Turn the corner

Turn the corner

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3664    Accepted Submission(s): 1534

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：

•   求出对应的函数表达式，y=-tan（q）*x+Lsin(q)+d/cos(q)
•   令y=X; 则求出来的-x和y进行比较   如果大于y 就不能通过  否则是可以的。
• 
  

#include<cstdio>
#include<string.h>
#include<cctype>
#include<cmath>
#define pi acos(-1)
double x,y,l,d;
const double eps=1e-7;
double cal(double t)
{
    return -tan(t)*x+l*sin(t)+d/cos(t);
}
int main()
{
    double low,high,mid,mmid;
    while(~scanf("%lf%lf%lf%lf",&x,&y,&l,&d))
    {
        if(x<d||y<d)   //如果x和y小于d则肯定不能通过
        {
            printf("no\n");
            continue;
        }
        low=0;      //利用三分求出他的最大值
        high=pi/2;
        while(high-low>eps)
        {
            mid = (low+high)/2;
            mmid = (mid+high)/2;
            if(cal(mid)>cal(mmid)) high = mmid;
            else low = mid -1e-9;
        }
        if(cal(mid)<y)
            printf("yes\n");
        else printf("no\n");
    }
    return 0;
}

Source

2008 Asia Harbin Regional Contest Online