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

Source

2008 Asia Harbin Regional Contest Online

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题解转一发：

http://www.cnblogs.com/wally/archive/2013/05/14/3077383.html

网上大牛思路：

可以根据边界，汽车已经转弯，设水平方向为x轴，垂直方向为y轴。

则汽车的内边界(靠近里面的边界)的直线方程式f(x)为：y=x*tan(a)+l*sin(a)+d/cos(a).其中a是汽车与x轴的夹角

当y=X时，求解出的-x即为汽车的内边界到y轴的距离h，若h小于Y即可转弯，若大于Y就不能转弯。
所以只需要利用方程式，求-x的最大值，即可判断能否通过。
由于f(x)是凸函数(随着x的增大y先增大后减小)，所以，需要借助三分求解。

图示：

第一道三分求极值题啊！！！

 1 #include <cstdio>
 2 #include <cmath>
 3
 4 using namespace std;
 5
 6 const double mi = 1e-7;
 7 const double eps = 1e-9;
 8 const double pi = 4.0 * atan(1.0);
 9
10 double x,y,l,d;
11
12 double cal(double a)
13 {
14     return (-x + l * sin(a) + d / cos(a)) / tan(a) ;
15 }
16
17 int main()
18 {
19     while(scanf("%lf%lf%lf%lf",&x,&y,&l,&d)!=EOF)
20     {
21         if(d>y || d>x){
22             printf("no\n");continue;
23         }
24         double low=0,high = pi / 2,mid,mmid;
25         double te1,te2;
26         while(high - low > mi)
27         {
28             mid = (low + high) / 2;
29             mmid = (low + mid) / 2;
30             te1 = cal(mid);
31             te2 = cal(mmid);
32             if(te1 > te2){
33                 low = mmid;
34             }
35             else{
36                 high = mid;
37             }
38         }
39         te1 = cal(low);
40         if(te1 < y){
41             printf("yes\n");
42         }
43         else{
44             printf("no\n");
45         }
46     }
47 }