Uva 11880 - Ball in a Rectangle

你为何这么叼！

标程简直丧心病狂！

Ball in a Rectangle

Time limit: 1.000 seconds

 

Input: Standard Input Output: Standard Output

� There is a rectangle on the cartesian plane, with bottom-left corner at (0,0) and top-right corner at (L, W). There is a ball centered at (x, y), with radius=R, shown below

At time 0, the ball starts to move along a ray with polar angle a (the angle from positive x-axis to the ray, rotating counter-clockwise). When hitting the rectangle boundary, the reflex angle always equals to the incidence angle. The ball's velocity is always v (i.e. it never changes when hitting the rectangle). Where is the center of the ball at time s?

Input 

There will be at most 25 test cases, each contains a line with 8 integers L,W,x,y,R,a,v,s (100L,W10⁹, 1R5, RxL - R, RyW - R, 0a < 360, 1v, s10⁹), as stated above. The input terminates with L = W = x = y = R = a = v = s = 0, which should not be processed.

Output 

For each test case, output a line containing two floating-point numbers x, y, rounded to two decimal points, indicating that the center of ball will be at (x, y) at time s.

Sample Input 

100 100 80 10 5 90 2 23
110 100 70 10 5 180 1 9999
0 0 0 0 0 0 0 0

Sample Output 

80.00 56.00
71.00 10.00

 

Problemsetter: Rujia Liu, Special Thanks: Yiming Li, Shamim Hafiz & Sohel Hafiz

 

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
using namespace std;
int n,m;
double l,w,s,a,v,x,y,r,t;
int main(){
    while(scanf("%lf%lf%lf%lf%lf%lf%lf%lf",&l,&w,&x,&y,&r,&a,&v,&t)){
        if(l+w+x+y+r+a+v+t==0)break;
        l-=2*r;w-=2*r;x-=r;y-=r;a*=acos(0.0)/90.0;
        double dx=fmod(fmod(v*t*cos(a),2*l)+2*l,2*l);
        double dy=fmod(fmod(v*t*sin(a),2*w)+2*w,2*w);
        if(x+dx<=l)x+=dx;else if(x+dx<=2*l)x=l-(x+dx-l);else x=x+dx-2*l;
        if(y+dy<=w)y+=dy;else if(y+dy<=2*w)y=w-(y+dy-w);else y=y+dy-2*w;
        printf("%.2lf %.2lf\n",x+r,y+r);
    }
    return 0;
}