HDU2438 Turn the corner

Turn the corner

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1906    Accepted Submission(s): 726

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 13.5 410 6 14.5 4

 

Sample Output

yesno

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

<span style="font-size:14px;">#include<stdio.h>#include<stdlib.h>#include<string.h>#include<math.h>#define PI acos(-1)double x,y,l,d;double h,s;double cal(double a){s=l*cos(a)+d*sin(a)-x;h=s*tan(a)+d*cos(a);return h;}int main(){double left,right,mid,midmid;while(scanf("%lf%lf%lf%lf",&x,&y,&l,&d)!=EOF){left=0;right=PI/2;if(x<d||y<d){printf("no\n");continue;}while(fabs(right-left)>1e-8){mid=(left+right)/2;midmid=(mid+right)/2;if(cal(mid)>=cal(midmid))right=midmid;elseleft=mid;}mid=(right+left)/2;if(cal(mid)<=y)printf("yes\n");elseprintf("no\n");}return 0;}</span><span style="font-size:12px;"></span>