ZOJ Light Bulb 3203【思维】

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本文介绍了一个有趣的问题：通过给定的灯泡高度、人物高度及灯泡与墙壁的距离，计算人物在房间内移动时的最大影子长度。文章提供了解题思路与具体算法实现。

Light Bulb

Time Limit: 1 Second Memory Limit: 32768 KB

Compared to wildleopard's wealthiness, his brother mildleopard is rather poor. His house is narrow and he has only one light bulb in his house. Every night, he is wandering in his incommodious house, thinking of how to earn more money. One day, he found that the length of his shadow was changing from time to time while walking between the light bulb and the wall of his house. A sudden thought ran through his mind and he wanted to know the maximum length of his shadow.

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

Author: GUAN, Yao
Source: The 6th Zhejiang Provincial Collegiate Programming Contest

题目大意：给你图中标明的H,h,D，h随意走动，求最大的L的长度。

思路：枚举+推导公式

思路：

首先求出两个相对值，一个是没有射在墙上的影子，一个是整个h的长度就是影子的长度。

然后我们枚举i来求在地上和墙上都有影子的情况。

对于这部分有图：

根据图片中给出的数据不难得出：

H-h/i==H-l/D；

辣么l=H-(H-h)*D/i;

因为数据范围不大，每个数的数据范围都是从10^-2到10^3,我们暴力枚举i即可，每一次i都加上0.001。如果这里数据比较大的话，我们需要使用三分算法。

AC代码：

#include<stdio.h>
#include<math.h>
using namespace std;
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        double H,h,D;
        scanf("%lf%lf%lf",&H,&h,&D);
        double v=D*h/H;
        double compare=h;
        double output=max(v,h);
        double i=D-v;
        while(1)
        {
            i+=0.001;
            if(i>D)break;
            double l=H-(H-h)*D/i;
            //printf("%.3lf\n",l+D-i);
            output=max(l+D-i,output);
        }
        printf("%.3lf\n",output);
    }
}