101498I - Rock Piles

[Maybe this is flawed, but it got AC. Nevertheless, here’s my reasoning.]

• Let’s define:
  1. Even state: 2 piles have an even amount of rocks.
  2. Odd state: not even state.
• If a player is on an even state, the second player can always force him to be on even states.
• So, since the end state has trivially an even state, then whoever starts on an uneven state can force the second player to lose by keeping him on even states.

Time complexity: $O(1)$

Memory complexity: $O(1)$

using namespace std;
using ll = long long;
using ii = pair<int, int>;
using vi = vector<int>;
int main(void)
{
    ios::sync_with_stdio(false), cin.tie(NULL);
    int t;
    cin >> t;
    string ans[2] = {"hasan", "abdullah"};
    while (t--)
    {
        ll n, m;
        cin >> n >> m;
        cout << ans[n % 2 == 0 and m % 2 == 0] << endl;
    }
    return 0;
}