[POJ 2461] Billiard

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Billiard

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 1362		Accepted: 842

Description

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：一个长宽分别为ab的矩形桌子中间有一个球，朝着某个方向以某个速度发射出去，在s时间恰好与水平方向碰撞n次,垂直方向碰撞m次,求发射角度与速度。

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cmath>
#include<cstring>
using namespace std;
#define INF 0x3f3f3f3f
#define PI acos(-1.0)

int main()
{
    int n,m;
    double a,b,s;
    while(scanf("%lf%lf%lf%d%d",&a,&b,&s,&m,&n)!=EOF)
    {
        if(a+b+s+m+n==0) break;
       double angel=atan2(b*n,a*m)*180/PI;
       double v=sqrt(a*a*m*m+b*b*n*n)/s;
       printf("%.2f %.2f
",angel,v);
    }
    return 0;
}