Example 16 (31st Moscow Mathematical Olympiad) There are two small piles of candies, and two people play a game. They take turns making moves, and the player whose turn it is eats one of the small piles and divides the other into two parts (which can be equal or unequal). If the other pile has only one candy and cannot be divided further, then the player eats this one candy and wins. If initially the two small piles have 33 and 35 candies respectively, who will win? The first player or the second player? How should the player move to win?

To be able to win, the player should:
Analysis: In order to win, the player who makes the first move must force the opponent to eventually split 2 or 3 candies into two piles. Therefore, the winning positions can be denoted as $(1,1)$ or $(2,1)$.

Solution: The player who makes the first move will win. In the first step, he should eat the pile with 33 candies, then split the other pile into 17 and 18 candies. In the subsequent process, he should always leave the opponent with two piles of candies, each with a number of $5k+2$ or $5k+3$ (he can achieve this). This way, his opponent will eventually be forced to split 2 or 3 candies into two piles, thus losing.