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

本题是个纯数学问题问一个直角弯能不能怪过来

一开始宽x转弯后宽y然后给了车的长宽

看懂题后把感觉需要算一下后来算了算要想拐过来所需要最大的长度是(-x + l * sin(a) + w/ cos(a)) / tan(a)

{a为车与y平线的夹角}

然后题目就变成了如何求一下一个角度a使所需最大长度小于这个拐弯后的宽度然后就可以得到最优的解如何求a则用的是三分

法

代码

#include<iostream>
#include<math.h>
using namespace std;
const double mi = 1e-7;
 const double eps = 1e-9;
const double pi = 4.0 * atan(1.0);
double x,y,l,w;
double sum(double a)
{
    return (-x + l * sin(a) + w/ cos(a)) / tan(a) ;

}
int main()
{

    while(cin>>x>>y>>l>>w)
    {
        if(w>y||w>x)
            cout<<"no"<<endl;
        else
            {
                 double low=0,high = pi / 2,mid,mmid;
                 double te1,te2;
                while(high - low > mi)
                    {
                        mid = (low + high) / 2;
                        mmid = (low + mid) / 2;
                        te1 = sum(mid);
                        te2 =sum(mmid);
                        if(te1 > te2)
                        {
                        low = mmid;
                    }
                        else
                        {
                        high = mid;
                            }
                        }
                te1 = sum(low);
                if(te1 < y)
                {
           cout<<"yes"<<endl;
         }
         else{
                cout<<"no"<<endl;
         }
    }
            }

    return 0;
}