# Find the player who wins the game by removing the last of given N cards

Given two integers N and K, where N represents the total number of cards present when game begins and K denotes the maximum number of cards that can be removed in a single turn. Two players A and B get turns to remove at most K cards, one by one starting from player A. The player to remove the last card is the winner. The task is to check if A can win the game or not. If found to be true, print ‘A’ as the answer. Otherwise, print ‘B’.

Examples:

Input: N = 14, K = 10
Output: Yes
Explanation:
Turn 1: A removes 3 cards in his first turn.
Turn 2: B removes any number of cards from the range [1 – 10]
Finally, A can remove all remaining cards and wins the game, as the number of remaining cards after turn 2 will be ≤ 10

Input: N = 11, K=10
Output: No

Approach: The idea here is to observe that whenever the value of N % (K + 1) = 0, then A will never be able to win the game. Otherwise A always win the game.
Proof:

1. If N ≤ K: The person who has the first turn will win the game, i.e. A.
2. If N = K + 1: A can remove any number of cards in the range [1, K]. So, the total number of cards left after the first turn are also in the range [1, K]. Now B gets the turn and number of cards left are in the range [1, K]. So, B will win the game.
3. If K + 2 ≤ N ≤ 2K + 1: A removes N – (K + 1) cards in his first turn. B can remove any number of cards in the range [1, K] in the next turn. Therefore, the total number of cards left now are in the range [1, K].Now, since the remaining cards left are in the range [1, K], so A can remove all the cards and win the game.

Therefore the idea is to check if N % (K + 1) is equal to 0 or not. If found to be true, print B as the winner. Otherwise, print A as the winner.
Below is the implementation of the above approach:

## C++

 `// C++ Program to implement ` `// the above approach ` ` `  `#include ` `using` `namespace` `std; ` ` `  `// Function to check which ` `// player can win the game ` `void` `checkWinner(``int` `N, ``int` `K) ` `{ ` `    ``if` `(N % (K + 1)) { ` `        ``cout << ``"A"``; ` `    ``} ` `    ``else` `{ ` `        ``cout << ``"B"``; ` `    ``} ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` ` `  `    ``int` `N = 50; ` `    ``int` `K = 10; ` `    ``checkWinner(N, K); ` `}`

## Java

 `// Java program to implement ` `// the above approach ` `import` `java.util.*; ` ` `  `class` `GFG{ ` ` `  `// Function to check which ` `// player can win the game ` `static` `void` `checkWinner(``int` `N, ``int` `K) ` `{ ` `    ``if` `(N % (K + ``1``) > ``0``)  ` `    ``{ ` `        ``System.out.print(``"A"``); ` `    ``} ` `    ``else` `    ``{ ` `        ``System.out.print(``"B"``); ` `    ``} ` `} ` ` `  `// Driver code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``int` `N = ``50``; ` `    ``int` `K = ``10``; ` `     `  `    ``checkWinner(N, K); ` `} ` `} ` ` `  `// This code is contributed by Amit Katiyar `

## Python3

 `# Python3 program to implement ` `# the above approach ` ` `  `# Function to check which ` `# player can win the game ` `def` `checkWinner(N, K): ` ` `  `    ``if``(N ``%` `(K ``+` `1``)): ` `        ``print``(``"A"``) ` `    ``else``: ` `        ``print``(``"B"``) ` ` `  `# Driver Code ` `N ``=` `50` `K ``=` `10` ` `  `# Function call ` `checkWinner(N, K) ` ` `  `# This code is contributed by Shivam Singh `

## C#

 `// C# program to implement ` `// the above approach ` `using` `System; ` ` `  `class` `GFG{ ` ` `  `// Function to check which ` `// player can win the game ` `static` `void` `checkWinner(``int` `N, ``int` `K) ` `{ ` `    ``if` `(N % (K + 1) > 0)  ` `    ``{ ` `        ``Console.Write(``"A"``); ` `    ``} ` `    ``else` `    ``{ ` `        ``Console.Write(``"B"``); ` `    ``} ` `} ` ` `  `// Driver code ` `public` `static` `void` `Main(String[] args) ` `{ ` `    ``int` `N = 50; ` `    ``int` `K = 10; ` `     `  `    ``checkWinner(N, K); ` `} ` `} ` ` `  `// This code is contributed by Amit Katiyar`

Output:

```A
```

Time Complexity: O(1)
Auxiliary Space: O(1)

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