Physics Experiment poj 3684 弹性碰撞

Language: Default Physics Experiment Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 1107 Accepted: 380 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：将N个半径为R的球放入一个圆桶（圆桶口径刚好放入一个球），将圆桶竖直放着，最下端距离地面H高度，让球每隔一秒自由下落，求T时刻各个球距离地面的高度。

思路：所有的球都一样可以忽视它们的碰撞，视为互相穿过继续运动。这样就可以分别单独求出每个球T时刻的高度后排序就是答案了。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#pragma comment (linker,"/STACK:102400000,102400000")
#define maxn 105
#define MAXN 2005
#define mod 1000000009
#define INF 0x3f3f3f3f
#define pi acos(-1.0)
#define eps 1e-6
#define lson rt<<1,l,mid
#define rson rt<<1|1,mid+1,r
typedef long long ll;
using namespace std;

int N,T;
double H,R;
double ans[maxn];

double solve(int T)
{
    if (T<0) return H;
    double t=sqrt((2*H)/10.0);
    int k=(int)T/t;
    if (k%2)
        return H-5.0*(k*t+t-T)*(k*t+t-T);
    else
        return H-5.0*(T-k*t)*(T-k*t);
}

int main()
{
    int c;
    scanf("%d",&c);
    while (c--)
    {
        scanf("%d%lf%lf%d",&N,&H,&R,&T);
        for (int i=0;i<N;i++)
            ans[i]=solve(T-i);
        sort(ans,ans+N);
        printf("%.2f",ans[0]);
        for (int i=1;i<N;i++)
            printf(" %.2f",ans[i]+2*R*i/100.0);
        printf("\n");
    }
    return 0;
}
/*
2
1 10 10 100
2 10 10 100
*/

代码：