Add: Solution for Nim game problem in C++

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The given code defines a solution to the Nim Game problem using the canWinNim function. Here's the breakdown of the code:

Function Purpose: The function canWinNim(int n) determines if the player who starts the game can win when there are n stones in a pile. In the Nim Game, each player can remove 1 to 3 stones from the pile on their turn, and the player who removes the last stone wins.

Logic Explanation:

If the number of stones, n, is divisible by 4 (i.e., n % 4 == 0), the player cannot win. This is because in such cases, no matter how many stones the player takes (1, 2, or 3), they will always leave a number of stones that is again divisible by 4 for the opponent. This ultimately forces the starting player to lose if the opponent plays optimally.
If n is not divisible by 4, then the player can make the opponent face a multiple of 4 on their next turn, allowing the player to win.
Return Values:

false: If n % 4 == 0, indicating that the starting player will lose if both play optimally.
true: If n % 4 != 0, meaning the starting player has a winning strategy.
Code Summary: The function returns true if n is not a multiple of 4, indicating the player can win; otherwise, it returns false, suggesting the player will lose if they start the game with n stones.