1. Two players, A and B, take stones from a pile of $n$ stones according to the following rules: A starts first, taking at least 1 stone but not all; then each player can take no more stones than the previous player took, and they cannot skip their turn. The player who takes the last stone wins. For each $n \geqslant 2$, determine who has a winning strategy.
For positive integers $n$ greater than 1, if $A$ (either player A or B) is guaranteed to win, then for the positive integer $2n$, $A$ is also guaranteed to win. This is because for the positive integer $2n$, each player will always take an even number of stones (if one leaves the other with an odd number of stones, the other wins). Therefore, every two stones can be merged into one, reducing it to the case of $n$. Thus, if $A$ is guaranteed to win when $n$, then $A$ is also guaranteed to win when $2n$.

When $n=2$, it is clear that B is guaranteed to win. Therefore, when $n$ is a power of 2, B is guaranteed to win.

When $n$ is odd, A is guaranteed to win. A only needs to take one stone at the beginning, and then each player can only take one stone each time. Since A takes the last stone, A is guaranteed to win. Therefore, when $n$ has an odd prime factor, A is guaranteed to win.

In summary, B is guaranteed to win if and only if $n$ is a power of 2. Otherwise, A is guaranteed to win.