二分---LIGHTOJ 1062

1062 - Crossed Ladders

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Time Limit: 2 second(s)

Memory Limit: 32 MB

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	Output for Sample Input
4 30 40 10 12.619429 8.163332 3 10 10 3 10 10 1	Case 1: 26.0328775442 Case 2: 6.99999923 Case 3: 8 Case 4: 9.797958971

大意：求底的长度；

找关系：设x与左边楼的交点为A，与右边楼的交点为C，y与左边楼的交点为B，与右边楼的交点为D；x与y的交点为E，由E向地面做垂线，设垂足为F。由三角形相似：EF/AB=CF/CB=1-(BF/BC)=1-(EF/CD)，也就是EF/AB=1-(EF/CD)，两边同时除以EF

注意：输出精度

#include<stdio.h>
#include<math.h>
#define eps 1e-9
#define min(a,b) ((a)<(b)?(a):(b))
int main(){
	int t;
	double a,b,c,high,low,mid;
	scanf("%d",&t);
//	cout<<min(t,a);
	for(int i=1;i<=t;i++){
		scanf("%lf%lf%lf",&a,&b,&c);
		 high=min(a,b);
		 low=0;

		while(high-low>eps){
			mid=(high+low)/2;
			if((1/sqrt(a*a-mid*mid)+1/sqrt(b*b-mid*mid))>1/c)  //说明mid偏大
				high=mid;
			else low=mid;
		}
		printf("Case %d: %.8lf\n",i,mid);
	}
	return 0;
}

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