[LeetCode]292.Nim Game

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。通过分析游戏规则及策略,得出结论:当石子总数不能被4整除时,先手玩家将赢得比赛。

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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:石子总数为N,每次可移动的最大值为K,若N小于等于K,则先移动石子者获胜,

  若N能整除K,则先移动石子者失败,反之获胜。

Java带测试版本

public class Solution {

 public boolean canWinNim(int n) {

                  if(n<4) return true;

                  else { if (n%4 == 0) return false;

                                    else return true;

                           }

                  }

                    public static void main(String args[]) {

                             System.out.println("please enter thenumber of the stones");

                             Scanner input = new Scanner(System.in);

                             int n = input.nextInt();

                             boolean win = new         Solution().canWinNim(n);

                             if(win == true) System.out.println("youwill win the game");

                             else System.out.println("you will losethe game");

                    }  

   

}


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