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

 思路：

三分題關鍵是要確定決定的結果的因變數f（x），和決定f(x)的自變數x。

在這個題中，老司機拐彎的時候，最佳拐彎路徑是右側兩個車角緊貼著街邊行駛（如下圖所示）。

這時候就要做一條輔助線PH，來時車道的左側延長線，交車左邊於P，右邊於Q，交有側車道沿於H。

我們會發現，當max（PH）>Y時車是不能拐過去的，會頂到拐角處。否則就可以過去。

這樣顯而易見，我們的f（x）就是PH的長度。而x我選用了角d，用三角函式分別算出PQ和QH的長度他們的和就是所求PH長。

f（d）=(x-l*sin(d))*tan(d)+(w-(x-l*sin(d))/cos(d))/sin(d)

然後就是常規三分了。　　　　

#include<iostream>
#include<cstdio>
#include<cmath>
#define pi 3.1415926
using namespace std;
double x,y,l,w;

 
double f(double d){
    return (x-l*sin(d))*tan(d)+(w-(x-l*sin(d))/cos(d))/sin(d);
}

int main(){
    while(~scanf("%lf%lf%lf%lf",&x,&y,&l,&w)){
    double low=0,high=pi/2.0;
    while(high-low>=0.0001){
        double temp=(high-low)/3.0;
        double lm=low+temp;
        double rm=high-temp;
        if(f(lm)>f(rm)) high=rm;
        else low=lm;        
        }
    if(f(low)>y) cout<<"no"<<endl;
    else cout<<"yes"<<endl;      
    }
    return 0;
}