Light Bulb (三分)

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本文针对一个具体的数学问题，即在给定灯的高度、人的高度及灯到墙的距离的情况下，求解人在墙上的影子的最大长度。文章通过分析不同情况下影子的长度变化，并采用三分法进行查找，最终给出了一个高效的解决方案。

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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

Sample Output

1.000
0.750
4.000

题意：给出灯的高度H，人的高度h，灯到墙的水平距离D，求人的最长的影子长度。

题解

设人到灯的水平距离为x，影长有三种情况：

1. 影子恰好到墙，根据三角形相似，得到 x=D-D*h/H，则影长为 D*h/H；

2. 影子一部分在地上，一部分在墙上，根据三角形相似，得到影长为 D-x+H-D*(H-h)/x ；

3. 影子全部在墙上，此时影长为 D。

在这里只说三分法，公式法链接：https://blog.youkuaiyun.com/Baiyi_destroyer/article/details/79718020

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <stack>
#include <algorithm>
#include <cmath>
#include <cstdlib>
#include <stack>
#include <vector>
#define MAX 0x3f3f3f3f
#define LL long long
using namespace std;
const double eps = 1e-9;
double H,h,D;
double L(double x)
{
       return D-x+H-(H-h)*D/x;
}
double ans(double l,double r)                                               // 三分查找
{
       double mid,mmid;
       mid=(l+r)/2;
       mmid=(mid+r)/2;
       if(r-l<eps)
           return L(mid);
       return L(mid)>=L(mmid)?ans(l,mmid):ans(mid,r);    
}
int main()
{
       int c;
       cin >> c;
       while(c--)
       {
           cin >> H >> h >> D;
           printf("%.3lf\n",ans((H-h)*D/H,D));   
       }
       return 0;
}