弹性碰撞 poj 3684

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 ： 每隔一秒会释放一个小球，问最终所有小球底端距离地面的位置高度。
思路 ：
　　1 . 由于是发生弹性碰撞，即小球发生碰撞后，速度反向，大小不变，因此当两个小球发生
　　2 .先考虑 小球的半径为 0 的时候，即所有的小球都从同一点释放出去，只是时间不同，计算出所有小球的高度，排序后即为所求
　　3 . 若有半径，上面的求一定比最下面的球多 2*r*i 的重力势能，因此只需要再加上此高度即可
　　
推荐博客 ：http://www.cnblogs.com/smilesundream/p/5134406.html

代码 ：

int n, h, r, t;

double arr[105];

double cal(int t){

    if (t < 0) return 1.0*h;

    double t1 = sqrt(2.0*h/10.0);

    int k = t / t1;

    if (k & 1){

        return  1.0*h - 5.0*((k+1)*t1 - t)*((k+1)*t1 - t);

    }

    else {

        return 1.0*h - 5.0*(t - k*t1)*(t - k*t1);

    }

}

int main() {

    //freopen("in.txt", "r", stdin);

    //freopen("out.txt", "w", sttout);

    int T;

    cin >> T;

    while(T--){

        scanf("%d%d%d%d", &n, &h, &r, &t);

        for(int i = 0; i < n; i++){

            arr[i] = cal(t - i);

        }

        sort(arr, arr+n);

        for(int i = 0; i < n; i++){

            printf("%.2lf", arr[i]+2.0*r*i/100.0);

            printf("%c", i+1 == n?'\n':' ');

        }

    }

    return 0;

}

巴特西

弹性碰撞 poj 3684

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