c++ - 【acm】关于完整遍历一个圈中所有点的问题，确定间隔和圈上点的个数，为什么判断gcd（m,n） == 1 就可以判断遍历完整?

Question

兔子的烦恼(一）
时间限制：1000 ms | 内存限制：65535 KB
描述
在一座山上有n个山洞（每个洞从0~n-1分别编号），这山上有一只兔子躲在山洞里，有一只狼，从编号为0的山洞开始入洞，每隔m-1个洞，再次入洞抓兔子，现在问，兔子能否避免这场灾难？
例如：m=2 and n=6, 狼将要进入的山洞号为：0,2,4,0;
输入
有多组测试数据，不超过1000组;

每组有两个整数m,n(0<m,n<=1000);
注意：程序以文件结束符“EOF”结束输入。
输出
兔子能避免这场灾难输出YES，否则输出NO;
样例输入
1 2
2 2
样例输出
NO
YES

Answer 1

Because if two numbers are not mutually prime, you can know that they will have a common factor $k$, there will be $n = ak, m = bk$, then all possible positions should be $cn- dm = cak - dbk = k (ca - bd) $ can know that all possible positions must be multiples of this common factor $k$. Obviously not all locations can be traversed.
If the two numbers are relatively prime, $k$ here is 1, and obviously all positions can be traversed.