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,):;}

 template<typename T>inline char MAX(T&A,T B){return A<B?(A=B,):;}

 template<typename T>inline void _swap(T&A,T&B){A^=B^=A^=B;}

 template<typename T>inline T read(T&x){

     x=;int f=;char c;while(!isdigit(c=getchar()))if(c=='-')f=;

     while(isdigit(c))x=x*+(c&),c=getchar();return f?x=-x:x;

 }

 const int N=+;

 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=;i<=n+;++i)for(register int j=;j<=n;++j)scanf("%Lf",&B[i][j]);

     for(register int i=;i<=n;++i){

         for(register int j=;j<=n;++j)A[i][j]=*(B[n+][j]-B[i][j]);

         for(register int j=;j<=n;++j)A[i][n+]+=B[n+][j]*B[n+][j]-B[i][j]*B[i][j];

     }

     for(register int i=;i<=n;++i){

         maxl=i;

         for(register int j=i+;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=;l<=n;++l)if(l^i)

             for(register int j=n+;j>=i;--j)

                 A[l][j]-=A[l][i]/A[i][i]*A[i][j];

     }

     for(register int i=;i<=n;++i)printf("%.3Lf\n",A[i][n+]/A[i][i]);

     return ;

 }