Easy-11

原创于 2017-05-03 17:32:14 发布 · 288 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#leetcode

Lancher's re-code life 专栏收录该内容

49 篇文章

订阅专栏

本文介绍了一种简单而有效的算法来判断Nim游戏中的胜负情况。通过分析游戏规则，可以得出只要剩余石头数量减去1、2或3后的结果对4取余不为0，则先手玩家将赢得游戏。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

leetcode 292. Nim Game

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.

AC:

bool canWinNim(int n) {
    if((n-1)%4==0)
    return true;
    if((n-2)%4==0)
    return true;
    if((n-3)%4==0)
    return true;
    return false;
}