通过删除最后一张给定的N张牌来找到赢得比赛的玩家

给定两个整数N和K ，其中N表示游戏开始时出现的总牌数， K表示单回合可以移除的最大牌数。两名玩家A和B轮流移除最多K张牌，从玩家A开始一张一张。移除最后一张牌的玩家为获胜者。任务是检查A是否可以赢得比赛。如果发现是真的，请打印“ A”作为答案。否则，打印‘B’ 。

例子：

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

方法：这里的想法是观察到，每当N％(K + 1)= 0的值时， A都将无法赢得比赛。否则， A总是赢球。
证明：

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.

因此，该想法是检查N％(K + 1)是否等于0。如果发现是真的，则将B打印为获胜者。否则，将A打印为获胜者。
下面是上述方法的实现：

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

输出：

A

时间复杂度： O(1)
辅助空间： O(1)