Nam Game

哪一方最终给对方留下4的倍数个石头则win，即想方设法的给对方留下4的倍数个石头。

例如： 9（B win）

A：3 　　　　　　　　　　　　　　　　　　　　

B：2（B取2，给对方余4，对方则lose）　

A：1 | 2 | 3

B：3 | 2 | 1(B win)

例如：9 （A win）

A：1 （留8，4的2倍）

B：1 | 2 | 3  (余下7 | 6 |5)

A：3 | 2 | 1 （余下4，4的1倍）

B：1 | 2 | 3 （余下3 | 2 | 1）

A：3 | 2 | 1 （A win）

当先手一方掌握了可以给对方留下4的倍数的权力，那么这一方就可以稳赢，否则则输

所以python代码实现：

"""
 You are playing the following Nim Game with your friend: There is a heap of stones on the table, each time one of you take turns to remove 1 to 3 stones. The one who removes the last stone will be the winner. You will take the first turn to remove the stones.

Both of you are very clever and have optimal strategies for the game. Write a function to determine whether you can win the game given the number of stones in the heap.

For example, if there are 4 stones in the heap, then you will never win the game: no matter 1, 2, or 3 stones you remove, the last stone will always be removed by your friend. 
"""
class Solution(object):
    def canWinNim(self, n):
        """
        :type n: int
        :rtype: bool
        """
        if n % 4 == 0 :
            return False
        else:
            return True