【二分+几何】E - Crossed Ladders

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本文探讨了一道经典的几何问题——交叉梯子问题。通过给定两梯子的长度及交叉点离地面的高度，求解街道宽度。利用数学方法进行求解，并通过编程实现了数值解的计算。

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E - Crossed Ladders

Time Limit:2000MS Memory Limit:32768KB 64bit IO Format:%lld & %llu

Submit Status

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

0<=wide<=min(x,y);

可供参考：http://blog.youkuaiyun.com/lyy289065406/article/details/6648562

代码：

#include<cstdio>
#include<cmath>
double x,y,c,w,maxn;
double judge(double mid)
{
	return 1-c/sqrt(x*x-mid*mid)-c/sqrt(y*y-mid*mid);
}
int main()
{
	int t,k=1;
	while(~scanf("%d",&t))
	{
		while(~scanf("%lf%lf%lf",&x,&y,&c))
		{
			maxn=x;
			if(maxn<y)
				maxn=y;
			double l=0.0,r=maxn,mid;
			while(r-l>1e-6)
			{
				mid=(l+r)/2;			
				if(judge(mid)>0)
				{
					l=mid;
				}
				else 
				{
					r=mid;
				}
			}
			printf("Case %d: %.7lf\n",k++,mid);    //Errors less than 10-6 will be ignored.  小数大于6位就满足题意 
		}		
	}
	return 0;
 }