10387 - Billiard_billiard table compute-优快云博客

本文链接：https://blog.youkuaiyun.com/sio__five/article/details/9565957

本文探讨了一种经典的物理反弹问题，在给定的长宽尺寸的台球桌上，球从中心出发并经历多次弹性碰撞后返回原点。通过数学计算确定球的发射角度和初始速度。假设所有碰撞为理想弹性，速度的水平和垂直分量保持不变。

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Problem A: Billiard

In a billiard table with horizontal side a inches and vertical side b inches, a ball is launched from the middle of the table. After s > 0 seconds the ball returns to the point from which it was launched, after having made m bounces off the vertical sides and n bounces off the horizontal sides of the table. Find the launching angle A (measured from the horizontal), which will be between 0 and 90 degrees inclusive, and the initial velocity of the ball.

Assume that the collisions with a side are elastic (no energy loss), and thus the velocity component of the ball parallel to each side remains unchanged. Also, assume the ball has a radius of zero. Remember that, unlike pool tables, billiard tables have no pockets.

Input

Input consists of a sequence of lines, each containing five nonnegative integers separated by whitespace. The five numbers are: a, b, s, m, and n, respectively. All numbers are positive integers not greater than 10000.

Input is terminated by a line containing five zeroes.

Output

For each input line except the last, output a line containing two real numbers (accurate to two decimal places) separated by a single space. The first number is the measure of the angle A in degrees and the second is the velocity of the ball measured in inches per second, according to the description above.

Sample Input

100 100 1 1 1
200 100 5 3 4
201 132 48 1900 156
0 0 0 0 0

Sample Output

45.00 141.42
33.69 144.22
3.09 7967.81

题意：一张长a，宽b的台球桌，从中心发射一枚台球，经过若干次弹性碰撞后，又回到了中心。其中与宽边碰了m次，与长边碰了n次，并且话费了s秒的时间。求发射角（与水平边，长边的夹角），以及发射速度。

由于每次碰撞之后，相对水平和垂直的速度绝对值不变，而每一次碰撞到下次碰撞走过的距离都是一个长边，或是一个宽边。所以这道题就可以直接化为简单的计算题

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#define PI acos(-1.0)
using namespace std;
int main ()
{
    long long a,b,s,m,n;
    double A,v;
    while(cin>>a>>b>>s>>m>>n)
    {
        if (a==0) break;
        long long L,H;
        double d;
        L=a*m;
        H=b*n;
        d=L*L+H*H;
        d=sqrt(d);
        v=d/s;
        A=atan(H/(L*1.0))*180/PI;
        printf("%.2lf %.2lf\n",A,v);
    }
    return 0;
}