Alberto and Barbara play the following game. Initially, there are some piles of coins on a table. Each player in turn, starting with Albert, performs one of the two following ways:
1) take a coin from an arbitrary pile;
2) select a pile and divide it into two non-empty piles.
The winner is the player who removes the last coin on the table. Determine which player has a winning strategy with respect to the initial state.
1. **Define the Game and Winning Strategy:**
   - Alberto and Barbara play a game with piles of coins.
   - Each player can either take one coin from any pile or split a pile into two non-empty piles.
   - The player who takes the last coin wins.

2. **Characterize the Piles:**
   - Let \( n_k \) be the number of piles with exactly \( k \) coins.
   - We need to determine the conditions under which Alberto has a winning strategy.

3. **Define Losing and Winning Positions:**
   - A "losing position" is defined as a position where both \( n_1 \) (the number of piles with exactly 1 coin) and \( n_2 + n_4 + n_6 + \ldots \) (the sum of the number of piles with an even number of coins) are even.
   - A "winning position" is any position that is not a losing position.

4. **Prove the Conditions for Winning and Losing Positions:**
   - If both \( n_1 \) and \( n_2 + n_4 + n_6 + \ldots \) are even, then the position is losing. This is because any move will change the parity of either \( n_1 \) or \( n_2 + n_4 + n_6 + \ldots \), thus moving to a winning position.
   - Conversely, if at least one of \( n_1 \) or \( n_2 + n_4 + n_6 + \ldots \) is odd, then the position is winning. This is because the player can always make a move that maintains the parity of \( n_1 \) or \( n_2 + n_4 + n_6 + \ldots \), thus moving to a losing position for the opponent.

5. **Conclusion:**
   - Alberto wins if and only if at least one of \( n_1 \) and \( n_2 + n_4 + n_6 + \ldots \) is odd.

The final answer is: Alberto wins if and only if at least one of \( \boxed{ n_1 } \) and \( n_2 + n_4 + n_6 + \ldots \) is odd.