Turn the corner

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?

the corner" title="Turn 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

题意：西先生开始出去，遇到一个弯，让你求能不能拐过去；

解题思路：假设过弯时l与底边夹角为s，用三分求最理想过弯角度（用过弯过程中与转弯之后的宽度y相比较），找到最合适的角度，如果最合适的角度，转弯的宽度还是比y大就肯定转不过去，反之就能转过去；

感悟：寻找过弯条件是最难的，找到之后就简单多了，记录第二道三分题；

代码（G++
0ms）：

#include

#include

#include

#define pi 3.141592653589793238

using namespace std;



double x,y,l,d;

double cal(double s)//进行三分的条件

{

    return
l*cos(s)+(d-x*cos(s))/(sin(s));

   
//l*cos(s)+(d-x*cos(s))/(sin(s))就是过弯过程中在y方向上的宽

}

int
main()

{

   
//freopen("in.txt", "r", stdin);

    double
left,right,mid,midmid;

   
while(scanf("%lf%lf%lf%lf",&x,&y,&l,&d)!=EOF)

    {

       
//printf("x=%.1f y=%.1f l=%.1f d=%.1f\n",x,y,l,d);

       
if(d>x||d>y)

       
{

           
printf("no\n");

           
continue;

       
}

       
left=0;

       
right=pi/2;//最大只能是pi/2，不可能倒着转弯吧

       
//printf("left=%.1f right=%.1f\n",left,right);

       
while (right-left>1e-10)

       
{

           
mid=(left+right)/2;

           
midmid=(mid+right)/2;

           
//printf("mid=%.4f mid mid=%.4f\n",mid,midmid);

           
if (cal(mid)>=cal(midmid))//过弯过程中在y方向上的宽，与y比较

               
right=midmid;

           
else left=mid;

       
}

       
//printf("right=%.1f\n",right);

       
if(cal(right)>y)//如果最大角度过弯时还是比y宽就肯定过不去

           
printf("no\n");

       
else

           
printf("yes\n");

    }

    return
0;

}