hdu5114 Collision (扩展欧几里德+思维)

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 ≤ 10 ⁵) 

The next line contains four integers x ₁, y ₁, x ₂, y ₂. The initial position of the two balls is (x ₁, y ₁) and (x ₂, y ₂). (0 ≤ x ₁, x ₂ ≤ x; 0 ≤ y ₁, y ₂ ≤ 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

310 101 1 9 910 100 5 5 1010 101 0 1 10

Sample Output

Case #1:6.0 6.0Case #2:Collision will not happen.Case #3:6.0 5.0

#include <iostream>#include <stdio.h>#include <stdlib.h>#include <string.h>#include <algorithm>#include <math.h>using namespace std;typedef long long LL;LL gcd(LL a,LL b){     return b?gcd(b,a%b):a;}void Ex_gcd(LL a, LL b, LL &x, LL &y){    if(b == 0)//递归出口    {        x = 1;        y = 0;        return;    }    LL x1, y1;    Ex_gcd(b, a%b, x1, y1);    x = y1;    y = x1-(a/b)*y1;}int main(){    LL  x,y,x1,x2,y1,y2;    int t;    scanf("%d",&t);    for(int cas=1;cas<=t;cas++)    {        scanf("%lld%lld",&x,&y);        scanf("%lld%lld%lld%lld",&x1,&y1,&x2,&y2);        double ansx,ansy;        printf("Case #%d:\n",cas);        if(x1==x2&&y1==y2)        {            ansx=x1;ansy=y1;        }        else if(x1==x2&&y1!=y2)        {            if(y1>y2)swap(y1,y2);            double t=2*y-y1-y2;            ansx=t/2+x1;            ansy=t/2+y1;        }        else if(x1!=x2&&y1==y2)        {            if(x1>x2)swap(x1,x2);            double t=2*x-x1-x2;            ansx=t/2+x1;            ansy=t/2+y1;        }        else        {            x*=2;y*=2;x1*=2;x2*=2;y1*=2;y2*=2;///用扩欧要保证整数，所有数据乘2            LL z=gcd(x,y);            LL quanx,quany;            LL c=(2*y-y1-y2-2*x+x1+x2)/2;            if(c%z!=0){printf("Collision will not happen.\n");continue;}            else            {                LL xx=x/z,yy=y/z;                c/=z;                Ex_gcd(xx,yy,quanx,quany);///用扩欧要保证整数，所有数据乘2                quanx*=c;                quanx=(quanx%yy+yy)%yy;///最小正整数解                LL shijian=quanx*x+(2*x-x1-x2)/2;///两周期重合时间+从（X1,Y1）到重合点时间               ansx=(x1+shijian)%(2*x);///一去一回2x算一个周期               ansy=(y1+shijian)%(2*y);               //printf("  time= %d  %.1lf   %.1lf\n",shijian,ansx,ansy);                 if(ansx > x) ansx = 2*x - ansx;///超过边界对称回来                if(ansy > y) ansy= 2*y - ansy;                ansx/=2;ansy/=2;            }        }        printf("%.1lf %.1lf\n",ansx,ansy);    }    return 0;}