UVA10387矩形内的无损碰撞

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这题利用分速度来理解，由于是无损碰撞所以速度大小不变，由于每次碰撞与水平方向的角度都是一样，所以分解为垂直方向速度与水平

方向速度，速度均不变，由于最后会回到起点，所以可以利用勾股定理求出实际行走的距离，水平方向距离为a*m，垂直距离方向为b*n;

然后由此求出距离，根据速度不变性原理和分解原理，速度初始角度即为行驶的水平距离和垂直距离的角度，利用atan即可求出然后注意转换成

角度。

#include<cstdio>#include<iostream>#include<vector>#include<queue>#include<algorithm>#include<string>#include<cstdlib>#include<map>#include<set>#include<cmath>#include<cstring>#include<cctype>#include<climits>#include<memory>#include<climits>#include<cstdlib>using namespace std;#define LL long long#define INT (1LL<<62);const double eps=1e-6;const double pi=4*atan(1.0);int dcmp(double x){    return fabs(x)<eps?0:(x>0?1:-1);}/*struct point{    double x,y;    point() {};    point(double x,double y):x(x),y(y){};    point operator + (point b)    {        return point(x+b.x,y+b.y);    }    point operator - (point b)    {        return point(x-b.x,y-b.y);    }    point operator / (double b)    {        return point(x/b,y/b);    }    point operator * (double b)    {        return point(x*b,y*b);    }    void in()    {        cin>>x>>y;    }};double dot(point a,point b){    return a.x*b.x+a.y*b.y;}double length(point a){    return sqrt(dot(a,a));}double dd(point a,point b){  return sqrt((b.x-a.x)*(b.x-a.x)+(b.y-a.y)*(b.y-a.y));}point zx(point a){   double d=length(a);   return point(-a.y/d,a.x/d);}*/int main(){    double a,b,s,m,n;    while(cin>>a>>b>>s>>m>>n&&(a||b||s||m||n))    {        double x=a*m;        double y=b*n;        printf("%.2f %.2f\n",atan(y/x)*180/pi,sqrt(x*x+y*y)/s);    }    return 0;}

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