Let $26 f$ be a mapping from the set of natural numbers $\mathbf{N}$ to $\mathbf{N}$. Denote $f^{[2]}=$ $f(f(n)), \cdots, f^{[m]}(n)=f^{[m-1]}[f(n)]$. Suppose $a, b, c$ are known natural numbers and 
$$
f(n)=\left\{\begin{array}{ll}
f(n+a) & \text { if } n > c ; \\
f(f(n+b) & \text { if } n \leqslant c .
\end{array}\right.
$$
(2) Find a necessary and sufficient condition for $f$ to have at least one fixed point;
(3) Express such a fixed point in terms of $a, b, c$.
(1990 Chinese National Training Team)
We can gradually find the expression for $f(x)$.
When $n>c$, we have $f(n)=n-a$. When $c \geqslant n>c-(b-a)$, we have
$$
f(n)=f(f(n+b))=f(n+b-a)=n+(b-a)-a;
$$

When $c-(b-a) \geqslant n>c-2(b-a)$, we have
$$
f(n)=f(f(n+b))=f(n+2(b-a))=n+2(b-a)-a \cdots
$$

In general, when $c-k(b-a) \geqslant n>c-(k+1)(b-a)$, we have
$$
f(n)=n+(k+1)(b-a)-a \quad(k=0,1, \cdots, q, q \in \mathbf{N}).
$$

The natural numbers $n$ that satisfy
$$
q(b-a) \leqslant c-k(b-a)
$$

are all fixed points.
The following is an example of the iteration of a rational function.