zoukankan      html  css  js  c++  java
  • [POJ3684]Physics Experiment

     
    Time Limit: 1000MS   Memory Limit: 65536K
    Total Submissions: 1363   Accepted: 476   Special Judge

    Description

    Simon is doing a physics experiment with N identical balls with the same radius of R centimeters. Before the experiment, all N balls are fastened within a vertical tube one by one and the lowest point of the lowest ball is H meters above the ground. At beginning of the experiment, (at second 0), the first ball is released and falls down due to the gravity. After that, the balls are released one by one in every second until all balls have been released. When a ball hits the ground, it will bounce back with the same speed as it hits the ground. When two balls hit each other, they with exchange their velocities (both speed and direction).

    Simon wants to know where are the N balls after T seconds. Can you help him?

    In this problem, you can assume that the gravity is constant: g = 10 m/s2.

    Input

    The first line of the input contains one integer C (C ≤ 20) indicating the number of test cases. Each of the following lines contains four integers NHRT.
    1≤ N ≤ 100.
    1≤ H ≤ 10000
    1≤ R ≤ 100
    1≤ T ≤ 10000

    Output

    For each test case, your program should output N real numbers indicating the height in meters of the lowest point of each ball separated by a single space in a single line. Each number should be rounded to 2 digit after the decimal point.

    Sample Input

    2
    1 10 10 100
    2 10 10 100

    Sample Output

    4.95
    4.95 10.20

    Source

     

    THINKING

      如同POJ1852(ANTS)一样,当R=0时,可认为两个球碰撞后是互相交错而继续运动,当R=0时,处理方法相同,对于从下方开始的第i个球,在R=0的计算上加上d=2*r就好了。

    const g:real=10;
    
    var a:array[0..1000] of real;
        n,r,h,t,i,s:longint;
    
    function calc(x:longint):real;
    var d,tt:real;k:longint;
    begin
        if x<0 then exit(h);
        tt:=sqrt(2*h/g);
        k:=trunc(x/tt);
        if (k mod 2=0) then
            begin
                d:=x-k*tt;
                exit(h-g*d*d/2);
            end
        else
            begin
                d:=k*tt+tt-x;
                exit(h-g*d*d/2);
            end;
    end;
    
    procedure sort(l,r:longint);
          var
             i,j:longint;x,y:real;
          begin
             i:=l;
             j:=r;
             x:=a[(l+r) div 2];
             repeat
               while a[i]<x do
                inc(i);
               while x<a[j] do
                dec(j);
               if not(i>j) then
                 begin
                    y:=a[i];
                    a[i]:=a[j];
                    a[j]:=y;
                    inc(i);
                    j:=j-1;
                 end;
             until i>j;
             if l<j then
               sort(l,j);
             if i<r then
               sort(i,r);
          end;
    
    procedure main;
    var i:longint;
    begin
        readln(n,h,r,t);
        for i:=0 to n-1 do
            a[i]:=calc(t-i);
        sort(0,n-1);
        for i:=0 to n-1 do
            begin
                write((a[i]+2*r*i/100):0:2,' ');
            end;
    end;
    
    begin
        readln(s);
        for i:=1 to s do main;
    end.
    View Code
  • 相关阅读:
    [计算机网络] 互联网协议栈(TCP/IP参考模型)各层的主要功能及相应协议
    [计算机网络-应用层] P2P应用
    [剑指Offer] 45.扑克牌顺子
    [剑指Offer] 44.翻转单词顺序列
    [STL] 如何将一个vector赋给另一个vector
    最近更新少,是因为在用框架做项目
    转收藏:Git常用命令速查表
    CentOS常用指令
    CentOS修改服务器系统时间
    Javascript定时跳转
  • 原文地址:https://www.cnblogs.com/yangqingli/p/4889580.html
Copyright © 2011-2022 走看看