Codeforce 515 B . Drazil and His Happy Friends 暴力

最新推荐文章于 2022-05-26 19:50:53 发布

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#Codeforces

Drazil有一个由男孩和女孩组成的社交圈，他通过安排每天邀请一对男女共进晚餐来试图让所有人都变得快乐。此博客介绍了如何通过简单的计划使所有朋友快乐，并提供了输入输出示例。

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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 $\text{[math]}$ -th boy and $\text{[math]}$ -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 $\text{[math]}$ 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.

一开始想的是并查集=。=考虑来考虑去。发现不会写。。=。=后来才想到用暴力解这道题。因为最多100*100。就直接可以暴力解出来这道题

#include<iostream>
#include<string>
#include<algorithm>
#include<cstdlib>
#include<cstdio>
#include<set>
#include<map>
#include<vector>
#include<cstring>
#include<stack>
#include<cmath>
#include<queue>
#define INF 0x0f0f0f0f
using namespace std;

int main()
{
    int x,y,i,j,k,m,n,l,temp,cnt=0,showgril,showboy,ccnt;
    int a[105],b[105];
    memset(a,0,sizeof(a));
    memset(b,0,sizeof(b));
    
    scanf("%d%d",&x,&y);
    scanf("%d",&n);
    for(i=0;i<n;i++)
    {
        scanf("%d",&temp);
        a[temp]=1;
    }
    scanf("%d",&m);
    for(i=0;i<m;i++)
    {
        scanf("%d",&temp);
        b[temp]=1;
    }
    showgril=0;showboy=0;
    while(cnt<1000005)
    {
        cnt++;
        showboy=showboy%x;
        showgril=showgril%y;
        if(a[showboy]||b[showgril])
        {
            a[showboy]=1;
            b[showgril]=1;
        }
        showboy++;
        showgril++;

    }

    for(i=0;i<x;i++)
    {
        if(!a[i])
        {
            printf("No\n");
            return 0;
        }
    }
    for(i=0;i<y;i++)
    {
        if(!b[i])
        {
            printf("No\n");
            return 0;
        }
    }
    printf("Yes\n");
    return 0;

}