Example 5 defines an "operation": cutting a rope of positive integer length into three ropes of positive integer length, with the longest rope being exactly one. Given a rope of length $a^{b}$ (where $a$ and $b$ are integers greater than 1), two players, A and B, take turns, with A starting first by performing one operation on the rope, then B performing one operation on the longest rope resulting from A's operation, and so on. Each player performs an operation on the longest rope from the three ropes resulting from the previous operation by the other player. If a player cannot perform an operation when it is their turn, the other player wins. Question: Who has a winning strategy?
【Analysis】Consider the general case. Let the initial length of the rope be $n\left(n \in \mathbf{Z}_{+}\right)$.
Obviously, if $1 \leqslant n \leqslant 3$, A must fail;
if $4 \leqslant n \leqslant 7=3+2+2$, then A has a winning strategy;
if $8 \leqslant n \leqslant 9$, the length of the longest rope after A's operation must satisfy $4 \leqslant m \leqslant 7$, then B has a winning strategy;
if $10 \leqslant n \leqslant 25=9+8+8$, then A has a winning strategy; if $26 \leqslant n \leqslant 27$, the length of the longest rope after A's operation must satisfy $10 \leqslant m \leqslant 25$, then B has a winning strategy;
Using mathematical induction, we can prove:
B has a winning strategy $\Leftrightarrow n=1,3^{k}-1,3^{k}\left(k \in \mathbf{Z}_{+}\right)$.
If $a$ is a power of 3, then $a^{b}$ is also a power of 3. Thus, B has a winning strategy.
Now, let's assume $a$ is not a power of 3, and
$$
a^{b}=3^{k}-1(k \geqslant 2) \text {. }
$$

Then $a^{b} \equiv 2(\bmod 3)$
$\Rightarrow b$ is odd (otherwise, $a^{b}$ is a perfect square,
$$
\begin{aligned}
& \left.a^{b} \equiv 0,1(\bmod 3)\right) \\
\Rightarrow & 3^{k}=a^{b}+1 \\
& =(a+1)\left(a^{b-1}-a^{b-2}+\cdots-a+1\right) \\
\Rightarrow & a+1=3^{i} \Rightarrow a=3^{i}-1 \\
\Rightarrow & a \equiv-1(\bmod 3)
\end{aligned}
$$
$$
\Rightarrow 3^{k-i}=a^{b-1}-a^{b-2}+\cdots-a+1 \equiv b(\bmod 3)
$$
$\Rightarrow b$ is a multiple of 3.
Let $c=a^{\frac{b}{3}}$. Then
$$
\begin{array}{l}
3^{k}=c^{3}+1=(c+1)\left(c^{2}-c+1\right) \\
=(c+1)((c+1)(c-2)+3) . \\
\text { Hence }(c+1,(c+1)(c-2)+3) \\
=(c+1,3)=3 .
\end{array}
$$

Let $c+1=3^{\alpha}, c^{2}-c+1=3^{\beta},(\alpha, \beta) \geqslant 1$.
Thus, $\alpha 、 \beta$ must have one that is 1.
If $\alpha=1$, then $c=2, a=2, b=3$;
If $\beta=1$, then
$$
\begin{array}{l}
c^{2}-c+1=3 \\
\Rightarrow(c+1)(c-2)=0 \\
\Rightarrow c=2, a=2, b=3 .
\end{array}
$$

In summary, when $a=2, b=3$ or $a$ is a power of 3, B has a winning strategy; otherwise, A has a winning strategy.