POJ3684 弹性碰撞

Physics Experiment

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 2758		Accepted: 987		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 N, H, R, T.
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

题意：n个半径为r的球从高度为h的平台依次掉下，求t时各个球的高度。

分析：这完全是物理题好吗？完全弹性碰撞，而且各个球相同。

首先分析一个球，设

t = sqrt(2h/g)

k = int(T/t)

如果k为偶数，这个球正在下降状态，h = H-g*(T-kt)*(T-kt)/2

如果k为奇数，这个球正在上升状态，h = H-g(kt+t-T)*(kt+t-T)/2

考虑到完全相同球弹性碰撞时交换速度，可以看作没有碰撞，因此T时可以直接根据时间求出个球的位置。

而球只是将运动状态传递了，本身是卡在中间不能动的。所以高度从下到上依次变大。排序输出就好了。

#include <cstdio>
#include <cstdlib>
#include <iostream>
#include <stack>
#include <queue>
#include <algorithm>
#include <cstring>
#include <string>
#include <cmath>
#include <vector>
#include <bitset>
#include <list>
#include <sstream>
#include <set>
#include <functional>
using namespace std;

#define INF 0x3f3f3f3f
#define MAX 100
typedef long long ll;
int C,N,H,R,T;
double y[105];

double cal(int T1){
	if(T1 < 0) return H;
	double t = sqrt(2*H/10.0);
	int k = int(T1/t);
	double tmp = ((k%2 == 0)?(T1-k*t):(k*t+t-T1));
	return H-tmp*tmp*10.0/2;
}

int main()
{
#ifndef ONLINE_JUDGE
	freopen("in.txt","r",stdin);
	freopen("out.txt","w",stdout);
#endif
	scanf("%d",&C);
	while(C--) {
	    scanf("%d%d%d%d",&N,&H,&R,&T);
	    for (int i = 0; i < N; ++i) y[i] = cal(T-i);
	    sort(y,y+N);
	    for (int i = 0; i < N; ++i) printf("%.2f%c",y[i]+2*R*i/100.0,((i+1 == N)?'\n':' '));
	}
}