H - Turn the corner hdu2438（三分）

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博主的圖

意思當車的右側按不同的角度angle靠着右側的牆走的時候，看是否能碰着拐角。即此時車身左側的直線當y=X時，此時的橫座標在-Y的左邊還是右邊，如果滿足情況下，-x<Y,證明車子可以過去，否則則不能。

因爲隨着angle角的增大，-x的值是一個形似開頭向下的拋物線，故可以利用三分來求此拋物線的極點的值

 

#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)//傳弧度，返回Y=X時的-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\n");
        else
            printf("no\n");
    }
    return 0;
}