Turn the corner

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2785    Accepted Submission(s): 1102

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

题目大意：有一个转角，给你它的x,y和汽车的长l和宽d，问你它能否成功转弯。

思路分析：现象很常见，但我们很少去思考它能不能转弯，对于这种题目，一定要

将其转化为代数运算，建立合适的坐标系是比较常用的想法，

图像可以参考这个博客http://blog.csdn.net/u013761036/article/details/24588987

这样做是假设汽车贴着外边走，判断最宽的地方是否<Y,感觉假设贴着内边走也是能够做

出来的，至于为什么采用三分法，稍微感觉一下趋

势就可以，很显然最宽的时候是在转动的过程中，因此在【0，pi/2]之间二分即可

代码：

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <queue>
#include <cmath>
using namespace std;
double x,y,l,w;
const double pi=acos(-1.0);
const double eps=1e-7;
double cal(double a)
{
   double k=x-l*sin(a)-w/cos(a);
   return -k/tan(a);
}
int main()
{
    while(scanf("%lf%lf%lf%lf",&x,&y,&l,&w)!=EOF)
    {
       double l=0.0,r=pi/2;
       if(x<=w|| y<=w) {printf("no\n");continue;}
       while(l+eps<=r)
       {
           double mid=(l+r)/2.0;
           double mmid=(mid+r)/2.0;
           if(cal(mid)<=cal(mmid)) l=mid;
           else r=mmid;
       }
       if(cal(r)<y) printf("yes\n");
       else printf("no\n");
    }
    return 0;
}

刚始那句特判必不可少，不太懂为什么，因为它弱wa了三发，

求高人指教。