Turn the corner

Problem Description

Mr. West bought a new car! So he is travelling around the city.<br><br>One day he comes to a vertical corner. The street he is currently in has a width x, the street he wants to turn to has a width y. The car has a length l and a width d.<br><br>Can Mr. West go across the corner?<br><img src=../../../data/images/2438-1.jpg><br>

 

Input

Every line has four real numbers, x, y, l and w.<br>Proceed to the end of file.<br>

 

Output

If he can go across the corner, print "yes". Print "no" otherwise.<br>

 

Sample Input

   
   
   
   

    
    
    
    10 6 13.5 4<br>10 6 14.5 4<br>
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    yes<br>no<br>
   
   
   
   

 

Source

2008 Asia Harbin Regional Contest Online

 简单题意：

Mr.west 买了一辆新车，他在城市中旅行。一天，他驾车来到了一个垂直的角落。他想通过此弯道，现在给出车子的长和宽，以及直角弯道的宽度和他现在没转弯之前的街道宽度。现在写出一个程序，求出他能不能通过该弯道。

解题思路形成过程：

  容易得到一个单调三角函数，所以这也是一个简单的二分法。

感想：

   数学函数的构建是难点。

AC 代码：

#include <iostream>
#include <cstring>
#include <cstdio>
#include <cmath>
#define pi 3.1415
using namespace std;

const double eps = 1e-4;

double l,x,y,w;

double calu(double a){
    return l*cos(a)+(w-x*cos(a))/sin(a);
}

double ternary_search(double l,double r){
    double ll,rr;
    while(r-l>eps){
        ll=(2*l+r)/3;
        rr=(2*r+l)/3;
        if(calu(ll)>calu(rr))
            r=rr;
        else
            l=ll;
    }
    return r;
}

int main()
{
    while(cin>>x>>y>>l>>w){
        double l=0,r=pi/2;
        double tmp=ternary_search(l,r);
        if(calu(tmp)<=y)
            puts("yes");
        else
            puts("no");
    }
    return 0;
}