$36 \cdot 66$ There are three piles of stones with numbers $2,3,4$. Two people take turns to take away stones, the rule is: each person must take at least 1 stone each time, no limit on how many more, but must take from the same pile, the one who takes the last 1 stone loses. Then the winning strategy is
(A) The first player has.
(B) The second player has.
(C) Both have.
(D) Neither has.
（3rd "Five Sheep Cup" Junior High School Mathematics Competition, 1991）
［Solution］The first player has a winning strategy, please follow the steps below:
(1) First, take 3 stones from the 3rd pile, leaving 1 stone, changing $(2,3,4)$ $\rightarrow(2,3,1)$;
(2) If the opponent leaves $(2,3,0)$, you should aim to leave two piles with an equal number of stones, i.e., you should leave $(2,2,0)$ after your turn;
If the opponent leaves $(1,3,1)$, you should leave $(1,1,1)$ after your turn.
In short, once you leave the opponent with a situation where the number of stones in the three piles is $(k, k, 0)(k \geqslant 2)$ or $(1,1,1)$, you will definitely win (the remaining steps only require you to take the same number of stones from the other pile as the opponent did).
Therefore, the answer is $(A)$.