Turn the cornerTime Limit: 3000/1000 MS (Java/Others)Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2229Accepted Submission(s): 856
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
题意：
已知汽车的长和宽，l和w，以及俩条路的宽为x和y，汽车所处道路宽为x ，问汽车能否顺利转弯？
分析：汽车能否顺利转弯取决于在极限情况下，，随着角度的变化，汽车离对面路的距离是否大于等于0
如图中
在上图中需要计算转弯过程中h 的最大值是否小于等于y很明显，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
代码：
#include<iostream>#include<algorithm>#include<math.h>#include<cstdio>using namespace std;#define pi 3.141592653double x,y,l,w;double cal(double a){double s=l*cos(a)+w*sin(a)-x;double h=s*tan(a)+w*cos(a);return h;}int main(){while(scanf("%lf %lf %lf %lf",&x,&y,&l,&w)!=EOF){double left=0.0,right=pi/2;double lm,rm;while(fabs(right-left)>1e-6){lm=(left*2.0+right)/3.0;rm=(left+right*2.0)/3.0;if(cal(lm)>cal(rm))right=rm;else left=lm;}if(cal(left)<=y)printf("yes\n");else printf("no\n");}return 0;}
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