Light Bulb

Input

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.

Output

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

Sample Input

3
2 1 0.5
2 0.5 3
4 3 4

Sample Output

1.000
0.750
4.000

参考博客：http://blog.csdn.net/pi9nc/article/details/9666627

    http://www.cnblogs.com/kuangbin/archive/2012/09/27/2705751.html

解题思路：人的影子在地面上最大时的点，继续向右走的时候，地面加墙上的影子长度先变大后边小。这好像是常识，凸函数。利用三角形相似计算。x为人到路灯的距离。

大神1解释：根据图，很容易发现当灯，人的头部和墙角成一条直线时(假设此时人站在A点)，此时的长度是影子全在地上的最长长度。当人再向右走时，影子开始投影到墙上，当人贴着墙，影子长度即为人的高度。所以当人从A点走到墙，函数是先递增再递减，为凸性函数，所以我们可以用三分法来求解。

大神2代码：

#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<string.h>
#include<math.h>
using namespace std;
const double eps=1e-9;
double D,H,h;

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

double solve(double l,double r)
{
    double mid,midmid;
    double d1,d2;
    do
    {
        mid=(l+r)/2;
        midmid=(mid+r)/2;
        d1=calc(mid);
        d2=calc(midmid);
        if(d1>=d2) r=midmid;//这里要注意
        else l=mid;
    }
    while(r-l>=eps);
    return d1;
}
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%lf%lf%lf",&H,&h,&D);
        printf("%.3lf\n",solve((H-h)*D/H,D));
    }
    return 0;
}

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