Zoj——3203 Light Bulb

最新推荐文章于 2019-08-05 16:00:03 发布

DlPF_C

最新推荐文章于 2019-08-05 16:00:03 发布

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分类专栏：简单算法日常

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本文介绍了一道关于计算最大阴影长度的数学问题及其解决方法。通过输入灯泡的高度、人的高度及灯泡与墙壁的距离，利用对勾函数或三分法找到最大阴影长度。

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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

Sample Output

1.000
0.750
4.000

一道数学题。。。。刚开始未知数设的不合适，导致推出来的式子连自己都不认识。。。。。

这是我后来的推导过程

最后图出的方程为 H+D-x-D*（H-h）/x；

在这里我们可以使用三分或者是对勾函数性质求最值

对勾函数

#include<stdio.h>
#include<math.h>
using namespace std;
int main()
{
	int T;
	while(scanf("%d",&T)!=EOF)
	{
		while(T--)
		{
			double H,h,D;
			scanf("%lf%lf%lf",&H,&h,&D);
			double t=sqrt((H-h)*D);
			double t1=D*(1-h/H);
			if(t<t1)
				printf("%0.3lf\n",D*h/H);
			else if(t>=D)
				printf("%0.3lf\n",h);
			else
				printf("%0.3lf\n",H+D-t-(D*(H-h))/t);
		}
	}
	return 0;
}

三分

#include<stdio.h>
#define esp 1e-10
double H,D,h;
double f(double x)
{
	return H+D-x-D*(H-h)/x;
}
double slove(double left,double right)
{
	while(right-left>esp)
	{
		double mid=(left+right)/2;
		double mmid=(mid+right)/2;
		if(f(mid)>f(mmid))
			right=mmid;
		else 
			left=mid;
	}
	return f(left)>f(right)?f(left):f(right);
}
int main()
{
	int T;
	scanf("%d",&T);
	while(T--)
	{
		scanf("%lf%lf%lf",&H,&h,&D);
		printf("%.3lf\n",slove(D*(1-h/H),D));
	}
 }