for(int j=i;j<=n+1;j++) a[i][j]/=del;
        for(k=1;k<=n;k++){
    
            if(k!=i){
    
                del=a[k][i];
                for(int j=i;j<=n+1;j++)
                    a[k][j] -= a[i][j]*del;
            }
        }
    }
    for(int i=1;i<=n;i++){
    
        printf("%.3f ",a[i][n+1]/a[i][i]);
    }
}
int main(void)
{
    
    scanf("%d",&n);
    for(int i=1;i<=n+1;i++){
    
        for(int j=1;j<=n;j++){
    
            scanf("%lf",&b[i][j]);
        }
    }
    for(int i=1;i<=n;i++){
    
        for(int j=1;j<=n;j++){
    
            a[i][j] = 2.0*(b[i][j]-b[i+1][j]);
            a[i][n+1] += (b[i][j]*b[i][j]-b[i+1][j]*b[i+1][j]);
        }
    }
    gauss();
    return 0;
}