Alice and Basilio are playing the following game; from a bag initially containing 1331 coins, they take turns taking coins, with Alice making the first move and taking 1 coin, and thereafter on each subsequent turn, a player takes either the same number of coins that the other player took on their last turn, or one more. The player who cannot make a move according to the rules loses. Which of the players can ensure a win regardless of the other's moves?
Let Alice, regardless of Basil's actions, take 1 coin on the first move, 2 coins on the second, 3 coins on the third, and so on, increasing the number of coins taken by 1 each time. This does not contradict the rules, as under such a strategy by Alice, Basil can take 1 or 2 coins on the first move, 2 or 3 coins on the second, and so on. Then after $k$ moves by Alice, the players will have taken no fewer than

$$
(1+2+\ldots+k)+(1+2+\ldots+(k-1))=k^{2}
$$

and no more than

$$
(1+2+\ldots+k)+(2+3+\ldots+k)=k(k+1)-1
$$

Since $36 \cdot 37-1=1331$, there will be enough coins for Alice's 36th move in any case. And since $36^{2}=1296$ and $1331-1296=35$, there will not be enough coins for Basil's 36th move in any case. Thus, Alice will win.

## Answer

Alice wins.