Drazil and His Happy Friends
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
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 
-th boy and 
-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
2 301 0
output
Yes
input
2 41 01 2
output
No
input
2 31 01 1
output
Yes
推了半天没推出规律，索性10W的暴力来一发，后来。。过了。。。。
#include <iostream>
#include <cmath>
#include <cstring>
#include <cstdio>
#include <string>
#include <algorithm>
#include <cctype>
#include <queue>
#include <map>
using namespace std;
const int SIZE = 105;
int main(void)
{
int n,m,box;
int h_n,h_m;
int s_n[SIZE] = {0};
int s_m[SIZE] = {0};
cin >> n >> m;
cin >> h_n;
for(int i = 0;i < h_n;i ++)
{
cin >> box;
s_n[box] = 1;
}
cin >> h_m;
for(int i = 0;i < h_m;i ++)
{
cin >> box;
s_m[box] = 1;
}
for(int i = 0;i < 100000;i ++)
if(s_n[i % n] || s_m[i % m])
s_n[i % n] = s_m[i % m] = 1;
int flag_1 = 1;
int flag_2 = 1;
for(int i = 0;i < n;i ++)
if(!s_n[i])
{
flag_1 = 0;
break;
}
for(int i = 0;i < m;i ++)
if(!s_m[i])
{
flag_2 = 0;
break;
}
if(flag_1 && flag_2)
cout << "Yes" << endl;
else
cout << "No" << endl;
return 0;
}