通过与给定范围内的任何值相乘，找到至少达到 N 的玩家

给定一个整数N ，两个玩家 A 和 B 的任务是通过交替轮换将X与范围[2, 9]中的任何数字相乘，使X的值（以1初始化）至少为N。假设两个玩家都发挥最佳。任务是找到玩家获得valuie≥N个第一。

例子：

Input: N = 12
Output: Player B
Explanation: 
Initially, X = 1.
A multiplies X with 9. Therefore, X = 1 * 9 = 9.
In second turn, B multiplies X with 2. Therefore, X = 9*2 = 18
Therefore, B wins.

Input: N = 10
Output: Player B

方法：这个想法是使用组合博弈论的概念。找出位置，如果一个数字乘以X导致胜利，以及导致失败的位置。以下是步骤：

• 在组合博弈论中，让我们定义N 位置是下一个移动的玩家如果他玩得最佳而获胜的位置，而P 位置是一个位置，如果他的对手玩得最佳，下一个移动的玩家总是输。
• 可以达到N的最低位置是res = ceil(N/9) 。因此，从[res, res + 1, res + 2, ….., N – 1] 的所有位置都是N 个位置。
• 唯一被迫移动到[res, res + 1, …, (N – 1)] 的位置是那些乘以2 的位置，并且它们位于该区间内，该区间由Y = ceil(res/2 ) ,
• 因此， [Y, Y + 1, …, (res – 1)]都是P 位置。
• 重复以上步骤直到找到包含1的区间，如果1是N位，则A胜，否则B胜。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to find the winner
char Winner(int N)
{
    bool player = true;
 
    // Backtrack from N to 1
    while(N > 1)
    {
        int den = (player) ? 9 : 2;
        int X = N/den, Y = N%den;
        N = (Y)? X + 1: X;
        player = !player;
    }
    if (player)
      return 'B';
    else
      return 'A';
}
 
// Driver Code
int main()
{
  int N = 10;
  cout << Winner(N);
  return 0;
}
 
// This code is contributed by mohit kumar 29.

Java

// Java program for the above approach
import java.util.*;
class GFG
{
 
  // Function to find the winner
  static char Winner(int N)
  {
    boolean player = true;
 
    // Backtrack from N to 1
    while(N > 1)
    {
      int den = (player) ? 9 : 2;
      int X = N / den;
      int Y = N % den;
      N = (Y > 0) ? X + 1: X;
      player = !player;
    }
    if (player)
      return 'B';
    else
      return 'A';
  }
 
  // Driver Code
  public static void main(String[] args)
  {
    int N = 10;
    System.out.print(Winner(N));
  }
}
 
// This code is contributed by 29AjayKumar

Python3

# Python program for the above approach
 
# Function to find the winner
def Winner(N):
    player = True
     
    # Backtrack from N to 1
    while N > 1:
                 
        X, Y = divmod(N, (9 if player else 2))
         
        N = X + 1 if Y else X
        player = not player
     
    if player:
      return 'B'
    else:
      return 'A'
 
# Driver Code
N = 10
print(Winner(N))

C#

// C# program for the above approach
using System;
class GFG
{
 
  // Function to find the winner
  static char Winner(int N)
  {
      bool player = true;
 
      // Backtrack from N to 1
      while (N > 1)
      {
          int den = (player) ? 9 : 2;
          int X = N / den;
          int Y = N % den;
          N = (Y > 0) ? X + 1 : X;
          player = !player;
      }
      if (player)
          return 'B';
      else
          return 'A';
  }
 
  // Driver Code
  static public void Main()
  {
    int N = 10;
    Console.WriteLine(Winner(N));
  }
}
 
// This code is contributed by Dharanendra L V

Javascript

// Javascript program for the above approach
function Winner(N)
{
    var player = Boolean(true);
 
    // Backtrack from N to 1
    while(N > 1)
    {
        var den = (player) ? 9 : 2;
        var X = parseInt(N/den), Y = parseInt(N%den);
        N = (Y)? X + 1: X;
        player = !player;
    }
    if (player)
      document.write( 'B');
    else
      document.write( 'A');
}
var N = 10;
Winner(N);
 
// This code is contributed by SoumikMondal

输出：

B

时间复杂度： O(log N)
辅助空间： O(1)