13.04.07 Turn the corner （三分）

Turn the corner

Time Limit : 3000/1000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other)

Total Submission(s) : 41   Accepted Submission(s) : 9

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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：

	1. 当x或y小于d时，一定不能够拐弯。

	2. 当x和y都大于d时，

其中：

s = l * cos(θ) + w * sin(θ) - x;

h = s * tan(θ) + w * cos(θ);

这是关于 θ的一个凸函数，我们要求其在 [0, π/2] 之间的最大值，故可以用三分法。

算法：

三分法

类似二分的定义Left和Right

mid = (Left + Right) / 2

midmid = (mid + Right) / 2;

如果mid靠近极值点，则Right = midmid；

否则(即midmid靠近极值点)，则Left = mid;

代码：

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
using namespace std;
double PI=acos(-1.0);
double x,y,l,d;
double f(double tha)
{
    double s,h;
    s=l*cos(tha)+d*sin(tha)-x;
    h=s*tan(tha)+d*cos(tha);
    return h;
}
int main ()
{
    double h,left,right,mid,midmid;
    while (cin>>x>>y>>l>>d)
    {
        if (x<=d || y<=d)
        {
            cout<<"no"<<endl;
            continue;
        }
        else
        {
            left=0.0;
            right=PI/2;
            while(fabs(right-left)>1.0e-8)
            {
                mid=(left+right)/2;
                midmid=(mid+right)/2;
                if (f(mid)>f(midmid))
                    right=midmid;
                else
                    left=mid;
            }
        }
        if ((f(mid))<=y)
            cout<<"yes"<<endl;
        else
            cout<<"no"<<endl;
    }
    return 0;
}