1、题目
Alice and Bob take turns playing a game, with Alice starting first.
Initially, there is a number N on the chalkboard. On each player’s turn, that player makes a move consisting of:
Choosing any x with 0 < x < N and N % x == 0. Replacing the number N on the chalkboard with N - x. Also, if a player cannot make a move, they lose the game.
Return True if and only if Alice wins the game, assuming both players play optimally.
Example 1:
Input: 2
Output: true
Explanation: Alice chooses 1, and Bob has no more moves.
Example 2:
Input: 3
Output: false
Explanation: Alice chooses 1, Bob chooses 1, and Alice has no more moves.
Note:
1 <= N <= 1000
2、思路
对于第一个开始的用户来说偶数会赢，计数会输，采用数学归纳法证明：
假设N<=n时，偶数会赢，奇数会输是成立的，证明当N = n + 1时，结论仍然成立
证明：当n是偶数，N为奇数，我们要想赢只能第一个数选偶数，但是N是一个奇数且要保证N % x == 0,所以第一个数只能选奇数，最后就会输
当n是奇数时，N为偶数，我们想赢只能让对方在选择时是个奇数，于是我们只要第一个数选1就能保证对方选的时候是奇数，对方就会输
综上所示，偶数会赢，奇数会输
3、实现
public class{
public boolean divisorGame(int N) {
return N % 2 == 0;
}
}
