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++ Program to implement
// the above approach

#include <bits/stdc++.h>
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 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 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# 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 

<script>
    // Javascript Program to implement
// the above approach

// Function to check which
// player can win the game
function checkWinner(N, K)
{
    if (N % (K + 1)) {
        document.write("A");
    }
    else {
        document.write("B");
    }
}

// Driver code

    let N = 50;
    let K = 10;
    checkWinner(N, K);

// This code is contributed by Saurabh Jaiswal
</script>

A

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