leetcode292.[Array] Nim Game

本文介绍了一种Nim游戏的胜负判断策略。在一个由一堆石头构成的游戏中,玩家轮流从堆中取走1到3块石头,拿走最后一块石头的人获胜。本文提供了一个简洁的函数来判断初始状态下玩家是否能够赢得游戏。

摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往体验 >

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):
        if n%4==0:
            return False
        else:
            return True
评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值