POJ 2507Crossed ladders（二分）

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

4

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

题意：在两个房子之间有两条线，告诉你两条线长度还有交点的高度问两栋房子之间距离多远。

思路：两个三角形求出1-c/sqrt(x*x-w*w)=c/sqrt(y*y-w*w);然后二分求出距离。（距离肯定小于x，y之间较小的数）。这里不要判断s（mid）==0，精度太小计算不出来，直接在后面输出最后的mid就行了

代码：

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<string.h>
#include<math.h>
#define eps 1e-12
using namespace std;
double x,y,c;
double s(double w)
{
return 1-c/sqrt(x*x-w*w)-c/sqrt(y*y-w*w);

}
int main()
{
int t;
scanf("%d",&t);
int k=1;
while(t--)
{
scanf("%lf %lf %lf",&x,&y,&c);
int p=0;
double l=0,r=min(x,y),mid,ans;
while(l<=r)
{
mid=(l+r)/2;
if(s(mid)>0)
{
l=mid+eps;
}
else
r=mid-eps;
}
printf("Case %d: %lf\n",k++,mid);
}
return 0;
}