# Judgement subgroup

Time Limit: 1000 ms Memory Limit: 65536 KiB

### Problem Description

Given a finite group s and one of its subsets S1, the binary operation * on S is defined as the modular M addition operation to determine whether the subset is a subgroup of S. (definition of subgroup: let H be a non empty subset of group G. If h constitutes a group under the operation of G, then h is a subgroup of G)

### Input

For multiple input groups, the first behavior is the number of elements of finite group S, n (0 < = n < 100), m (0 < = m < 100), and the number of elements of its subset, m (0 < = m < 100) (elements may be repeated)
The number of n in the second row is the element X (0 < = x < 100) contained in the finite group, and the number of m in the third row is the element Y (0 < = y < 100) contained in the subset.
output
If the subset is a subgroup of S, output "YES", otherwise output "NO" (without quotation marks)

### Output

If the subset is a subgroup of S, output "YES", otherwise output "NO" (without quotation marks)

### Sample Input

```8 7 7
0 1 2 3 4 5 6 6
0 1 2 3 4 5 6
10 7 7
0 1 2 3 4 5 6 7 8 9
0 1 2 3 4 5 9```

### Sample Output

```YES
NO```
```#include<stdio.h>
#include <string.h>
#define ll long long
int arr[110];
int vis[110];
int num[110];
int b[110];
int main()
{
int n, m, q;
while(~scanf("%d %d %d", &n, &m, &q))
{
int flag = 1;
memset(b,0,sizeof(b));
memset(vis,0,sizeof(vis));
memset(num,-1,sizeof(num));
memset(arr,-1,sizeof(arr));
for(int i = 1; i <= n; i++)
{
scanf("%d", &arr[i]);
}
for(int i = 1; i <= q; i++)
{
scanf("%d", &b[i]);
vis[b[i]] = 1;
}
if(vis[0]==0)
{
flag = 0;
}
num[0] = 0;
for(int i = 1; i <= n; i++)
{
for(int j = i+1; j <= n; j++)
{
if(vis[(b[i] + b[j]) % m] && (b[i] + b[j]) % m == 0)
{
num[b[i]] = b[j];
num[b[j]] = b[i];
break;
}
}
}
for(int i = 1; i <= n; i++)
{
if(num[b[i]] == -1)
{
flag = 0;
}
}
if(!flag)
{
printf("NO\n");
}
else
{
printf("YES\n");
}
}
return 0;
}
```

Tags: Programming

Posted on Thu, 07 Nov 2019 11:29:06 -0500 by jphilapy