ural 1286

题目：http://acm.timus.ru/problem.aspx?space=1&num=1286

题目要求 p * x1 + q * y1 = abs(ex - sx) && p * x2 + q * y2 = abs(ey - sy)同时成立，设 d = gcd(p,q) ,dx = abs(ex - sx), dy = (ey - sy)，如果（dx % d || dy % d ）成立，那么从开始点一定到不了终点，否则 dx /= d, dy /= d, p /= d, q /= d；分情况讨论，详见此出http://shaidaima.com/source/view/7789

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 1 ll gcd(ll a,ll b)
 2 {
 3     if(!b) return a;
 4     else return gcd(b,a % b);
 5 }
 6 ll tfab(ll x)
 7 {
 8     if(x < 0) return -x;
 9     else return x;
10 }
11 int main()
12 {
13     ll p,q;
14     ll sx,sy,ex,ey;
15     //freopen("data.txt","r",stdin);
16     while(cin>>p>>q>>sx>>sy>>ex>>ey)
17     {
18         ll d = gcd(p,q);
19         ll dx = tfab(sx - ex);
20         ll dy = tfab(sy - ey);
21         if(dx == 0 && dy == 0)
22         {
23             printf("YES\n");continue;
24         }
25         if(!p && !q)
26         {
27             if(dx || dy) printf("NO\n"); continue;
28         }
29         if(dx % d || dy % d)
30         {
31             printf("NO\n"); continue;
32         }
33         dx /= d, dy /= d, p  /= d, q /= d;
34         if((p % 2 == 0 && q % 2) || (p % 2 && q % 2 == 0))
35         {
36             printf("YES\n");continue;
37         }
38         if(p % 2 && q % 2)
39         {
40             if((dx % 2 == 0 && dy % 2 == 0) || (dx % 2 && dy % 2)) printf("YES\n");
41             else printf("NO\n");
42             continue;
43         }
44     }
45     return 0;
46 }