hdu Turn the corner

Turn the corner

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 151 Accepted Submission(s): 61

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

汽车拐弯问题，给定X, Y, l, d判断是否能够拐弯。首先当X或者Y小于d，那么一定不能。
其次我们发现随着角度θ的增大，最大高度ｈ先增长后减小，即为凸性函数，可以用三分法来求解。

这里的Calc函数需要比较繁琐的推倒公式：
s = l * cos(θ) + w * sin(θ) - x;
h = s * tan(θ) + w * cos(θ);
其中s为汽车最右边的点离拐角的水平距离, h为里拐点最高的距离, θ范围从0到90。

3分搜索法

#include <iostream>
#include <stdio.h>
#include <math.h>
using namespace std;
double pi = acos(-1.0);
double x,y,l,w,s,h;
double cal(double a)
{
    s = l*cos(a)+w*sin(a)-x;
    h = s*tan(a)+w*cos(a);
    return h;
}
int main()
{
    double left,right,mid,midmid;
    while(scanf("%lf%lf%lf%lf",&x,&y,&l,&w)!=EOF)
    {
        left = 0.0;
        right = pi/2;
        while(fabs(right-left)>1e-8)
        {
            mid = (left+right)/2;
            midmid = (mid+right)/2;
            if(cal(mid)>=cal(midmid))right = midmid;
            else left = mid;
        }
        if(cal(mid)<=y)printf("yes\n");
        else printf("no\n");
    }
    return 0;
}