Bzoj 1013 [JSOI2008] spherical space Generator Sphere | Gaussian elimination element

Topic:

http://www.lydsy.com/JudgeOnline/problem.php?id=1013

Exercises

Consider the two-dimensional we can understand a truth:

Two points to the left can represent an equation, and then subtract from the 22 equation to get a one-time equation

So we can do it with Gaussian elimination.

#include <cstdio>#include<algorithm>#include<cstring>#include<cmath>#defineN 13using namespacestd;intn,m;DoubleC[n][n],f[n][n],ans[n];inlinevoidGauss () { for(intI=1; i<=n;i++)    {    intL=i;  for(intj=l+1; j<=n;j++)        if(Fabs (F[l][i]) < Fabs (F[j][i]) l=J; if(l!=i) for(intj=i;j<=m;j++) Swap (f[l][j],f[i][j]);  for(intj=i+1; j<=n;j++)    {        Doubletemp=f[j][i]/F[i][i];  for(intk=i;k<=m;k++) F[j][k]=f[j][k]-f[i][k]*temp; }    }     for(inti=n;i>=1; i--)    {    Doublet=F[i][m];  for(intj=n;j>i;j--) T-=ans[j]*F[i][j]; Ans[i]=t/F[i][i]; }}intMain () {scanf ("%d", &n); m=n+1;  for(intI=0; i<=n;i++)     for(intj=1; j<=n;j++) scanf ("%LF",&C[i][j]);  for(intI=1; i<=n;i++)    {    intj=i-1;DoubleD=0;  for(intk=1; k<=n;k++) {F[i][k]= (C[i][k]-c[j][k]) *2; D+=c[i][k]*c[i][k]-c[j][k]*C[j][k]; } F[i][m]=D;    } Gauss ();  for(intI=1; i<=n;i++)    if(i<n) printf ("%.3LF", Ans[i]); Elseprintf"%.3lf\n", Ans[i]); return 0;}

Bzoj 1013 [JSOI2008] spherical space Generator Sphere | Gaussian elimination element