POJ3684-Physices Experiment【弹性碰撞】

原题链接
Physics Experiment
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 2182 Accepted: 756 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
Source

POJ Founder Monthly Contest – 2008.08.31, Simon
题意：有一个竖直的管子内有n个小球，小球的半径为r，最下面的小球距离地面h高度，让小球每隔一秒自由下落一个，小球与地面，小球与小球之间可视为弹性碰撞，让求t时间后这些小球的分布
思路：这是一个弹性碰撞的题，有一个中心思想就是他们的碰撞可以视为没有碰撞，只是穿过了对方，而这种思路适用于没有半径的时候，其实有半径的时候也可以这样来解决。详细见博客。博客链接
下面附上我的代码

#include <algorithm>#include <iostream>#include <utility>#include <sstream>#include <cstring>#include <cstdio>#include <vector>#include <queue>#include <stack>#include <cmath>#include <map>#include <set>using namespace std;typedef long long ll;const int MOD = int(1e9) + 7;//int MOD = 99990001;const int INF = 0x3f3f3f3f;const ll INFF = (~(0ULL)>>1);const double EPS = 1e-9;const double OO = 1e20;const double PI = acos(-1.0); //M_PI;const int fx[] = {-1, 1, 0, 0};const int fy[] = {0, 0, -1, 1};const int maxn=10000 + 5;const double g=10.0;int n,h,r,t;double loc[maxn];//记录最终的位置double calc(int T){    if(T<0) return h;//如果他还没有下落那么就返回第一个小球的位置    double temp=sqrt(2*h/g);//算出一次完整的下落所需的时间    int k=(int)(T/temp);//得到经历了几次完整的下落或者上升的过程    if(k&1){//此时属于上升的过程        double temp2=k*temp + temp -T;        return h-g/2*temp2*temp2;    }    else{//此时属于下降的过程        double temp2=T-k*temp;        return h-g/2*temp2*temp2;    }}int main(){        int kase;        cin >> kase;        while(kase--){            cin >> n >> h >> r >> t;            for(int i=0;i<n;i++)                loc[i]=calc(t-i);//算出t时刻第i个小球的位置            sort(loc,loc+n);//按照最后的位置进行排序来确定这些小球的位置            for(int i=0;i<n;i++)                printf("%.2f%c",loc[i] + 2 * r * i / 100.0,i+1==n ? '\n' : ' ' );        }        return 0;}/*关于r>0的情况下互换小球位置的分析http://www.bubuko.com/infodetail-1315939.html*/