Collision（hdu5114）

Collision

Time Limit: 15000/15000 MS (Java/Others)    Memory Limit: 512000/512000 K (Java/Others)
Total Submission(s): 846    Accepted Submission(s): 200

Problem Description

Matt is playing a naive computer game with his deeply loved pure girl.

The playground is a rectangle with walls around. Two balls are put in different positions inside the rectangle. The balls are so tiny that their volume can be ignored. Initially, two balls will move with velocity (1, 1). When a ball collides with any side of the rectangle, it will rebound without loss of energy. The rebound follows the law of refiection (i.e. the angle at which the ball is incident on the wall equals the angle at which it is reflected).

After they choose the initial position, Matt wants you to tell him where will the two balls collide for the first time.

 

Input

The first line contains only one integer T which indicates the number of test cases.

For each test case, the first line contains two integers x and y. The four vertices of the rectangle are (0, 0), (x, 0), (0, y) and (x, y). (1 ≤ x, y ≤ 10⁵)

The next line contains four integers x₁, y₁, x₂, y₂. The initial position of the two balls is (x₁, y₁) and (x₂, y₂). (0 ≤ x₁, x₂ ≤ x; 0 ≤ y₁, y₂ ≤ y)

 

Output

For each test case, output “Case #x:” in the first line, where x is the case number (starting from 1). 

In the second line, output “Collision will not happen.” (without quotes) if the collision will never happen. Otherwise, output two real numbers x_c and y_c, rounded to one decimal place, which indicate the position where the two balls will first collide.

 

Sample Input

3

10 10

1 1 9 9

10 10

0 5 5 10

10 10
1 0 1 10

 

Sample Output

Case #1:
6.0 6.0

Case #2:
Collision will not happen.

Case #3:
6.0 5.0

Hint

In first example, two balls move from (1, 1) and (9, 9) both with velocity (1, 1), the ball starts from (9, 9) will rebound at point (10, 10) then move with velocity (−1, −1). The two balls will meet each other at (6, 6).

 思路：扩展欧几里德；

 首先可以知道，要相遇肯定在整数点，和半数点处，那么我们先将所有的都乘以2，为了防止小数。

 当x1 = x2的时候，那么无论啥时候，x1=x2；那么这个时候，只要求y1=y2的时刻，

  ty =(m-(y2+(m-y1)))/2+m-y1 = (2*n-(y1+y2))/2;那么根据ty 可以求得坐标；

 同理y1=y2;

 那么当x1！=x2&&y1!=y2的时候，

 得到t1 = tx+k1*n;t2  =ty+k2*m;

  那么t2 = t1;所以解两个同余方程即可，取最小的t；

  1 #include<stdio.h>
  2 #include<algorithm>
  3 #include<iostream>
  4 #include<stdlib.h>
  5 #include<string.h>
  6 #include<math.h>
  7 #include<queue>
  8 using namespace std;
  9 typedef long long LL;
 10 pair<LL,LL>exgcd(LL n,LL m);
 11 LL solve(LL x, LL y,LL n,LL m);
 12 LL gcd(LL n,LL m);
 13 int main(void)
 14 {
 15         LL n,m,k;
 16         int N;
 17         int __ca = 0;
 18         scanf("%d",&N);
 19         while(N--)
 20         {
 21                 scanf("%lld %lld",&n,&m);
 22                 LL x1,y1,x2,y2;
 23                 n*=2;
 24                 m*=2;
 25                 scanf("%lld %lld %lld %lld",&x1,&y1,&x2,&y2);
 26                 x1*=2;
 27                 y1*=2,x2*=2;
 28                 y2*=2;
 29                 printf("Case #%d:\n",++__ca);
 30                 if(x1==x2&&y1==y2)
 31                 {
 32                         printf("%.1f %.1f\n",x1/2,y1/2);
 33                 }
 34                 else if(x1==x2)
 35                 {
 36                         LL ty = (2*m-(y1+y2))/2;
 37                         LL yy = min(y1,y2)+ty;
 38                         LL xx = min(x1,x2)+ ty;
 39                         if((xx/n)%2==0)
 40                         {
 41                                 xx = xx%n;
 42                         }
 43                         else
 44                         {
 45                                 xx = ((n-xx)%n+n)%n;
 46                                 if(xx == 0)xx = n;
 47                         }
 48                         if((yy/m)%2==0)
 49                         {
 50                                 yy = yy%m;
 51                         }
 52                         else
 53                         {
 54                                 yy = ((m-yy)%m+m)%m;   if(yy == 0)yy = m;
 55                         }
 56                         printf("%.1lf %.1lf\n",xx/2.0,yy/2.0);
 57                 }
 58                 else if(y1 == y2)
 59                 {
 60                         LL tx = (2*n-(x1+x2))/2;
 61                         LL yy = min(y1,y2)+tx;
 62                         LL xx = min(x1,x2)+tx;
 63                         if((yy/m)%2==0)
 64                         {
 65                                 yy = yy%m;
 66                         }
 67                         else
 68                         {
 69                                 yy = ((m-yy)%m+m)%m;
 70                                 if(yy == 0)yy = m;
 71                         }
 72                         if((xx/n)%2==0)
 73                         {
 74                                 xx = xx%n;
 75                         }
 76                         else
 77                         {
 78                                 xx = ((n-xx)%n+n)%n;
 79                                 if(xx == 0)xx = n;
 80                         }
 81                         printf("%.1lf %.1lf\n",xx/2.0,yy/2.0);
 82                 }
 83                 else
 84                 {
 85                         LL ty = (2*m-(y1+y2))/2;
 86                         LL tx = (2*n-(x1+x2))/2;
 87                         LL ask = solve(tx,ty,n,m);
 88                         if(ask==1e18)
 89                                 printf("Collision will not happen.\n");
 90                         else
 91                         {
 92                                 LL yy = y1+ask;
 93                                 if((yy/m)%2==0)
 94                                 {
 95                                         yy = yy%m;
 96                                         //if(yy == 0)yy = m;
 97                                 }
 98                                 else
 99                                 {
100                                         yy = ((m-yy)%m+m)%m;
101                                         if(yy == 0)yy = m;
102                                 }
103                                 LL xx = x1+ask;
104                                 if((xx/n)%2==0)
105                                 {
106                                         xx = xx%n;
107                                 }
108                                 else
109                                 {
110                                         xx = ((n-xx)%n+n)%n;
111                                         if(xx == 0)xx = n;
112                                 }
113                                 printf("%.1lf %.1lf\n",xx/2.0,yy/2.0);
114                         }
115                 }
116
117         }
118         return 0;
119 }
120 pair<LL,LL>exgcd(LL n,LL m)
121 {
122         if(m==0)
123                 return make_pair(1,0);
124         else
125         {
126                 pair<LL,LL>ak = exgcd(m,n%m);
127                 return make_pair(ak.second,ak.first-(n/m)*ak.second);
128         }
129 }
130 LL solve(LL x, LL y,LL n,LL m)
131 {
132         LL cc = n;
133         LL c = x-y;
134         LL gc = gcd(n,m);
135         if(c%gc)return 1e18;
136         else
137         {
138                 c/=gc;
139                 n/=gc;
140                 m/=gc;
141                 pair<LL,LL>ak = exgcd(n,m);
142                 LL x0 = (ak.first*c%m+m)%m;
143                 LL lcm = (LL)m*cc;
144                 x = x-cc*x0;
145                 x = x%lcm+lcm;
146                 x%=lcm;
147                 return x;
148         }
149 }
150 LL gcd(LL n,LL m)
151 {
152         if(m==0)return n;
153         else return gcd(m,n%m);
154 }