zoukankan      html  css  js  c++  java
  • ZOJ 3203 Light Bulb (三分+计算几何)

    B - Light Bulb
    Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%lld & %llu

    Description

    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
    
    题目意思:
    给你H,h,D三个数字
    问你影子的长度
    随着h的变化,影子的长度也会变化(影子的长度包括在地上和墙上的)
    可以预想一下
    跟h有关的影子的长度的函数是一个由峰值的函数
    所以三分解决
     
    #include<cstdio>
    #include<string>
    #include<cstdlib>
    #include<cmath>
    #include<iostream>
    #include<cstring>
    #include<set>
    #include<queue>
    #include<algorithm>
    #include<vector>
    #include<map>
    #include<cctype>
    #include<stack>
    #include<sstream>
    #include<list>
    #include<assert.h>
    #include<bitset>
    #include<numeric>
    #define max_v 35
    #define eps 10e-6
    using namespace std;
    double H,h,D;
    double f(double l)
    {
        return D*(h-l)/(H-l)+l;
    }
    int main()
    {
        int t;
        scanf("%d",&t);
        while(t--)
        {
            scanf("%lf %lf %lf",&H,&h,&D);
            double l=0,r=h;
            double mid,mmid;
            while(fabs(l-r)>eps)
            {
                mid=(l+r)/2;
                mmid=(mid+r)/2;
                if(f(mid)<=f(mmid))
                {
                    l=mid;
                }
                else
                {
                    r=mmid;
                }
            }
            printf("%0.3lf
    ",f(l));
        }
        return 0;
    }
    /*
    题目意思:
    给你H,h,D三个数字
    问你影子的长度
    
    随着h的变化,影子的长度也会变化(影子的长度包括在地上和墙上的)
    可以预想一下
    跟h有关的影子的长度的函数是一个由峰值的函数
    所以三分解决
    
    */
     
  • 相关阅读:
    Linux常用命令大全(非常全!!!)
    洛谷 P3379 【模板】最近公共祖先(LCA)
    POJ 3259 Wormholes
    POJ 1275 Cashier Employment
    POJ 3169 Layout
    POJ 1201 Intervals
    洛谷 P5960 【模板】差分约束算法
    洛谷 P3275 [SCOI2011]糖果
    POJ 2949 Word Rings
    POJ 3621 Sightseeing Cows
  • 原文地址:https://www.cnblogs.com/yinbiao/p/9504080.html
Copyright © 2011-2022 走看看