Codeforces Round #484 (Div. 2) E. Billiard（exgcd）

题目链接：http://codeforces.com/contest/982/problem/E

https://www.cnblogs.com/zhouzhendong/p/9055728.html

代码：

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
ll exgcd(ll a,ll b,ll &x,ll &y)//扩展gcd
{
	ll r,t;
	if(b==0)
	{
		x=1;y=0;
		return a;
	}
	r=exgcd(b,a%b,x,y);
	t=x;x=y;
	y=t-a/b*y;
	return r;
}

ll cal(ll a,ll b,ll c)//求解ax+by=c的x的最小整数解
{
    ll x,y,ret,gcd;
    gcd=exgcd(a,b,x,y);
    if(c%gcd!=0)
	{
		puts("-1");
		exit(0);
	}
    x*=c/gcd;b/=gcd;
    if(b<0) b=-b;
    ret=x%b;
    if(ret<=0) ret+=b;
    return ret;
}

int main()
{
	//freopen("in.txt","r",stdin);
	//freopen("out.txt","w",stdout);
	int n,m,x,y,vx,vy;
	scanf("%d%d%d%d%d%d",&n,&m,&x,&y,&vx,&vy);
	if(vx==0)
	{
		if(x==0||x==n) return 0*printf("%d %d\n",x,vy>0?m:0);
		else return 0*puts("-1");
	}
	if(vy==0)
	{
		if(y==0||y==m) return 0*printf("%d %d\n",vx>0?n:0,y);
		else return 0*puts("-1");
	}
	int fx=0,fy=0;
	if(vx==-1) fx=1,x=n-x;
	if(vy==-1) fy=1,y=m-y;
	ll a=cal(n,m,(x-y));
	ll b=-((x-y)-a*n)/m;
	ll ansx=n,ansy=m;
	if(a%2==0) ansx=n-ansx;
	if(b%2==0) ansy=m-ansy;
	if(fx) ansx=n-ansx;
	if(fy) ansy=m-ansy;
	printf("%lld %lld\n",ansx,ansy);
	return 0;
}