Ural 1286. Starship Travel

1286. Starship Travel

Time limit: 1.0 second
Memory limit: 64 MB

It is well known that a starship equipped with class B hyperengine is able to travel from any planet to any other planet. But your starship got severe damage in the last travel and now its movement ability is limited. The starship’s technician determined that with the damaged hyperengine the vehicle can move from a point with coordinates (i,j) only to a point from the following list: (i+q, j+p), (i−q, j+p), (i+q, j−p), (i−q, j−p), (i+p, j+q), (i−p, j+q), (i+p, j−q), (i−p, j−q) (all the coordinates here are integers and are given in the standard intergalaxy system). Help the captain of your ship to find out if the ship is able to reach the destination planet on its own or a repair ship must be called.

Input

The first line contains two integers p and q (the two remaining discrete power rates of the damaged hyperengine) separated with a space. The second line contains the coordinates of the point where the spaceship is now. The third line contains the coordinates of the destination planet. The numbers in the second and third lines are also separated with spaces. All the numbers are integers and do not exceed 2·109 in absolute value.

Output

If the commander can move the damaged starship to the destination planet, write ‘YES’. Write ‘NO’ if a repair ship must be called.

Samples

input	output
4 6 0 0 10 10	YES
4 6 0 0 9 9	NO

Problem Author: Alexander Klepinin
Problem Source: USU Personal Contest 2004

一个任意大的棋盘，一个Knight每次的移动可以从(i,j)到(i+q, j+p), (i−q, j+p),(i+q, j−p), (i−q, j−p),
 (i+p, j+q), (i−p, j+q),(i+p, j−q), (i−p, j−q)  中的任意一个。给定初始坐标(sX,sY) 问Knight是否可以移动到(dX,dY)？(所有的整数的绝对值均<=2·10^9 )。

数形结合

#include <iostream>
#include <cstdio>
using namespace std;
long long gcd(long long a,long long b)
{
    return b?gcd(b,a%b):a;
}
int main()
{
    long long p,q,x,y,x1,x2,y1,y2;
    while(cin>>p>>q>>x1>>y1>>x2>>y2)
    {
        x=x2-x1;
        y=y2-y1;
        if(x==0&&y==0)
            puts("YES");
        else if(p==0&&q==0)
            puts("NO");
        else
        {
            long long n=p*x-q*y;
            long long g=gcd(2*p*q,gcd(p*p-q*q,p*p+q*q));
            if(n%g==0)
                puts("YES");
            else
                puts("NO");
        }
    }
    return 0;
}