HDU 2438 Turn the corner 三角函数+三分查找法

题目地址：点击打开链接

Turn the corner

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3276    Accepted Submission(s): 1346

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】：车子要成功拐过这个弯道，最好的办法是使得车子的一条侧边靠墙走（然而现实中不存在）；

【解析】：

根据图示，车子右侧沿着内拐角，左外侧沿着左侧墙壁。根据图中角度关系，即可得出关于h的函数，角度变化范围（0~90度）。

易得，函数是先增后减的。

用三分查找法找到函数h的最大值。最大值和y比较一下，会不会撞墙

【代码】：

#include<stdio.h>#include<math.h>double x,y,l,d;const double PI=3.14159;double h(double a){return d*cos(a)+d*sin(a)*sin(a)/cos(a)+l*cos(a)*tan(a)-x*tan(a);}int main(){while(~scanf("%lf%lf%lf%lf",&x,&y,&l,&d)){double m=0;double n=PI/2;double P,Q;while(fabs(m-n)>1e-6){P=m+(n-m)/3;Q=n-(n-m)/3;if(h(P)>h(Q))//左边大{n=Q;}else{m=P;}}if(h(m)<=y)puts("yes");elseputs("no");}return 0;}