Turn the corner（三分）

Turn the corner

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

思路：
题目是问车子能否拐过这个拐角。能的话就打印“yes”，不能就打印“no”。在通过分析后可以得到，随着汽车由x进入y，图中h的高度先增大后减小，而解几何图形后可得：
s = l * cos(θ) + w * sin(θ) - x ;
h =s * tan(θ) + w * cos(θ);
这个函数是一个先增后减的凸函数，所以用三分法。
刚开始的时候我的三分是写成这样的
m1 = (a+b) / 2;
m2 = (m1+b) / 2;
但是WA了，后来又想了想改成了：
m1 = (2a+b) / 3;
m2 = (a+2b) / 3;
见图：

完整代码：

#include<iostream>
#include<cmath>
using namespace std;
double x,y,l,w;
const double eps=1e-10;
double pi=acos(-1.0);

double f(double n)      //函数关系
{
    double s=l*cos(n)+w*sin(n)-x;
    double h=s*tan(n)+w*cos(n);
    return h;
}
void thrp() //三分
{
    double m1,m2;
    double a=0.0;
    double b=pi/2;
    while(b-a>=eps)
    {
        m1=(2*a+b)/3;
        m2=(a+2*b)/3;
        if(f(m2)-f(m1)>=eps)
        {
            a=m1;
        }
        else
        {
            b=m2;
        }
    }
    if(f(m1)<=y)
    {
        cout<<"yes"<<endl;
    }
    else
    {
        cout<<"no"<<endl;
    }
}
int main()
{
    while(cin>>x>>y>>l>>w)
    {
        thrp();
    }
    return 0;
}