[JSOI2008 sphere]

[关键字]：高斯消元

[题目大意]：给出一个n维空间中球面上n+1个点，求出球心坐标。

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[分析]：看他给出的条件n+1个点，要求出n个值，感觉像解方程租。因为，每个点到球心的坐标相等。设圆心坐标为(x1,x2,x2......xn)，每个点到圆心的距离为di，然后就可以列出n个式子d1=d2\d2=d3......dn=dn+1，然后化简一下利用高斯消元解出就行了。

[代码]：

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;

const int MAXN=100;
const double ZERO=1e-8;

int n;
double f[MAXN],a[MAXN][MAXN],b[MAXN][MAXN];

int main()
{
    freopen("in.txt","r",stdin);
    freopen("out.txt","w",stdout);
    scanf("%d",&n);
    for (int i=1;i<=n+1;++i)
        for (int j=1;j<=n;++j)
            scanf("%lf",&a[i][j]);
    /*for (int i=1;i<=n+1;++i)
    {
        for (int j=1;j<=n;++j)
            printf("%.3lf ",a[i][j]);
        printf("\n");
    }*/
    for (int i=1;i<=n+1;++i)
    {
        double temp=0;
        for (int j=1;j<=n;++j)
        {
            temp+=a[i][j]*a[i][j];
            a[i][j]=2*a[i][j];
        }
        a[i][n+1]=temp;
    }
    for (int i=1;i<=n;++i)
        for (int j=1;j<=n+1;++j)
            b[i][j]=a[i][j]-a[i+1][j];
    /*for (int i=1;i<=n+1;++i)
    {
        for (int j=1;j<=n+1;++j)
            printf("%.3lf ",a[i][j]);
        printf("\n");
    }*/
    for (int i=1;i<=n;++i)
    {
        int j=i;
        while (fabs(b[j][i])<ZERO) ++j;
        for (int x=1;x<=n+1;++x)
            swap(b[i][x],b[j][x]);
        for (int j=i+1;j<=n;++j)
            if (fabs(b[j][i])>ZERO)
            {
                                   double temp=b[j][i]/b[i][i];
                                   for (int k=i;k<=n+1;++k)
                                       b[j][k]-=temp*b[i][k];
            }
    }
    for (int i=n;i>=1;--i)
    {
        double temp=b[i][n+1];
        for (int j=i+1;j<=n;++j) temp-=f[j]*b[i][j];
        f[i]=temp/(b[i][i]);
    }
    for (int i=1;i<=n;++i)
        printf("%.3lf ",f[i]);
    fclose(stdin);
    fclose(stdout);
    return 0;
}