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?

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 ",x,y,l,d);
        if(d>x||d>y)
        {
            printf("no ");
            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 ",left,right);
        while (right-left>1e-10)
        {
            mid=(left+right)/2;
            midmid=(mid+right)/2;
            //printf("mid=%.4f mid mid=%.4f ",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 ",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 ");
        else
            printf("yes ");
    }
    return 0;
}