Example 2 Let $n, k (n \geqslant k \geqslant 1)$ be positive integers. There are $n$ lamps placed on a circle, all of which are initially off. Each time, you can change the on or off state of any $k$ consecutive lamps. For the following three cases:
(1) $k$ is an odd prime,
(2) $k$ is an odd number,
(3) $k$ is an even number,

How many of the $2^{n}$ possible states can be achieved through a series of operations?
Consider $n$ lamps arranged in a circle, numbered as 1, 2, ..., $n$.

In a state that can be achieved through several operations, assign a value $a_{i} \in \{0,1\}$ to the $i$-th lamp ($i=1,2, \cdots, n$), where 0 indicates the lamp is off and 1 indicates the lamp is on.
Assign a value $b_{i} (i \in \{1,2, \cdots, n\})$ to each operation:
$b_{i}=1$ indicates that the lamps from the $i$-th to the $(i+k-1)$-th are toggled (i.e., their states are changed); $b_{i}=0$ indicates that no operation is performed on the lamps from the $i$-th to the $(i+k-1)$-th.

Clearly, performing the operation $b_{i}=1$ twice is equivalent to $b_{i}=0$, i.e., no operation.
Thus, the operation $b_{i}=1 (i=1,2, \cdots, n)$ is performed at most once.
The following arithmetic rules apply: $1+1=0, 0-1=1$, i.e., operations are performed modulo 2.
By the problem statement,
$$
a_{i}=\sum_{j=0}^{k-1} b_{i-j} (i=1,2, \cdots, n),
$$

where the indices are taken modulo $n$.
Thus, each set of operations $\left(b_{1}, b_{2}, \cdots, b_{n}\right)$ corresponds to a state $\left(a_{1}, a_{2}, \cdots, a_{n}\right)$, denoted as
$$
f\left(b_{1}, b_{2}, \cdots, b_{n}\right)=\left(a_{1}, a_{2}, \cdots, a_{n}\right).
$$

For a set of operations $\left(b_{1}, b_{2}, \cdots, b_{n}\right)$, we calculate the number of sets $\left(b_{1}^{\prime}, b_{2}^{\prime}, \cdots, b_{n}^{\prime}\right)$ different from $\left(b_{1}, b_{2}, \cdots, b_{n}\right)$ that satisfy
$$
f\left(b_{1}, b_{2}, \cdots, b_{n}\right)=f\left(b_{1}^{\prime}, b_{2}^{\prime}, \cdots, b_{n}^{\prime}\right).
$$

It is easy to see that equation (1) is equivalent to
$$
\sum_{j=0}^{k-1} b_{i-j}=\sum_{j=0}^{k-1} b_{i-j}^{\prime} (i=1,2, \cdots, n).
$$

From equation (2), we have
$$
\sum_{j=1}^{k} b_{i-j}=\sum_{j=1}^{k} b_{i-j}^{\prime}
$$
(2) - (3) gives $b_{i}-b_{i-k}=b_{i}^{\prime}-b_{i-k}^{\prime}$, i.e.,
$$
b_{i}^{\prime}-b_{i}=b_{i-k}^{\prime}-b_{i-k}.
$$

Let $c_{i}=b_{i}^{\prime}-b_{i}$. Then
$$
c_{i}=c_{i+k}=c_{i+k} (i=1,2, \cdots, n ; t=1,2, \cdots).
$$
$$
\text{Let } (n, k)=d, n=n_{0} d, k=k_{0} d, (n_{0}, k_{0})=1.
$$

Consider $\left\{n_{0} s+1 \mid s=1,2, \cdots, k_{0}\right\}$, which are distinct modulo $k_{0}$.
Thus, there exists $s \in \left\{1,2, \cdots, k_{0}\right\}$ such that
$k_{0} \mid (n_{0} s+1)$, i.e., $k_{0} t=n_{0} s+1$, then
$k t=n s+d$.
Substituting the above equation into equation (4) gives
$$
c_{i}=c_{i+4}=c_{i+n+d}=c_{i+d} (i=1,2, \cdots, n).
$$

Thus, each set $\left(b_{1}^{\prime}, b_{2}^{\prime}, \cdots, b_{n}^{\prime}\right)$ corresponds one-to-one to $\left(c_{1}, c_{2}, \cdots, c_{d}\right)$,
where $c_{i} \in \{0,1\}$, $c_{i}$ are not all 0, and
$$
k_{0} \sum_{i=1}^{d} c_{i}=0.
$$

Clearly, the number of $\left(c_{1}, c_{2}, \cdots, c_{d}\right)$ satisfying the above conditions is
$$
F\left(k_{0}\right)=\left\{\begin{array}{ll}
2^{d}-1, & k_{0} \text{ is even; } \\
2^{d-1}-1, & k_{0} \text{ is odd. }
\end{array}\right.
$$

Thus, each state satisfying the conditions corresponds to $F\left(k_{0}\right)+1$ sets of operations.

Since there are $2^{n}$ different sets of operations, the number of states satisfying the conditions is
$$
\frac{2^{n}}{F\left(k_{0}\right)+1}=\left\{\begin{array}{ll}
2^{n-d}, & k_{0} \text{ is even; } \\
2^{n-d+1}, & k_{0} \text{ is odd, }
\end{array}\right.
$$

where $d=(n, k), k_{0}=\frac{k}{d}$.