【leetcode】【c++】292. Nim Game

版权声明：来自 T2777 ，请访问 https://blog.csdn.net/T2777 一起交流进步吧 https://blog.csdn.net/T2777/article/details/87885596

Nim Game 是一种回合制游戏，要求游戏双方轮流从一堆棋子中取棋子，当不能再取时所轮到的一方为输家，即取到最后一颗棋子的玩家获胜。

题目描述：

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.

Example:

Input: 4
Output: false 
Explanation: 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.

每个玩家可以从这堆石子中取 1-3 个石子，谁取到了最后的棋子就是胜利者，你先取，判断你能获得胜利的棋子个数。

对游戏进行分析，当 有 0 个棋子时， 你必败，有 1,2,3个棋子时你必胜；当有4个棋子时，你必败，当有 5,6,7 个棋子时你必胜；故被 4 整除时必输，不被 4 整除时必胜。或者说，当 n (石子总数) 存在一种方式一步到达必输状态，当 存在 x ，使得f(n-x)为必输时，x取[1,3]，则你必胜，即别人必输的情况，你必胜；当不存在这种 x 那么你必输。

代码为：

class Solution {
public:
    bool canWinNim(int n) {
        return (n % 4 != 0);
    }
};