补题之路：HDU2438 Turn the corner_mr. west bought a new car! so he is travelling aro-优快云博客

本文探讨了一辆汽车在垂直角落转弯时是否会遇到障碍的问题。通过对车身与街道宽度的关系进行数学建模，使用直线方程及函数图像的分析方法，确定了汽车能否顺利通过角落的条件。代码实现中采用三分法查找函数最大值，最终判断汽车是否能够跨越角落。

题目描述

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?

输入：

Every line has four real numbers, x, y, l and w.
Proceed to the end of file.

输出：

If he can go across the corner, print “yes”. Print “no” otherwise.

样例：
10 6 13.5 4
10 6 14.5 4
输出：
yes
no

题意
车转弯过程会不会被卡住~
思路

一开始想简单了，只判断45度角时的特殊情况，连样例都没过x
正常思路是车身贴边，建系求直线方程，通过函数找函数图像最高点是否满足条件。方程：y = tan(t)x+l*sin(t)+d/cos(t)。显然先增后减，最大值不一定在t=45°取到，验证了前面的想法是错的。然后标准三分找最大值，比较-x和y就可以啦

代码

#include <iostream>
#include <cmath>
using namespace std;
double x,y,l,w;
double finda(double t){
 return (l*sin(t)+w/cos(t)-x)/tan(t);
}
int main(){
 
 while(cin >> x >> y >> l >> w){
  double left = 0,right = 3.1415926/2.0,ml,mr;
 // cout << right << endl;
  while(right-left>1e-6){
   ml = (right-left)*1/3+left;
   mr = (right-left)*2/3+left;
   if(finda(ml)<=finda(mr)){
    left = ml;
   }
   else right = mr;
  }
  if(finda(right)<=y)cout << "yes" << endl;
  else cout <<"no" << endl; 
 }
 return 0;
}