LeetCode 292. Nim Game

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.

问题大意：有一堆石头，一共有n个，你和你朋友一起玩一个游戏：每次你或者你朋友可以取1个，2个或者3个石头，取到最后一个石头的玩家胜利。给定一个n，请问最后是谁的胜利。假设双方都总是做出对自己最有利的决策。你本人先手。

这是一个很简单的博弈题。只要判断石头的个数是否是4的倍数即可。如果是4的倍数，则无论你取几个石头，你的朋友总是能取相对应的数量的石头使得剩下的石头的个数依然是4的倍数，这样的话，他总能取到最后一个石头。而如果不是4的倍数，则你可以取一定的数量使得剩下4的倍数个石头。这样就变成第一种情况，而先手的变成你的朋友，这样你必胜。

代码非常简单，如下：

class Solution {public:    bool canWinNim(int n) {        return n%4 ;    }};