Solution:

Let us first assume that we start from a situation in which there is at least one token present on the table, otherwise the game cannot even begin.
Let $C_{0}, C_{1}, C_{2}, \ldots$ be the combinations of piles present at the start of the game, after the first move, after the second move, and so on. For each combination $C_{i}$ let $r_{i}$ be the number of piles that have a number of tokens $a_{1}, a_{2}, \ldots, a_{r_{i}}$ greater than 1 and $s_{i}$ the number of piles that have one token. Let then $m_{i}=\left(a_{1}-1\right)+\cdots+\left(a_{r_{i}}-1\right)$. We show that Barbara has a winning strategy if $m_{0}$ and $s_{0}$ are both even and that Alberto has a winning strategy in all other cases. The winning strategy is to leave the opponent, at each move, a combination $C_{i}$ such that $m_{i}$ and $s_{i}$ are both even.

Indeed, observe that if a player is facing a combination in which $m_{i}$ and $s_{i}$ are both even, every move they make must lead to a combination in which at least one of $m_{i+1}$ and $s_{i+1}$ is odd. Conversely, suppose a player is facing a combination in which $m_{i}$ and $s_{i}$ are not both even:
- if $m_{i}$ is even and $s_{i}$ is odd, they can remove a pile with 1 token;
- if $m_{i}$ is odd and $s_{i}$ is even, they can remove a token from a pile with at least 3 tokens (if one exists), or split a pile with 2 tokens into two piles with one token;
- if $m_{i}$ and $s_{i}$ are both odd, they can split a pile with $a \geq 3$ tokens (if one exists) into two piles with $a-1$ and 1 token, or they can remove a token from a pile containing 2.

In all cases, the player manages to produce again a combination in which $m_{i+1}$ and $s_{i+1}$ are both even.

Since during the game the number of tokens present on the table can be left unchanged at most $m_{0}$ times, the game certainly ends, and clearly the victory will go to whoever manages to propose the combination $C_{n}$ in which $m_{n}=s_{n}=0$, both even.

Solution:

Let $s_{i}$ be the number of piles with a single token and $m_{i}$ the number of piles with an even number of tokens. Also with this definition Barbara has a winning strategy if and only if $s_{0}$ and $m_{0}$ are even, and the winning strategy consists in always leaving the opponent a configuration in which $s_{i}$ and $m_{i}$ are even. Indeed if a player is facing a combination in which $s_{i}$ and $m_{i}$ are even, every move they make leads to a combination in which at least one of $s_{i+1}$ and $m_{i+1}$ is odd. Conversely, if a player is facing a combination in which $s_{i}$ and $m_{i}$ are not both even, by moving as indicated above they manage to produce again a configuration in which $s_{i+1}$ and $m_{i+1}$ are both even.