$36 \cdot 47 A B C D-A_{1} B_{1} C_{1} D_{1}$ is a unit cube, and two ants, one black and one white, both start from point $A$ and crawl forward along the edges. Each time they finish walking along one edge, it is called “finishing one segment”.
The white ant's route is $A A_{1} \rightarrow A_{1} D_{1} \rightarrow \cdots$, and the black ant's route is $A B \rightarrow B B_{1} \rightarrow \cdots$. They both follow the rule that the line of the $(i+2)$-th segment must be a skew line to the line of the $i$-th segment $(i$ is a natural number). After both ants finish the 1990th segment, they each stop at a certain vertex of the cube. At this point, the distance between the black and white ants is
(A) 1.
(B) $\sqrt{2}$.
(C) $\sqrt{3}$.
(D) 0.
(1st “Hope Cup” National Mathematics Invitational Competition, 1991)
[Solution] The route taken by the white ant according to the given rules is
$$
A A_{1} \rightarrow A_{1} D_{1} \rightarrow D_{1} C_{1} \rightarrow C_{1} C \rightarrow C B \rightarrow B A \rightarrow A A_{1},
$$

with a period of 6. Since $1990 \equiv 4(\bmod 6)$, it is known that it finally lands on point $C$.
The route taken by the black ant according to the given rules is
$$
A B \rightarrow B B_{1} \rightarrow B_{1} C_{1} \rightarrow C_{1} D_{1} \rightarrow D_{1} D \rightarrow D A \rightarrow A B,
$$

with a period also of 6, so it finally lands on point $D_{1}$.
At this point, the distance between the two ants is $\sqrt{1^{2}+1^{2}}=\sqrt{2}$.
Therefore, the answer is $(B)$.