# Group decision

Time Limit: 1000 ms Memory Limit: 65536 KiB

### Problem Description

Let d be a set of nonnegative subnumbers, and let binary operation + be modular M addition. Given D and M, we ask if the algebraic system v = < D, + > constitutes a group. If so, then we find the inverse element of the given element.

### Input

For each group of test data, the first row has three numbers n (1 < = n < = 100), m (1 < = m < = 100) and Q (1 < = 100 < = q). Where n is the number of elements in S (the elements may be repeated, please remove the duplicates by yourself), M is as mentioned above, and Q is the number of queries. Next, n numbers x (0 < = x < 100) represent the elements in S. Next row Q, a number y (0 < = y < 100) in each row, represents the inverse element of the query y.

### Output

If V does not form a group, just output a - 1.
Otherwise, output Q lines, one number per line, corresponding to the inverse element of a query.

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

```5
-1```

### Hint

The example of modular M addition is as follows: for example, modular 7 addition, 3 + 6 = 2, 5 + 2 = 0, that is, the answer is the remainder with division by remainder.

```#include <bits/stdc++.h>
#define ll long long
using namespace std;
int arr;
int vis;
int num;
int b;
int main()
{
int n, m, q;
while(cin >> 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++)
{
cin>>arr[i];
vis[arr[i]] = 1;
}
for(int i = 1; i <= q; i++)
{
cin>>b[i];
}
if(vis==0)
{
flag = 0;
}
num = 0;
for(int i = 1; i <= n; i++)
{
for(int j = 1; j <= n; j++)
{
if(vis[(arr[i] + arr[j]) % m] && (arr[i] + arr[j]) % m == 0)
{
num[arr[i]] = arr[j];
num[arr[j]] = arr[i];
break;
}
}
}
for(int i = 1; i <= n; i++)
{
if(num[arr[i]] == -1)
{
flag = 0;
}
}
if(!flag)
{
cout << "-1" << endl;
}
else
{
for(int i = 1; i <= q; i++)
{
cout << num[b[i]] << endl;
}
}
}
return 0;
}```

Tags: Programming

Posted on Thu, 07 Nov 2019 16:32:50 -0500 by aiwebs