leetcode:292 Nim Game－每日编程第一题

最新推荐文章于 2020-06-28 09:30:33 发布

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最新推荐文章于 2020-06-28 09:30:33 发布

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本文探讨了Nim游戏中的胜利策略，通过分析石子数量与胜利的关系，揭示了游戏胜负的关键因素。当石子数量为4的倍数时，玩家将不可避免地失败；反之，非4的倍数情况下，玩家可以通过特定策略确保获胜。

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Nim Game

Total Accepted: 24894 Total Submissions: 49816 Difficulty: Easy

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.

思路：

1).当石子数量为4*n时，无论Ａ取1〜3个石头，Ｂ都可取3〜1个石头，使得每一轮石子总数－4，因此Ｂ可以确保取胜，Ａ输。

2).当石子总数为其他情况时，Ａ可取1~3个石头，使得石子总数变成4*n，跟据1).的分析，Ｂ此时必输，Ａ胜。

因此，当石子数量为 4*n时，Ａ输，其他情况Ａ胜。

<span style="font-weight: normal;">class Solution {
public:
    bool canWinNim(int n) {
        if(0==n%4) 
            return false;
        else
            return true;
    }
};</span>