Codeforces Round #292 (Div. 2) -- B. Drazil and His Happy Friends

浏览数：30 / 时间：2015年06月11日

Drazil has many friends. Some of them are happy and some of them are unhappy. Drazil wants to make all his friends become happy. So he invented the following plan.

There are n boys and m girls among his friends. Let‘s number them from 0 to n?-?1 and 0 to m?-?1 separately. In i-th day, Drazil invites $技术分享$ -th boy and $技术分享$ -th girl to have dinner together (as Drazil is programmer, i starts from 0). If one of those two people is happy, the other one will also become happy. Otherwise, those two people remain in their states. Once a person becomes happy (or if he/she was happy originally), he stays happy forever.

Drazil wants to know whether he can use this plan to make all his friends become happy at some moment.

Input

The first line contains two integer n and m (1?≤?n,?m?≤?100).

The second line contains integer b (0?≤?b?≤?n), denoting the number of happy boys among friends of Drazil, and then follow b distinct integers x1,?x2,?...,?xb (0?≤?xi?<?n), denoting the list of indices of happy boys.

The third line conatins integer g (0?≤?g?≤?m), denoting the number of happy girls among friends of Drazil, and then follow g distinct integers y1,?y2,?... ,?yg (0?≤?yj?<?m), denoting the list of indices of happy girls.

It is guaranteed that there is at least one person that is unhappy among his friends.

Output

If Drazil can make all his friends become happy by this plan, print "Yes". Otherwise, print "No".

Sample test(s)

input
2 3
0
1 0

output
Yes

input
2 4
1 0
1 2

output
No

input
2 3
1 0
1 1

output
Yes

Note

By $技术分享$ we define the remainder of integer division of i by k.

In first sample case:

On the 0-th day, Drazil invites 0-th boy and 0-th girl. Because 0-th girl is happy at the beginning, 0-th boy become happy at this day.
On the 1-st day, Drazil invites 1-st boy and 1-st girl. They are both unhappy, so nothing changes at this day.
On the 2-nd day, Drazil invites 0-th boy and 2-nd girl. Because 0-th boy is already happy he makes 2-nd girl become happy at this day.
On the 3-rd day, Drazil invites 1-st boy and 0-th girl. 0-th girl is happy, so she makes 1-st boy happy.
On the 4-th day, Drazil invites 0-th boy and 1-st girl. 0-th boy is happy, so he makes the 1-st girl happy. So, all friends become happy at this moment.

思路：求最大公约数，如果是1，只要看是否有一个为快乐的，有则yes否则no，如果大于1，则看每个长度为t（最大公约数）（b和g可以分为好多个长度为t的区间）的区间中每个对应位置至少有一个是快乐的

AC代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

int gcd(int a, int b) { 
    if(a > b) return gcd(b, a);  
    if(b % a == 0) return a;  
    return gcd(b % a, a);  
}  

int main() {
	int a[105], c[105], tmp[105];
	int n, m, b, g, t, x;
	while(scanf("%d %d", &n, &m) != EOF) {
		memset(a, 0, sizeof(a));
		memset(c, 0, sizeof(c));
		memset(tmp, 0, sizeof(tmp));
		
		scanf("%d", &b);
		for(int i = 0; i < b; i++) {
			scanf("%d", &x); a[x] = 1;
		}
		scanf("%d", &g);
		for(int i = 0; i < g; i++) {
			scanf("%d", &x); c[x] = 1;
		}
		if(gcd(n, m) == 1) {
			if(b + g >= 1) printf("Yes\n");
			else printf("No\n");
			continue;
		}
		
		if((t = gcd(n, m)) != 1){
			for(int i = 0; i < n; i++)
				if(a[i]) tmp[i%t] = 1;
			for(int i = 0; i < m; i++)
				if(c[i]) tmp[i%t] = 1;
		}
		
		int cnt = 0;
		for(int i = 0; i < t; i++) if(tmp[i]) cnt++;
		if(cnt == t) printf("Yes\n");
		else printf("No\n");
	}
	return 0;
}