hdu Turn the corner

Turn the corner

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 151 Accepted Submission(s): 61

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 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 #include <iostream> #include <stdio.h> #include <math.h> using namespace std; double pi = acos(-1.0); double x,y,l,w,s,h; double cal(double a) { s = l*cos(a)+w*sin(a)-x; h = s*tan(a)+w*cos(a); return h; } int main() { double left,right,mid,midmid; while(scanf("%lf%lf%lf%lf",&x,&y,&l,&w)!=EOF) { left = 0.0; right = pi/2; while(fabs(right-left)>1e-8) { mid = (left+right)/2; midmid = (mid+right)/2; if(cal(mid)>=cal(midmid))right = midmid; else left = mid; } if(cal(mid)<=y)printf("yes\n"); else printf("no\n"); } return 0; }