| | #include <iostream> |
| | #include <unordered_set> |
| | using namespace std; |
| |
|
| | int num[10002]; |
| | int N, K; |
| |
|
| | unordered_set<int> proc(int ini, int fin) { |
| | if (ini < fin) { |
| | unordered_set<int> s1 = proc(ini, (ini + fin)/2), s2 = proc((ini + fin)/2 + 1, fin), r; |
| | for (auto it1 = s1.begin(); it1 != s1.end(); ++it1) { |
| | for (auto it2 = s2.begin(); it2 != s2.end(); ++it2) { |
| | r.insert((*it1 + *it2)%K); |
| | r.insert((*it1 - *it2 + K)%K); |
| | } |
| | } |
| | return r; |
| | } else { |
| | unordered_set<int> r; |
| | r.insert(((num[ini]%K)+K)%K); |
| | return r; |
| | } |
| | } |
| |
|
| | int main() { |
| | int M; |
| | cin >> M; |
| | while (M--) { |
| | cin >> N >> K; |
| | for (int i = 0; i < N; i++) { |
| | cin >> num[i]; |
| | } |
| | unordered_set<int> r = proc(0, N-1); |
| | if (r.count(0)) printf("Divisible\n"); |
| | else printf("Not divisible\n"); |
| | } |
| | return 0; |
| | } |
| |
|