BZOJ1013 [JSOI2008]球形空间产生器sphere[高消]

数论进度开的好慢啊。我整天做的都是什么鬼题啊。

简单的高消题，用一个式子把另外$n$个有二次项和距离的式子全消掉就行了。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#define dbg(x) cerr << #x << " = " << x <<endl
using namespace std;
typedef long long ll;
typedef long double db;
typedef pair<int,int> pii;
template<typename T>inline T _min(T A,T B){return A<B?A:B;}
template<typename T>inline T _max(T A,T B){return A>B?A:B;}
template<typename T>inline char MIN(T&A,T B){return A>B?(A=B,1):0;}
template<typename T>inline char MAX(T&A,T B){return A<B?(A=B,1):0;}
template<typename T>inline void _swap(T&A,T&B){A^=B^=A^=B;}
template<typename T>inline T read(T&x){
    x=0;int f=0;char c;while(!isdigit(c=getchar()))if(c=='-')f=1;
    while(isdigit(c))x=x*10+(c&15),c=getchar();return f?x=-x:x;
}
const int N=10+4;
db A[N][N],B[N][N];
int n,maxl;

int main(){//freopen("test.in","r",stdin);//freopen("test.out","w",stdout);
    read(n);
    for(register int i=1;i<=n+1;++i)for(register int j=1;j<=n;++j)scanf("%Lf",&B[i][j]);
    for(register int i=1;i<=n;++i){
        for(register int j=1;j<=n;++j)A[i][j]=2*(B[n+1][j]-B[i][j]);
        for(register int j=1;j<=n;++j)A[i][n+1]+=B[n+1][j]*B[n+1][j]-B[i][j]*B[i][j];
    }
    for(register int i=1;i<=n;++i){
        maxl=i;
        for(register int j=i+1;j<=n;++j)if(fabs(A[j][i])>fabs(A[maxl][i]))maxl=j;
        if(maxl^i)swap(A[maxl],A[i]);
        for(register int l=1;l<=n;++l)if(l^i)
            for(register int j=n+1;j>=i;--j)
                A[l][j]-=A[l][i]/A[i][i]*A[i][j];
    }
    for(register int i=1;i<=n;++i)printf("%.3Lf\n",A[i][n+1]/A[i][i]);
    return 0;
}