12. Let $a_{1}, a_{2}, \cdots, a_{n}$ be a permutation of the integers $1,2, \cdots, n$, and satisfy
(1) $a_{1}=1$;
(2) $\left|a_{i}-a_{i+1}\right| \leqslant 2, i=1,2, \cdots, n-1$.

The number of such permutations is denoted by $f(n)$. Determine whether $f(2008)$ can be divisible by 3 (fill in "can" or "cannot").
12. It can be verified that $f(1)=1 \cdot f(2)=1 \cdot f(3)=2$.

Let $n \geqslant 4$, then it must be that $a_{1}=1, a_{2}=2$ or 3.
For $a_{2}=2$, the number of permutations is $f(n-1)$. This is because by removing the first term and reducing all subsequent terms by 1, we can establish a one-to-one correspondence of sequences.
For $a_{2}=3$, if $a_{1}=2$, then $a_{1}=4$. The number of such permutations is $f(n-3)$.
Therefore, $f(n)=f(n-1)+f(n-3)+1$.
Let $r(n)$ be the remainder when $f(n)$ is divided by 3, then $r(1)=r(2)=1, r(3)=2$, and for $n \geqslant 4$, $r(n) \equiv [r(n-1)+r(n-3)+1] \pmod{3}$.
Thus, $\{r(n)\}$ forms a sequence with a period of 8: 1, 1, 2, 1, 0, 0, 2, 0.
Since $2008 \equiv 0 \pmod{8}$, we have $r(2008)=0$, which means $f(2008)$ is divisible by 3.