# 282.Nim Game

**292.Nim Game**

难度:Easy

> 你和你的朋友，两个人一起玩 Nim游戏：桌子上有一堆石头，每次你们轮流拿掉 1 - 3 块石头。 拿掉最后一块石头的人就是获胜者。你作为先手。

你们是聪明人，每一步都是最优解。 编写一个函数，来判断你是否可以在给定石头数量的情况下赢得游戏。

示例:

```
输入: 4
输出: false 
解释: 如果堆中有 4 块石头，那么你永远不会赢得比赛；
因为无论你拿走 1 块、2 块 还是 3 块石头，最后一块石头总是会被你的朋友拿走。
```

方法：直接递归是不行的，肯定会超时。解决方法就是保持先手，只要保证最后一个自己拿，那么前面的保证两个人每次拿到的都是(min+max)个数，让对方拿第四块即可。

```
class Solution {
public:
    bool canWinNim(int n) {
        if(!n) return false;
        if(n<4) return true;
        if(!(n%4)) return false;
        else 
            return true;
    }
};
```


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