Let $A=\{1,2, \cdots, 1993\}$, and denote
$$
f^{(1)}(x)=f(x), \quad f^{(n)}(x)=f^{(n-1)}(f(x)) .
$$

Then the number of mappings from $A$ to $A$ that satisfy $f^{(1993)}(x)=f(x)$ is ( ).
A. 1993
B. 1994
C. 3985
D. 3986.
The original solution constructs two types of $f(x)$ that meet the requirements, totaling 3985, and selects C.