Judgement subgroup SDUT discrete mathematics OJ4172

Judgement subgroup

Time Limit: 1000 ms Memory Limit: 65536 KiB

Submit Statistic

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