ZOJ_3203_LightBulb
来源:互联网 发布:自动化设备 图形 编程 编辑:程序博客网 时间:2024/06/02 06:31
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
32 1 0.52 0.5 34 3 4
Sample Output
1.0000.7504.000
Author: GUAN, Yao
Source: The 6th Zhejiang Provincial Collegiate Programming Contest
题目意思求如图所示的影子最大长度
显然当影子到达墙壁弯折之前的任意状态长度不如刚好到达墙壁时
影子到达墙壁后 墙壁上的影子变长
而地上的影子变短
因此影子长度应该为一个凸函数,三分求最大值就可以了
这里选取了人与灯的水平距离x为变量
这个题目精度要设置的高些 1e-8WA了一发……
#include <iostream>#include <stdio.h>#include <math.h>using namespace std;const double AC=1e-9;double H,h,D;double f(double x){ double s1=H*x/(H-h); double s2=(s1-D)*H/s1; return D-x+s2;}double ts(){ double lo=(H-h)*D/H,hi=D; double mid,midmid; while(hi-lo>AC) { mid=(lo+hi)/2; midmid=(mid+hi)/2; //cout<<"ts "<<mid<<" "<<midmid<<endl; if(f(mid)>f(midmid)) hi=midmid; else lo=mid; //cout<<endl; } return f(lo);}int main(){ int t; scanf("%d",&t); while(t--) { scanf("%lf%lf%lf",&H,&h,&D); printf("%.3lf\n",ts()); } return 0;}
- ZOJ_3203_LightBulb
- 愿以真心换真心—不妄求(2016年终总结)
- Android M Dialer完全总结
- iOS多线程学习-NSThread、Cocoa NSOperation、GCD
- MySQL Replication常见错误整理
- [POJ 3164][最小树形图]
- ZOJ_3203_LightBulb
- 生成函数
- 「译」通过Fragment处理配置变化
- JZOJ 1503.体育场
- Unity3d 开发(八)复制内容到剪切板
- android简单实例---------------SD卡(外部)存储的使用与讲解
- 关于MVC的理解
- C++学习笔记之 函数重载和函数指针在一起
- Java基本数据类型