zoukankan      html  css  js  c++  java
  • POJ 3122, Pie

    Time Limit: 1000MS  Memory Limit: 65536K
    Total Submissions: 3965  Accepted: 1505  Special Judge


    Description
    My birthday is coming up and traditionally I'm serving pie. Not just one pie, no, I have a number N of them, of various tastes and of various sizes. F of my friends are coming to my party and each of them gets a piece of pie. This should be one piece of one pie, not several small pieces since that looks messy. This piece can be one whole pie though.

    My friends are very annoying and if one of them gets a bigger piece than the others, they start complaining. Therefore all of them should get equally sized (but not necessarily equally shaped) pieces, even if this leads to some pie getting spoiled (which is better than spoiling the party). Of course, I want a piece of pie for myself too, and that piece should also be of the same size.

    What is the largest possible piece size all of us can get? All the pies are cylindrical in shape and they all have the same height 1, but the radii of the pies can be different.

     

    Input
    One line with a positive integer: the number of test cases. Then for each test case:
    One line with two integers N and F with 1 ≤ N, F ≤ 10 000: the number of pies and the number of friends.
    One line with N integers ri with 1 ≤ ri ≤ 10 000: the radii of the pies.

    Output
    For each test case, output one line with the largest possible volume V such that me and my friends can all get a pie piece of size V. The answer should be given as a floating point number with an absolute error of at most 10−3.

     

    Sample Input
    3
    3 3
    4 3 3
    1 24
    5
    10 5
    1 4 2 3 4 5 6 5 4 2

    Sample Output
    25.1327
    3.1416
    50.2655

     

    Source
    Northwestern Europe 2006


    // POJ3122.cpp : Defines the entry point for the console application.
    //

    #include 
    <iostream>
    #include 
    <cmath>

    int main(int argc, char* argv[])
    {
        
    double pie[10000];
        
    const double PI = acos(-1.0);

        
    int c, n, f;
        scanf(
    "%d",&c);
        
    while(c--)
        {
            scanf(
    "%d%d",&n,&f);
            
    double minv = 0,maxv;
            
    for(int i = 0;i < n; ++i)
            {
                scanf(
    "%lf",&pie[i]);
                
    if(PI * pie[i] * pie[i] > maxv)
                    maxv 
    = PI*pie[i]*pie[i];
            }

            
    for(int i = 0;i < 100++i)
            {
                
    double midv = (minv+maxv)/2;
                
    int cnt = 0;
                
    for(int i = 0;i < n; ++i)
                {
                    
    double V = pie[i] * pie[i] * PI;
                    cnt 
    += (int)(V/midv);
                }
                
    if(cnt > f)
                    minv 
    = midv;
                
    else
                    maxv 
    = midv;
            }
            printf(
    "%.4lf\n",minv);
        }
        
    return 0;
    }

  • 相关阅读:
    fiddler 保存请求数据并发送到自己的服务器接口
    入门chrome插件开发教程和经验总结,一篇就搞掂!
    文件图标SVG
    银行开发平台
    银联号
    数学好玩 沛沛猜想
    Jenkins安装 maven插件
    jQuery的on绑定click和直接绑定click区别
    PC与移动端都可以使用的横向导航,可点击可滑动
    layui加载层(有文字提示版)跳出iframe父级
  • 原文地址:https://www.cnblogs.com/asuran/p/1593736.html
Copyright © 2011-2022 走看看