问题描述
题解
设答案为((p_1,p_2,p_3,...,p_n))
因为是一个球体，令其半径为(r)，则有
[sum_{i=1}^{n}{(a_i-p_i)}^2={
m dis}^2
]
拆式子可得
[sum_{i=1}^{n}a_i^2-2 imessum_{i=1}^{n}{a_ip_i}=sum_{i=1}^{n}p_i^2-{
m dis}^2
]
于是可以构造出新的方程矩阵：
[f_{i,j}=2 imes (a_{i+1,j}-a_{i,j})
]
[f_{i,n+1}=sum_{j=1}^n a_{i+1,j}^2-a_{i,j}^2
]
(mathrm{Code})
#include<bits/stdc++.h>
using namespace std;
void read(int &x){
x=0;char ch=1;int fh;
while(ch!='-'&&(ch<'0'||ch>'9')) ch=getchar();
if(ch=='-') fh=-1,ch=getchar();
else fh=1;
while(ch>='0'&&ch<='9'){
x=(x<<1)+(x<<3)+ch-'0';
ch=getchar();
}
x*=fh;
}
#define maxn 107
int n;
double a[maxn][maxn],bf[maxn][maxn];
int pla;
int main(){
ios::sync_with_stdio(0);
cin>>n;
for(register int i=1;i<=n+1;i++){
for(register int j=1;j<=n;j++) cin>>bf[i][j];
}
for(register int i=1;i<=n+1;i++){
for(register int j=1;j<=n;j++){
a[i][j]=2*(bf[i+1][j]-bf[i][j]);
a[i][n+1]+=bf[i+1][j]*bf[i+1][j]-bf[i][j]*bf[i][j];
}
}
for(register int i=1;i<=n;i++){
pla=i;
while(pla<=n&&a[pla][i]==0) pla++;
if(pla==n+1){
puts("No Solution");return 0;
}
for(register int j=1;j<=n+1;j++) swap(a[i][j],a[pla][j]);
double tmp=a[i][i];
for(register int j=1;j<=n+1;j++) a[i][j]=a[i][j]/tmp;
for(register int j=1;j<=n;j++){
if(i==j) continue;
double rp=a[j][i];
for(register int k=1;k<=n+1;k++){
a[j][k]=a[j][k]-rp*a[i][k];
}
}
}
for(register int i=1;i<=n;i++){
cout<<fixed<<setprecision(3)<<a[i][n+1]<<" ";
}
return 0;
}