The first line of the input contains an integer T (T <= 100), indicating the number of cases.

Each test case contains three real numbers H, h and D in one line. H is the height of the light bulb while h is the height of mildleopard. D is distance between the light bulb and the wall. All numbers are in range from 10^-2 to 10³, both inclusive, and H - h >= 10^-2.

For each test case, output the maximum length of mildleopard‘s shadow in one line, accurate up to three decimal places..

3
2 1 0.5
2 0.5 3
4 3 4

1.000
0.750
4.000


求影子长度的最大值，哦，最大值，那就是三分喽。



根据相似三角形边的关系，推导出公式影长：D-x+H-(H-h)*D/x；

#include<bits/stdc++.h>
using namespace std;
const double esp = 1e-10;
double H,h,D,L,R,mid,midd;

double f(double x){
	return D-x+H-(H-h)*D/x;	
}

int main(){
	int T;
	scanf("%d",&T);
	while(T--){
		scanf("%lf %lf %lf",&H,&h,&D);
		mid = midd = 0.0;
		L=0.0; R=D;
		while(R-L>esp){
			mid=L+(R-L)/2.0;
			midd=mid+(R-mid)/2.0;
			if(f(mid)<f(midd))
				L=mid;
			else
				R=midd;	
		}
		printf("%.3lf\n",f(L));	
	}
	return 0;
}