zoukankan      html  css  js  c++  java
  • 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?'
    ':' ');
            }
        }
    
        return 0;
    }
    


    东北日出西边雨 道是无情却有情
  • 相关阅读:
    (难)Codeforces Round #406 (Div. 2) C题Berzerk(有向图博弈)解题报告
    jquery清空kindEditor
    处理用户误输入html标签引起的网站布局混乱
    自动闭合所有标签的方法,用于获得textBox等值的时候,标签都是闭合的
    .NET一个页面多个Button按钮事件避免数据验证控件RequiredFieldValidator
    基类包括字段(),但其类型()与控件()的类型不兼容
    关于ajax回调数据类型为Json,如何获得他的值
    如何查找Repeater控件中嵌套的控件
    网站功能
    WIN7系统IIS上发布站点后水印效果失效的解决方法
  • 原文地址:https://www.cnblogs.com/ccut-ry/p/7858337.html
Copyright © 2011-2022 走看看