HOJ 2438 Turn the corner_turn the corner hoj c语言代码-优快云博客

本文链接：https://blog.youkuaiyun.com/weidaoRan/article/details/47956625

本文探讨了汽车能否通过垂直拐角的问题，通过计算汽车宽度、长度和街道宽度之间的关系来判断是否可行。使用三分法优化算法提高了解决方案的效率。

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简单三分

Turn the corner

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2272 Accepted Submission(s): 876

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

主要是算出拐弯时高度H。写成函数。

#include<cstdio>
#include<cmath>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdlib>

using namespace std;

#define fi acos(-1.0)

double l,d,x,y;

//核心代码
double cal(double a)
{
    return (l*cos(a)-x)*tan(a)+d/cos(a);
}

int main()
{
  while(scanf("%lf%lf%lf%lf",&x,&y,&l,&d)!=EOF)
  {
      double left=0.0,mid,midmid,right=fi/2;
      if(d>=y||d>=x)//    减少时间复杂度
      {
          cout<<"no"<<endl;
          continue;
      }
      while(fabs(right-left)>1e-7)//三分代码
      {
          mid=(right+left)/2;
          midmid=(mid+right)/2;
          if(cal(mid)>=cal(midmid))
            right=midmid;
          else
            left=mid;
      }
      if(cal(midmid)>y)
        //puts("no");
        cout<<"no"<<endl;
      else
        //puts("yes");
        cout<<"yes"<<endl;
  }
  return 0;
}