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 ), 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 =  m/s2.

Input

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

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  digit after the decimal point.

Sample Input


Sample Output

4.95
4.95 10.20

AC代码:

 #include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
#define N 106
#define g 10.0
int n,h,r,T;
double y[N];
double calc(int T){
if(T<) return h;
double t = sqrt(*h/g);
int k = (int)(T/t);
if(k%==){
double d = T-k*t;
return h-g*d*d/;
}else{
double d = k*t+t-T;
return h-g*d*d/;
}
}
void solve(){
for(int i=;i<n;i++){
y[i]=calc(T-i);
}
sort(y,y+n);
for(int i=;i<n;i++){
printf("%.2lf ",y[i]+*r*i/100.0);
}
printf("\n");
}
int main()
{
int t;
scanf("%d",&t);
while(t--){
scanf("%d%d%d%d",&n,&h,&r,&T);
solve();
}
return ;
}

poj 3684 Physics Experiment(数学,物理)的更多相关文章

  1. POJ 3684 Physics Experiment(弹性碰撞)

    Physics Experiment Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 2936   Accepted: 104 ...

  2. poj 3684 Physics Experiment 弹性碰撞

    Physics Experiment Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 1489   Accepted: 509 ...

  3. POJ 3684 Physics Experiment

    和蚂蚁问题类似. #include<cstdio> #include<cstring> #include<cmath> #include<vector> ...

  4. POJ3684 Physics Experiment 【物理】

    Physics Experiment Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 1031   Accepted: 365 ...

  5. Physics Experiment(POJ 3684)

    原题如下: Physics Experiment Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 3583   Accepte ...

  6. poj 3684

    Physics Experiment Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 784   Accepted: 266 ...

  7. [POJ3684]Physics Experiment

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

  8. sqlserver 行转列 语文,数学,物理,化学

    数据库查询行转列 1.原数据库值 stdname stdsubject result 张三 语文 张三 数学 张三 物理 李四 语文 李四 数学 李四 物理 李四 化学 李四 化学 2.要得到如下表 ...

  9. POJ:3684-Physics Experiment(弹性碰撞)

    Physics Experiment Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 3392 Accepted: 1177 Sp ...

随机推荐

  1. hdu 4586 Play the Dice(概率dp)

    Problem Description There is a dice with n sides, which are numbered from 1,2,...,n and have the equ ...

  2. java 正则表达式获取值

    @Test public void testtest() { String test = "hahahhehe sendCode\":\"12367890123rsdfs ...

  3. pyqt下拉菜单和打开指定的内容(或者exe,doc,ppt,url等内容)

    #下拉菜单及显示 def _action(self): menu=QMenu(self) menu1=menu.addMenu(u'文件') menu11=menu1.addAction(u'新建任务 ...

  4. DOS命令大全--具体解释

    在Linux和Windows下都能够用nslookup命令来查询域名的解析结果 DOS命令大全一)MD--建立子文件夹 1.功能:创建新的子文件夹 2.类型:内部命令 3.格式:MD[盘符:][路径名 ...

  5. DB2查询当前时间与指定时间的时间差(相隔的秒数)

    DB2查询当前时间与指定时间的时间差(相隔的秒数). 例子:“拍品表 auct_item”中有个“结束时间 end_date”的字段,求结束时间与当前时间的间隔秒数. select  (DAYS(a. ...

  6. [RxJS] Basic DOM Rendering with Subscribe

    While frameworks like Angular 2 and CycleJS provides great ways to update the DOM and handle subscri ...

  7. Sublime text3配置LiveReload

    Tip: LiveReload是很棒的插件,可以在浏览器中实时预览,但是在Sublime text3里,从Package Control中安装的LiveReload是无法使用的,但是可以选择手动安装解 ...

  8. android sqlite数据库封装 实现crud

    android常用的数据保存方式有文件.sharepreferences.数据库.网络.contentprovider集中方式. 文件存储方式,经常使用在缓存整个页面数据,比如电子书内容.html数据 ...

  9. mp4文件格式之fragment mp4

    目前网络上对mp4文件格式的总结已经相当多了,我就不在这里抄别人总结的东西了,想入门的话百度一下会有许多这方面的文章.这里介绍一下其他文章中很少涉及的一种mp4文件结构,即fragment mp4,也 ...

  10. asp.net mvc 页面缓存

    在任务中需要实现点击浏览器back按钮,加载的前一页面需要强制刷新. 想要在前端通过js来绑定数据实现,但是觉得太麻烦,还是用另一种方式来解决: 不缓存该页面. 简单易懂: Response.Cach ...