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：
用下https://blog.csdn.net/u013761036/article/details/24588987博主的图
意思当车的右侧按不同的角度angle靠着右侧的墙走的时候，看是否能碰着拐角。即此时车身左侧的直线当y=X时，此时的横坐标在-Y的左边还是右边，如果满足情况下，-x<Y,证明车子可以过去，否则则不能。
因为随着angle角的增大，-x的值是一个形似开头向下的抛物线，故可以利用三分来求此抛物线的极点的值
#include<cstdio>
#include<algorithm>
#include<cmath>
using namespace std;
const double wc=1e-6;
const double pi=acos(-1);
double X,Y,l,w;
double f(double x)//传弧度，返回Y=X时的-x的值
{
return (l*sin(x)+w/cos(x)-X)/tan(x);
}
int main()
{
while(~scanf("%lf%lf%lf%lf",&X,&Y,&l,&w))
{
double ll=0,rr=pi/2.0;
while(rr-ll>wc)
{
double M1=ll+(rr-ll)/3;
double M2=rr-(rr-ll)/3;
if(f(M1)<f(M2))
ll=M1;
else
rr=M2;
}
if(f(ll)<Y)
printf("yes
");
else
printf("no
");
}
return 0;
}