C. Round Corridor

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本文详细解析了 Codeforces 平台上编号为 1200C 的算法题目，该题目涉及在一个由内外两圈构成的圆形走廊中进行路径判断。文章通过分析走廊的结构和墙的位置，提出了一种有效的算法来判断从一个扇区到另一个扇区是否可达。使用 C++ 实现了解题代码，并展示了如何通过计算最大公约数和除以最大公约数后的商来判断路径可行性。

链接：https://codeforces.com/problemset/problem/1200/C

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

input

Copy

output

Copy

YES
NO
YES

Note

Example is shown on the picture in the statement.

题解：

#include<bits/stdc++.h>
using namespace std;
long long n,s=0,a,b,a1,a2,b1,b2;
int main()
{
	cin>>a>>b>>n;
	long long x=__gcd(a,b);
	long long p,q;
	p=a/x;
	q=b/x;
	while(n--)
	{
		cin>>a1>>a2>>b1>>b2;
		if(a1==1)
		{
			s=(a2-1)/p;
		}
		else
		{
			s=(a2-1)/q;
		}
		if(b1==1)
		{
			if((b2-1)/p==s)
			cout<<"YES"<<endl;
			else
			cout<<"NO"<<endl;
		}
		else
		{
			if((b2-1)/q==s)
			cout<<"YES"<<endl;
			else
			cout<<"NO"<<endl;
		}
	}
	    
}