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=,r=1e9,m1,m2;
    for(int i=;i<;i++)
     {
        m1=l+(r-l)/;
        m2=r-(r-l)/;
        if(f(m1)<f(m2))
             r=m2;
        else
         l=m1;
     }
     printf("%f\n",f(l));
     return ;
 }