hdu5114Collision
Collision
Time Limit: 15000/15000 MS (Java/Others) Memory Limit: 512000/512000 K (Java/Others)Total Submission(s): 26 Accepted Submission(s): 8
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.
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)
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.
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 5.0HintIn 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).
最后求一个k1>0的最小正整数解。用一下扩展欧几里得即可。
时间t=k1*x-（x0+x1）/2，有了t就可以算最后的坐标x了，注意模上2x，然后如果大于x，还要用2x减去这个数。同理可以算得y。
代码：
#include<cstdio>
#include<iostream>
#include<cstring>
#include<cmath>
#define ll long long
#define eps 1e-6
using namespace std;
ll myabs(ll x){
return x>=0?x:-x;
}
double fabs(double x){
return x>0?x:-x;
}
void extend_gcd(ll a,ll b,ll &x,ll &y,ll &q){
if(b==0){x=1;y=0;q=a;}
else{
extend_gcd(b,a%b,x,y,q);
ll tmp=x;x=y;y=tmp-a/b*y;
}
}
int main()
{
int t,cas=1;
ll x,y,x0,y0,x1,y1;
double tt,ansx,ansy;
cin>>t;
while(t--){
cin>>x>>y>>x0>>y0>>x1>>y1;
printf("Case #%d:\n",cas++);
if(x0==x1&&y0==y1) printf("%.1f %.1f\n",1.0*x0,1.0*y0);
else if(x0==x1){
tt=y-1.0*(y0+y1)/2;
ansy=y0+tt>y?2*y-y0-tt:y0+tt;
ansx=x0+tt>x?2*x-x0-tt:x0+tt;
printf("%.1f %.1f\n",ansx,ansy);
}
else if(y0==y1){
tt=x-1.0*(x0+x1)/2;
ansy=x0+tt>x?2*x-x0-tt:x0+tt;
ansx=y0+tt>y?2*y-y0-tt:y0+tt;
printf("%.1f %.1f\n",ansx,ansy);
}
else{
if((x0+x1-y0-y1)%2) {puts("Collision will not happen.");continue;}
ll k=(x0+x1-y0-y1)/2;
ll a,b,xt,yt,q;
a=x,b=-y;
extend_gcd(a,b,xt,yt,q);
if(k%q) {puts("Collision will not happen.");continue;}
else{
xt*=k/q;
ll z=xt/(y/q);
xt-=y/q*z;
if(xt<1) xt+=myabs(y/q);
tt=xt*x-(x0+x1)*1.0/2;
ll d=floor((x0+tt)/(2*x)+eps);
double tx,ty;
tx=x0+tt-2*x*d;
d=floor((y0+tt)/(2*y)+eps);
ty=y0+tt-2*y*d;
ansx=tx>x?2*x-tx:tx;
ansy=ty>y?2*y-ty:ty;
printf("%.1f %.1f\n",ansx,ansy);
}
}
}
return 0;
}