HDU2438 Turn the corner(三分)

Problem Description

Mr. West bought a new car! So he is travelling around the city.

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.

Can Mr. West go across the corner?

 

Input

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

 

Output

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

 

Sample Input

   
   
   
   

    
    
    
    10 6 13.5 4
10 6 14.5 4
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    yes
no
   
   
   
   
   
   
   
   

    
    
    
    
 
     要使汽车能转过此弯道，那么就是汽车的左边尽量贴着那个直角点，而汽车的右下后方的点尽量贴着最下面的边。 
    
    
    
    
    
 
     我们以O点为原点建立直角坐标系，我们可以根据角a给出P点横坐标的函数F(a) 
    
    
    
    
    
 
       
    
    
    
    
    
 
     那么很容易得到： 
    
    
    
    
    
 
      
    
    
    
    
    

        
    
    
    
    
    
 
     其中有条件：，可以很容易证明是一个单峰函数，所以接下来就是三分了，如果的最大值小于等于 
    
    
    
    
    
 
       
    
    
    
    
    
 
     y，那么就能通过此直角弯道，否则就通不过。 
    
代码如下：
   
   
   
   
   
   
   
   

    
    
    
    
#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;
}