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.
Sample Input
8 7 1 0 6 3 6 1 2 4 5 2 3 4 2 0 2 3 0 3Sample Output
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[110]; int vis[110]; int num[110]; int b[110]; 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]==0) { flag = 0; } num[0] = 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; }