一个高斯消元的好题...感谢@JCarlson 带我找题

设目标点的坐标,套公式,得到n+1个方程

n+1个方程减一下,得到n个一元一次方程

高斯消元可解

#include<set>
#include<map>
#include<list>
#include<queue>
#include<stack>
#include<string>
#include<time.h>
#include<math.h>
#include<memory>
#include<vector>
#include<bitset>
#include<fstream>
#include<stdio.h>
#include<utility>
#include<sstream>
#include<string.h>
#include<iostream>
#include<stdlib.h>
#include<algorithm>
using namespace std;
double c[15][15];
double a[15][15];
double b[15];
int main()
{
    #ifdef absi2011
    freopen("input.txt","r",stdin);
    freopen("output.txt","w",stdout);
    #endif
    int n;
    scanf("%d",&n);
    int i;
    for (i=0;i<n;i++)
    {
        scanf("%lf",&c[n][i]);
    }
    for (i=0;i<n;i++)
    {
        int j;
        for (j=0;j<n;j++)
        {
            scanf("%lf",&c[i][j]);
            a[i][j]=2*c[i][j]-2*c[n][j];
            b[i]+=c[i][j]*c[i][j]-c[n][j]*c[n][j];
        }
    }
    for (i=0;i<n;i++)
    {
        int j;
        for (j=i;j<n;j++)
        {
            if (a[i][i]!=0) break;
        }
        int k;
        for (k=0;k<n;k++)
        {
            swap(a[i][k],a[j][k]);
        }
        for (j=i+1;j<n;j++)
        {
            for (k=i+1;k<n;k++)
            {
                a[j][k]-=a[i][k]*a[j][i]/a[i][i];
            }
            b[j]-=b[i]*a[j][i]/a[i][i];
            a[j][i]=0;
        }
    }
    for (i=n-1;i>=0;i--)
    {
        b[i]/=a[i][i];
        a[i][i]=1;
        int j;
        for (j=0;j<i;j++)
        {
            b[j]-=b[i]*a[j][i];
            a[j][i]=0;
        }
    }
    for (i=0;i<n-1;i++)
    {
        printf("%.3lf ",b[i]);
    }
    if (i==n-1) printf("%.3lf\n",b[i]);
    return 0;
}