【最大公约数】Round Corridor

题目：

Amugae is in a very large round corridor. The corridor consists of two areas. The inner area is equally divided by nn sectors, and the outer area is equally divided by mm sectors. A wall exists between each pair of sectors of same area (inner or outer), but there is no wall between the inner area and the outer area. A wall always exists at the 12 o’clock position.

The inner area’s sectors are denoted as (1,1),(1,2),…,(1,n)(1,1),(1,2),…,(1,n) in clockwise direction. The outer area’s sectors are denoted as (2,1),(2,2),…,(2,m)(2,1),(2,2),…,(2,m) in the same manner. For a clear understanding, see the example image above.

Amugae wants to know if he can move from one sector to another sector. He has qq questions.

For each question, check if he can move between two given sectors.

Input
The first line contains three integers nn, mm and qq (1≤n,m≤10181≤n,m≤1018, 1≤q≤1041≤q≤104) — the number of sectors in the inner area, the number of sectors in the outer area and the number of questions.

Each of the next qq lines contains four integers sxsx, sysy, exex, eyey (1≤sx,ex≤21≤sx,ex≤2; if sx=1sx=1, then 1≤sy≤n1≤sy≤n, otherwise 1≤sy≤m1≤sy≤m; constraints on eyey are similar). Amague wants to know if it is possible to move from sector (sx,sy)(sx,sy) to sector (ex,ey)(ex,ey).

Output
For each question, print “YES” if Amugae can move from (sx,sy)(sx,sy) to (ex,ey)(ex,ey), and “NO” otherwise.

You can print each letter in any case (upper or lower).

Example
inputCopy
4 6 3
1 1 2 3
2 6 1 2
2 6 2 4
outputCopy
YES
NO
YES

#include
#include
#include
#include
#include
#include
#include
using namespace std;
long long n,m,q;

long long gcd(long long a,long long b)
{
    return a==0?b:gcd(b%a,a);
}

int main()
{
    cin>>n>>m>>q;
    long long num=gcd(n,m);
    long long a1=0,b1=0;
    a1=n/num;          //一个联通的内外圈包含的内块
    b1=m/num;          //一个联通的内外圈包含的外块
    while(q--)
    {
        long long sx,sy,ex,ey;
        cin>>sx>>sy>>ex>>ey;
        sy--;                    //注意这里，联通圈求时需要
        ey--;
        long long a2=0;
        long long b2=0;
        if(sx==1)                //谁为1谁在内圈，查找它在第几个联通圈里，圈不同则不联通
            a2=sy/a1;
        else
            a2=sy/b1;
        if(ex==1)
            b2=ey/a1;
        else
            b2=ey/b1;
        if(a2==b2)
            cout<<"YES"<