Leetcode-Nim Game

本文介绍了一种Nim游戏的胜负判断策略。该策略通过分析游戏规则,得出结论:当初始石子数量对4取余不为0时,先手玩家必胜;反之则必败。文章提供了一个简洁的函数实现此策略。

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.

bool canWinNim(int n) {
    if (n % 4){
        return true;
    }
    else
    {
        return false;
    }
}

分析 :
n在区间[1,3]时,先取的人必定赢
当n为4时,先取的必输
n为5时,先取的必然可以找到一个赢的方法,能够化为n为4的情况,相当于后取的必然输

总结为:
当n % 4不为0时,先取者必胜;否则先取者必负。

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