Ellipsoid

Problem Description

Given a 3-dimension ellipsoid(椭球面)

your task is to find the minimal distance between the original point (0,0,0) and points on the ellipsoid. The distance between two points (x

₁,y

₁,z

₁) and (x

₂,y

₂,z

₂) is defined as

Input

There are multiple test cases. Please process till EOF.

For each testcase, one line contains 6 real number a,b,c(0 < a,b,c,< 1),d,e,f

(0 ≤ d,e,f < 1), as described above.

It is guaranteed that the input data forms a ellipsoid. All numbers are fit in double.

Output

For each test contains one line. Describes the minimal distance. Answer will be considered as correct if their absolute error is less than 10

^-5.

Sample Input

      1 0.04 0.01 0 0 0 

Sample Output

      1.0000000 

Source

Recommend

hujie

题目大意：

求一个椭球面上的一个点到原点的最短距离。

解题思路：

模拟退火，不多解释了。

解题代码：

#include 
    
     #include 
     
      #include 
      
       using namespace std;const double eps=1e-8;const double INF=1e100;const int offx[]={1,0,-1,0,1,-1,-1,1};const int offy[]={0,1,0,-1,1,1,-1,-1};double a,b,c,d,e,f;double getAns(double x,double y){    double A=c,B=d*y+e*x,C=a*x*x+b*y*y+f*x*y-1.0;    double delta=B*B-4*A*C;    if(delta<0) return INF+10;    delta=sqrt(delta);    double z1=(-B+delta)/(2*A),z2=(-B-delta)/(2*A);    return min( sqrt(x*x+y*y+z1*z1) , sqrt(x*x+y*y+z2*z2)   );}double tosolve(double sx,double sy){    double x=sx,y=sy,ans=getAns(sx,sy),step=1e6;    while(step>eps){        double sx=x,sy=y;        bool flag=false;        for(int i=0;i<8;i++){            double dx=x+offx[i]*step,dy=y+offy[i]*step;            double tmp=getAns(dx,dy);            if(tmp>=INF) continue;            if(tmp