题意大概是给一个椭球方程,求椭球面上到原点(0,0,0)的最小距离。
三分x然后三分y,用求根公式求出z返回距离..虽说是过来,但还是感觉有很多不科学的地方...比如三分区间(我感觉可以直接算出来啊,但直接算出区间结果不对..),还有eps的大小...感觉最后都是调参数调过去的......
/*=============================================================================
# Author:Erich
# FileName:
=============================================================================*/
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <cstring>
#include <string>
#include <vector>
#include <map>
#include <queue>
#include <stack>
#define lson id<<1,l,m
#define rson id<<1|1,m+1,r
using namespace std;
typedef long long ll;
const double inf=1e9;
const double PI=acos(-1.0);
const double eps=1e-12;
int n,m;
double a,b,c,d,e,f;
inline int dcmp(double x)
{
if (fabs(x)<eps) return 0;
if (x<0) return -1;
return 1;
}
double slove(double x,double y)
{
double z;
double aa,bb,cc;
aa=c;
bb=e*x+d*y;
cc=a*x*x+b*y*y+f*x*y-1;
double dlt=bb*bb-4*aa*cc;
if (dcmp(dlt)==-1) return 1e10+x*x+y*y;
else if (dcmp(dlt)==0)
{
z=-bb/(2.0*a);
return (x*x+y*y+z*z);
}
else
{
z=-bb+sqrt(dlt);
z/=(2.0*aa);
double mi=x*x+y*y+z*z;
z=-bb-sqrt(dlt);
z/=(2.0*aa);
mi=min(mi,x*x+y*y+z*z);
return mi;
}
}
double search(double x)
{
double L,R;
double tc=1-a*x*x;
L=-sqrt(tc);
R=sqrt(tc);
L=-inf; R=inf;
for (int i=0; i<200; i++)
{
double m1=L+(R-L)/3;
double m2=R-(R-L)/3;
if (slove(x,m1)>slove(x,m2)) L=m1; else R=m2;
}
return slove(x,(L+R)/2.0);
}
int main()
{
// freopen("in.txt","r",stdin);
// freopen("out.txt","w",stdout);
while(~scanf("%lf",&a))
{
scanf("%lf%lf%lf%lf%lf",&b,&c,&d,&e,&f);
// double L=-sqrt(1.0/a),R=sqrt(1.0/a);
double L=-inf,R=inf;
for (int i=0; i<200; i++)
{
double m1=L+(R-L)/3;
double m2=R-(R-L)/3;
// printf("%.10f %.10lf\n",L,R);
// printf("%.10f %.10lf\n",search(m1),search(m2));
// puts("----------");
if (search(m1)>search(m2)) L=m1; else R=m2;
}
printf("%.10lf\n",sqrt(search((L+R)/2.0)));
}
return 0;
}