IX Samara Regional Intercollegiate Programming Contest F 三分

F. Two Points

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

There are two points (x₁, y₁) and (x₂, y₂) on the plane. They move with the velocities (v_x1, v_y1) and (v_x2, v_y2). Find the minimal distance between them ever in future.

Input

The first line contains four space-separated integers x₁, y₁, x₂, y₂ ( - 10⁴ ≤ x₁,  y₁,  x₂,  y₂ ≤ 10⁴) — the coordinates of the points.

The second line contains four space-separated integers v_x1, v_y1, v_x2, v_y2 ( - 10⁴ ≤ v_x1,  v_y1,  v_x2,  v_y2 ≤ 10⁴) — the velocities of the points.

Output

Output a real number d — the minimal distance between the points. Absolute or relative error of the answer should be less than 10^- 6.

Examples

Input

1 1 2 2
0 0 -1 0

Output

1.000000000000000

Input

1 1 2 2
0 0 1 0

Output

1.414213562373095

题意：给你两个点 以及这两个点的x，y两个方向的速度 问运动过程中两个点间的最小距离

题解：第一次三分处理 类比二分 二分是针对单调函数的数值确定
     三分不仅适用于单调函数，而且适用于凸函数
     运动的两点间的距离的图像是不确定的  三分处理..

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<queue>
#include<stack>
#include<cmath>
using namespace std;
double x1,x2,y1,y2;
double vx1,vx2,vy1,vy2;
double f(double t)
{
    double xx1,xx2,yy1,yy2;
    xx1=x1+t*vx1;
    xx2=x2+t*vx2;
    yy1=y1+t*vy1;
    yy2=y2+t*vy2;
    return sqrt((xx1-xx2)*(xx1-xx2)+(yy1-yy2)*(yy1-yy2));
}
int main()
{
    scanf("%lf %lf %lf %lf",&x1,&y1,&x2,&y2);
    scanf("%lf %lf %lf %lf",&vx1,&vy1,&vx2,&vy2);
    double l=0,r=1e9,m1,m2;
   for(int i=0;i<200;i++)
    {
       m1=l+(r-l)/3;
       m2=r-(r-l)/3;
       if(f(m1)<f(m2))
            r=m2;
       else
        l=m1;
    }
    printf("%f
",f(l));
    return 0;
}