莫兹金数

在数学中，给定数字n的Motzkin数是在圆的n个点之间绘制非相交和弦的不同方式的数目(不一定要通过一个弦接触每个点)。
例如，对于n = 3，M ₄ = 9。

^第N^个Motzkin数的递推关系为：

莫兹金数可用于查找：

• 长度为n – 1的正整数序列的数量，其中开始和结束元素为1或2，任何两个连续元素之间的差为-1、0或1。
• 如果只允许在每一步中仅向右移动(上，下或直线)，但在网格的右上象限上，从坐标(0，0)到坐标(n，0)的步数(以n步为单位)不会浸入y = 0轴以下。

  例如 –

  下图显示了从(0，0)到(4，0)的9条有效Motzkin路径。
  

  例子 ：

  Input : n = 4
  Output : 9
  
  Input : n = 5
  Output : 21
  

  以下是查找第n个Motzkin数的程序：

  C++

  // CPP Program to find Nth Motzkin Number.
  #include 
  using namespace std;
    
  // Return the nth Motzkin Number.
  int motzkin(int n)
  {
      // Base Case
      if (n == 0 || n == 1)
          return 1;
    
      // Recursive step
      return ((2 * n + 1) * motzkin(n - 1) +
              (3 * n - 3) * motzkin(n - 2)) / (n + 2);
  }
    
  // Driven Program
  int main()
  {
      int n = 8;
      cout << motzkin(n) << endl;
      return 0;
  }

  Java

  // Java Program to find Nth Motzkin Number.
  import java.util.*;
    
  class Digits
  {
      // Return the nth Motzkin Number.
      public static int motzkin(int n)
      {
          // Base Case
          if (n == 0 || n == 1)
              return 1;
    
          // Recursive step
          return ((2 * n + 1) * motzkin(n - 1) +
                  (3 * n - 3) * motzkin(n - 2)) / (n + 2);
      } 
        
      // driver code    
      public static void main(String[] args)
      {
          int n = 8;
          System.out.print( motzkin(n) );
      }
  }
    
  // This code is contributed by rishabh_jain

  Python3

  # Python3 program to find Nth Motzkin Number.
    
  # Return the nth Motzkin Number.
  def motzkin(n) :
        
      # Base Case
      if (n == 0 or n == 1) :
          return 1
    
      # Recursive step
      return ((2 * n + 1) * motzkin(n - 1) + 
              (3 * n - 3) * motzkin(n - 2)) / (n + 2)
    
  # Driver code
  n = 8
  print( motzkin(n) )
    
  # This code is contributed by rishabh_jain

  C#

  // C# Program to find Nth Motzkin Number.
  using System;
    
  class GFG {
        
      // Return the nth Motzkin Number.
      public static int motzkin(int n)
      {
            
          // Base Case
          if (n == 0 || n == 1)
              return 1;
    
          // Recursive step
          return ((2 * n + 1) * motzkin(n - 1) +
              (3 * n - 3) * motzkin(n - 2)) / (n + 2);
      } 
        
      // driver code 
      public static void Main()
      {
          int n = 8;
            
          Console.WriteLine( motzkin(n) );
      }
  }
    
  // This code is contributed by vt_m

  输出 ：

  323
  

  使用动态编程：

  以下是找到第n个Motzkin数的动态编程解决方案：

  C++

  // CPP Program to find Nth Motzkin Number.
  #include 
  using namespace std;
    
  // Return the nth Motzkin Number.
  int motzkin(int n)
  {
      int dp[n + 1];
    
      // Base case
      dp[0] = dp[1] = 1;
    
      // Finding i-th Motzkin number.
      for (int i = 2; i <= n; i++)
          dp[i] = ((2 * i + 1) * dp[i - 1] + 
                    (3 * i - 3) * dp[i - 2]) / (i + 2);
    
      return dp[n];
  }
  // Driven Program
  int main()
  {
      int n = 8;
      cout << motzkin(n) << endl;
      return 0;
  }
  

  Java

  // Java Program to find Nth Motzkin Number.
  import java.util.*;
    
  class Digits
  {
      // Return the nth Motzkin Number.
      public static int motzkin(int n)
      {
          int[] dp = new int[n+1];
    
          // Base case
          dp[0] = dp[1] = 1;
    
          // Finding i-th Motzkin number.
          for (int i = 2; i <= n; i++)
              dp[i] = ((2 * i + 1) * dp[i - 1] + 
                  (3 * i - 3) * dp[i - 2]) / (i + 2);
    
          return dp[n];
      }
        
      // driver code    
      public static void main(String[] args)
      {
          int n = 8;
          System.out.print( motzkin(n) );
      }
  }
    
  // This code is contributed by rishabh_jain
  

  Python3

  # Python3 program to find Nth Motzkin Number.
    
  # Return the nth Motzkin Number.
  def motzkin(n) :
        
      dp = [None] * (n+1)
    
      # Base case
      dp[0] = dp[1] = 1;
    
      i = 2
      # Finding i-th Motzkin number.
      while i <= n :
          dp[i] = ((2 * i + 1) * dp[i - 1] + 
                  (3 * i - 3) * dp[i - 2]) / (i + 2);
          i = i + 1
      return dp[n];
    
  # Driver code
  n = 8
  print( motzkin(n) )
    
  # This code is contributed by rishabh_jain
  

  C＃

  // C# Program to find Nth Motzkin Number.
  using System;
    
  class GFG {
        
      // Return the nth Motzkin Number.
      public static int motzkin(int n)
      {
          int[] dp = new int[n+1];
    
          // Base case
          dp[0] = dp[1] = 1;
    
          // Finding i-th Motzkin number.
          for (int i = 2; i <= n; i++)
              dp[i] = ((2 * i + 1) * dp[i - 1] + 
               (3 * i - 3) * dp[i - 2]) / (i + 2);
    
          return dp[n];
      }
        
      // driver code 
      public static void Main()
      {
          int n = 8;
            
          Console.WriteLine( motzkin(n) );
      }
  }
    
  // This code is contributed by vt_m
  

  输出 ：

  323