专题二 1005

一. 题目编号

1005

Turn the corner

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

Source

2008 Asia Harbin Regional Contest Online

二.简单题意

Mr. West驾车来到了一个垂直的角落。Mr. West想通过此弯道，现在给出车子的长和宽，以及直角弯道的宽度和他现在没转弯之前的街道宽度。求出他能不能通过该弯道。

三、解题思路形成过程

要使汽车能转过此弯道，那么就是汽车的左边尽量贴着那个直角点，而汽车的右下后方的点尽量贴着最下面的边。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。

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。

四、感想

必须得把题目构建为数学模型来解决，

五、AC代码

#include <iostream>
 #include <cmath>
 using namespace std;
 double p= acos(-1.0);
 double x,y,l,w,s,h;
 double calculate(double a)
 {
     s = l*cos(a)+w*sin(a)-x;
     h = s*tan(a)+w*cos(a);
     return h;
 }
 int main()
 {
     double min,max,mid,midmid;
     while(cin>>x>>y>>l>>w)
     {
         min = 0.0;
         max = p/2;
         while(fabs(max-min)>1e-8)
         {
             mid = (min+max)/2;
             midmid = (mid+max)/2;
             if(calculate(mid)>=calculate(midmid))max = midmid;
             else min = mid;
         }
         if(calculate(mid)<=y)
         cout<<"yes"<<endl;
         else
         cout<<"no"<<endl;
     }
     return 0;
 }