# Nim Game Solutions in C++

Number 292

Difficulty Easy

Acceptance 54.9%

Link LeetCode

Other languages —

## Solutions

### C++ solution by haoel/leetcode

// Source : https://leetcode.com/problems/nim-game/// Author : Calinescu Valentin// Date : 2015-10-19/** Solutions* =========** Let's look at the example:** 0 stones - false* 1 stone - true* 2 stones - true* 3 stones - true* 4 stones - false** We notice that all we need for a position to be true is to get the opponent in a position* that is false. With 1, 2 and 3 you can take 1, 2 and 3 stones respectively to force your* opponent into having 0 stones, a position where he cannot win. No matter how many stones* we take from 4 we cannot** force the opponent into a losing positon, so position 4 becomes a losing position.* Let's take a look at the next 4 positions:** 5 stones - true* 6 stones - true* 7 stones - true* 8 stones - false** With 5, 6 and 7 stones we can take 1, 2 and 3 stones respectively to force the opponent into* position 4. Position 8 is a losing one because we can only force the opponent into winning* positions. We notice that this group of 4 positions can repeat itself indefinitely, because* we only need the previous 3 positions to judge whether a position is winning or losing.** Thus we can see the pattern:** n % 4 == 0 - false* n % 4 != 0 - true**/class Solution {public:bool canWinNim(int n) {return !(n % 4 == 0);}};