[leetcode] 292. Nim Game

最新推荐文章于 2025-09-11 19:19:55 发布

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最新推荐文章于 2025-09-11 19:19:55 发布

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分类专栏： leetcode 文章标签： leetcode Brainteaser

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本文介绍了一个简单的Nim游戏胜负判断算法。通过分析游戏规则，得出结论：当石头总数为4的倍数时，先手玩家必败；反之，则必胜。文章提供了简洁的C++实现代码。

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.

Hint:

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?

这道题是小游戏Nim Game，题目难度为Easy。

在我们回合如果剩1、2或3个石头则我们胜出，如果剩4个石头则对方胜出。超过4个石头时，如果石头总数不是4的倍数，我们能够保证拿完石头后石头总数变为4的倍数，这样石头总数会在若干轮后在对方回合达到4个，我们就赢了。相反，如果石头总数是4的倍数，在我们回合时对方总能保证石头总数是4的倍数，我们就输了，所以总数为4的倍数时我们必定输，否则我们必定赢。具体代码：

class Solution {
public:
    bool canWinNim(int n) {
        return n%4;
    }
};