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.

博弈论中极为经典的尼姆游戏。有总数为n的石头，每个人可以拿1~m个石头，两个人交替拿，拿到最后一个的人获胜。究竟是先手有利，还是后手有利？

1. 如果最后只剩下1个石子，先手全部拿走，后手败。
2. 如果最后只剩下2个石子，先手全部拿走；后手败。
3. 如果最后只剩下3个石子，先手全部拿走；后手败。
4. 如果最后只剩下4个石子，先手无论拿1个还是2个还是3个，最后最少剩一个，最多剩3个，后面拿的可以全部拿完，导致，前面的人没有东西拿就输了，所以如果我要赢，剩下石头不能是4的倍数，我要对方输，需要先保证我拿掉的石头是4的倍数，每次让对手拿石头的时候，都是4的倍数，我就可以赢了

public class Solution {public boolean canWinNim(int n) {return n % 4 == 0 ? false : true; }
}