1013. [JSOI2008] Spherical space generator sphere [Gaussian elimination]

Description

　　There is a spherical space generator capable of generating a hard sphere in n-dimensional space. Now, you are trapped in this n-dimensional sphere, you only know
the coordinates of n+1 points on the sphere, you need to determine the center coordinates of this n-dimensional sphere as quickly as possible in order to destroy this spherical space generator.

Input

　　The first line is an integer n (1<=N=10). The next n+1 lines, each with n real numbers, represent the n-dimensional coordinates of a point on the sphere. Each real number is accurate to
6 decimal places, and its absolute value does not exceed 20000.

Output

　　There is one and only one line, which gives the n-dimensional coordinates of the center of the sphere (n real numbers) in turn, and the two real numbers are separated by a space. Each real number is accurate to
3 decimal places. The data is guaranteed to be solvable. Your answer must be exactly the same as the standard output to be awarded a point.

Sample Input

2
0.0 0.0
-1.0 1.0
1.0 0.0

Sample Output

0.500 1.500

HINT

 

 

 1 #include<iostream>
 2 #include<cstring>
 3 #include<cstdio>
 4 #include<cmath>
 5 #define N  (101)
 6 using namespace std;
 7 
 8 double p[N][N],f[N][N],ans[N];
 9 int n;
10 
11 void Gauss()
12 {
13     for (int i=1; i<=n; ++i)
14     {
15         int num=i;
16         for (int j=i+1; j<=n; ++j) if (fabs(f[j][i])>fabs(f[num][i])) num=j;
17         if (num!=i) for (int j=1; j<=n+1; ++j) swap(f[i][j],f[num][j]);
18         
19         for (int j=i+1; j<=n+1; ++j)
20         {
21             double t=f[j][i]/f[i][i];
22             for (int k=i; k<=n+1; ++k)
23                 f[j][k]-=t*f[i][k];
24         }
25     }
26     for (int i=n; i>=1; --i)
27     {
28         for (int j=i+1; j<=n; ++j) f[i][n+1]-=f[i][j]*ans[j];
29         ans[i]=f[i][n+1]/f[i][i];
30     }
31 }
32 
33 int main()
34 {
35     scanf("%d",&n);
36     for (int i=1; i<=n+1; ++i)
37         for (int j=1; j<=n; ++j)
38             scanf("%lf",&p[i][j]);
39     for (int i=1; i<=n; ++i)
40         for (int j=1; j<=n; ++j)
41         {
42             f[i][n+1]+=p[i][j]*p[i][j]-p[i+1][j]*p[i+1][j];
43             f[i][j]=2*(p[i][j]-p[i+1][j]);
44         }
45     Gauss();
46     for (int i=1; i<=n-1; ++i)
47         printf("%.3lf ",ans[i]);
48     printf("%.3lf",ans[n]);
49 }