【leetcode】1025. Divisor Game

題目以下：

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. 1 <= N <= 1000

解題思路：假設當前操做是Bob選擇，咱們能夠定義一個集合dic = {}，裏面存儲的元素Bob必輸的局面。例如當前N=1，那麼Bob沒法作任何移動，是必輸的場面，記dic[1] = 1。那麼對於Alice來講，在輪到本身操做的時候，只有選擇一個x,使得N-x在這個必輸的集合dic裏面，這樣就是必勝的策略。所以對於任意一個N，只要存在 N%x == 0 而且N-x in dic，那麼這個N對於Alice來講就是必勝的。只要計算一遍1~1000全部的值，把必輸的N存入dic中，最後判斷Input是否在dic中便可獲得結果。

代碼以下：

class Solution(object):
    dic = {1:1}
    def init(self):
        for i in range(2,1000+1):
            flag = False
            for j in range(1,i):
                if i % j == 0 and i - j in self.dic:
                    flag = True
                    break
            if flag == False:
                self.dic[i] = 1

    def divisorGame(self, N):
        """
        :type N: int
        :rtype: bool
        """
        if len(self.dic) == 1:
            self.init()
        return N not in self.dic