# hdu 2438 Turn the corner（3分）

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hdu 2438 Turn the corner（三分）

# Turn the corner

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2010    Accepted Submission(s): 765

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```

给出街道在x轴的宽度X，y轴的宽度Y，还有车的长l和宽w，判断是否能够转弯成功。

盗网上大牛一张图，画的很详细

尽可能让车贴着外面的墙璧转弯，也就是图中的x轴和y轴，此时红线的方程就是图中的方程，此时p点的位置就是让y=X时解得的x值，要保证p点在Y内，也就是-x<y,假若在转弯的所有角度中都满足这个条件，那么就能转弯，分析得，-x先增大后减小，所以用三分求最大-x值。

```#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
const double pi=acos(-1.0);
const double eps=1e-6;
double x,y,l,w;
double solve(double angle)
{
return (-x+l*sin(angle)+w/cos(angle))/tan(angle);
}
int main()
{
while(~scanf("%lf%lf%lf%lf",&x,&y,&l,&w))
{
double l=0,r=pi/2,mid1,mid2;
while(l+eps<r)
{
mid1=l+(r-l)/3;
mid2=r-(r-l)/3;
if(solve(mid1)>solve(mid2))
r=mid2;
else
l=mid1;
}
if(solve(l)<y)
printf("yes\n");
else
printf("no\n");
}
return 0;
}
```