Crossed Ladders<poj2507题目一样>

最新推荐文章于 2019-04-24 21:51:25 发布

原创最新推荐文章于 2019-04-24 21:51:25 发布 · 782 阅读

0 ·

CC 4.0 BY-SA版权

二分同时被 2 个专栏收录

14 篇文章

订阅专栏

poj

14 篇文章

订阅专栏

本文探讨了一道有趣的几何问题：两条不同长度的梯子分别靠在街道两侧的建筑物上并交叉于一定高度，如何计算街道的宽度。通过迭代逼近的方法找到解决方案，并提供了具体的实现代码。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Description

A narrow street is lined with tall buildings. An x foot long ladder is rested at the base of the building on the right side of the street and leans on the building on the left side. A y foot long ladder is rested at the base of the building on the left side of the street and leans on the building on the right side. The point where the two ladders cross is exactly c feet from the ground. How wide is the street?

Input

Input starts with an integer T (≤ 10), denoting the number of test cases.

Each test case contains three positive floating point numbers giving the values of x, y, and c.

Output

For each case, output the case number and the width of the street in feet. Errors less than 10^-6 will be ignored.

Sample Input

30 40 10

12.619429 8.163332 3

10 10 3

10 10 1

Sample Output

Case 1: 26.0328775442

Case 2: 6.99999923

Case 3: 8

Case 4: 9.797958971

求两个房子之间的距离

#include<cstdio>   
#include<cmath>
#define  eps 1e-8 
int main()
{
  int t;
  int cut=0;
  scanf("%d",&t);
  while(t--)
  {
  	cut++;
  	double x,y,h;
  	 scanf("%lf%lf%lf",&x,&y,&h);
  	  int S=100;
  	  double l=0.0,r=1000.0; 
			double ans;
  	  while(S--)
  	  {
  	  	double mid=(l+r)/2.0;
  	  	double jiao1=acos(mid/y);
  	  	double jiao2=acos(mid/x);
  	  	double L=h/tan(jiao1)+h/tan(jiao2);
  	 
  	  	if(mid>L)
			{
				l=mid+eps;
				ans=mid;
			}
			else
			{
				r=mid-eps;
			}
		}
  	   printf("Case %d: %.7lf\n",cut,ans);
	  }	
	return 0;
}