[LeetCode]292. Nim Game(轮流拿掉石头)

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.

谷歌翻译：你和你的朋友一起玩下面的Nim游戏：在桌子上有一堆石头，每次你轮流去除1到3块石头。 去除最后一块石头的人将是赢家。 你会采取第一个回合去除石头。

你们都很聪明，并且有最好的策略。 编写一个函数来确定是否可以赢得游戏给定的堆数量的石头。

例如，如果在堆中有4块石头，那么你永远不会赢得比赛：无论你删除1，2，或3块石头，最后的石头将永远被你的朋友删除。

Hint:

1. If there are 5 stones in the heap, could you figure out a way to remove the stones such that you will always be the winner?

如果堆中有5块石头，你能找出一种方法去除石头，让你永远是赢家吗？

C++

class Solution {
public:
bool canWinNim(int n) {
if(n%4 == 0 && n)
return false;
return true;
}
};

注意：

很明显，n大于0时，自己拿过后给另一个人留4的倍数的石头，你就一定会赢。多注意观察规律就好