Subtask 1

Consider the sum of ~\log_2(A_i) - 1~ where ~A_i~ are the numbers currently written on the board. This sum decreases by exactly ~1~ on each move. Thus, we can find the winner by considering the parity of this sum.

Subtask 2

The described game is an impartial game, and the remainder of the editorial will assume knowledge of the Sprague-Grundy theorem.

Observe that we only need to care about the prime factorizations of the numbers on the board. More specifically, for ~X = p_1^{e_1} \cdots p_k^{e_k}~, we only care about the multiset ~\left\{ e_1, \ldots, e_k \right\}~.

It turns out there are around ~7000~ such multisets for blackboard numbers between ~2~ and ~10^{12}~. To reduce the number of states, observe that multisets ~S~ and ~S + \left\{ 1 \right\}~ (adding one ~1~ to ~S~) will always have the same Grundy number. This can be proved inductively, but the intuition is that if prime factor ~p~ of ~X~ has multiplicity ~1~, we cannot do anything with ~p~ when splitting ~X~ into ~Y~ and ~Z~.

With this observation, the number of states is cut down to around ~1500~. Then we can compute Grundy values in ~\mathcal{O}(M^2 \log A)~, where ~M \approx 1500~ is the number of such states. Since we do not care for prime factors of multiplicity ~1~, we can calculate the multisets corresponding to each ~A_i~ in ~A^{1/3}~ time. Hence we have solved this problem in ~\mathcal{O}(NA^{1/3} + M^2 \log A)~.

Note from the problem author: Thanks to mangoqwq for preparing the problem :)