数学--数论--HDU 2104 丢手绢（离散数学 mod N+ 剩余类 生成元）+（最大公约数）

The Children’s Day has passed for some days .Has you remembered something happened at your childhood? I remembered I often played a game called hide handkerchief with my friends.

Now I introduce the game to you. Suppose there are N people played the game ,who sit on the ground forming a circle ,everyone owns a box behind them .Also there is a beautiful handkerchief hid in a box which is one of the boxes .

Then Haha(a friend of mine) is called to find the handkerchief. But he has a strange habit. Each time he will search the next box which is separated by M-1 boxes from the current box. For example, there are three boxes named A,B,C, and now Haha is at place of A. now he decide the M if equal to 2, so he will search A first, then he will search the C box, for C is separated by 2-1 = 1 box B from the current box A . Then he will search the box B ,then he will search the box A.

So after three times he establishes that he can find the beautiful handkerchief. Now I will give you N and M, can you tell me that Haha is able to find the handkerchief or not. If he can, you should tell me “YES”, else tell me “POOR Haha”.

Input

There will be several test cases; each case input contains two integers N and M, which satisfy the relationship: 1<=M<=100000000 and 3<=N<=100000000. When N=-1 and M=-1 means the end of input case, and you should not process the data.

Output

For each input case, you should only the result that Haha can find the handkerchief or not.

Sample Input

3 2

-1 -1

Sample Output

YES

每次加MmodN，如果每次加N能遍历整个集合的话，那他一定是mod N+剩余类的生成元。生成元的条件是M与N互质。

import java.util.Scanner;

public class Main {
	static int gcd(int a, int b) {
		if (b == 0)
			return a;
		else
			return gcd(b, a % b);
	}

	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		while (in.hasNext()) {
			int a = in.nextInt();
			int b = in.nextInt();
			if (a == -1 && b == -1) {
				break;
			}
			if (gcd(a, b) == 1)
				System.out.println("YES");
			else System.out.println("POOR Haha");
		}
		in.close();
	}
}