报名号码

恩特林格数E(n，k)是{1，2，…，n + 1}的排列数，从k + 1开始，在最初下降后交替下降然后上升。恩特林格是由：

例如，对于n = 4和k = 2，E(4，2)为4。
他们是：
3 2 4 1 5
3 2 5 1 4
3 1 4 2 5
3 1 5 2 4
例子 ：

Input : n = 4, k = 2
Output : 4

Input : n = 4, k = 3
Output : 5

以下是查找Entringer编号E(n，k)的程序。该程序基于上述简单的递归公式。

C++

// CPP Program to find Entringer Number E(n, k)
#include 
using namespace std;
 
// Return Entringer Number E(n, k)
int zigzag(int n, int k)
{
    // Base Case
    if (n == 0 && k == 0)
        return 1;
 
    // Base Case
    if (k == 0)
        return 0;
 
    // Recursive step
    return zigzag(n, k - 1) +
           zigzag(n - 1, n - k);
}
 
// Driven Program
int main()
{
    int n = 4, k = 3;
    cout << zigzag(n, k) << endl;
    return 0;
}

Java

// JAVA Code For Entringer Number
import java.util.*;
 
class GFG {
     
    // Return Entringer Number E(n, k)
    static int zigzag(int n, int k)
    {
        // Base Case
        if (n == 0 && k == 0)
            return 1;
      
        // Base Case
        if (k == 0)
            return 0;
      
        // Recursive step
        return zigzag(n, k - 1) +
               zigzag(n - 1, n - k);
    }
     
    /* Driver program to test above function */
    public static void main(String[] args)
    {
        int n = 4, k = 3;
        System.out.println(zigzag(n, k));
         
    }
}
 
// This code is contributed by Arnav Kr. Mandal.

Python3

# Python Program to find Entringer Number E(n, k)
 
# Return Entringer Number E(n, k)
def zigzag(n, k):
 
    # Base Case
    if (n == 0 and k == 0):
        return 1
 
    # Base Case
    if (k == 0):
        return 0
 
    # Recursive step
    return zigzag(n, k - 1) + zigzag(n - 1, n - k);
 
# Driven Program
n = 4
k = 3
print(zigzag(n, k))
 
# This code is contributed by
# Smitha Dinesh Semwal

C#

// C# Code For Entringer Number
using System;
 
class GFG {
 
    // Return Entringer Number E(n, k)
    static int zigzag(int n, int k)
    {
        // Base Case
        if (n == 0 && k == 0)
            return 1;
 
        // Base Case
        if (k == 0)
            return 0;
 
        // Recursive step
        return zigzag(n, k - 1) +
               zigzag(n - 1, n - k);
    }
 
    /* Driver program to test above function */
    public static void Main()
    {
        int n = 4, k = 3;
        Console.WriteLine(zigzag(n, k));
    }
}
 
// This code is contributed by vt_m.

输出 ：

5

以下是使用动态编程查找Entringer编号的实现：

C++

// CPP Program to find Entringer Number E(n, k)
#include 
using namespace std;
 
// Return Entringer Number E(n, k)
int zigzag(int n, int k)
{
    int dp[n + 1][k + 1];
    memset(dp, 0, sizeof(dp));
 
    // Base cases
    dp[0][0] = 1;
    for (int i = 1; i <= n; i++)
        dp[i][0] = 0;   
 
    // Finding dp[i][j]
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= i; j++)
            dp[i][j] = dp[i][j - 1] +
                       dp[i - 1][i - j];
 
    return dp[n][k];
}
 
// Driven Program
int main()
{
    int n = 4, k = 3;
    cout << zigzag(n, k) << endl;
    return 0;
}

Java

// JAVA Code For Entringer Number
import java.util.*;
 
class GFG {
     
    // Return Entringer Number E(n, k)
    static int zigzag(int n, int k)
    {
        int dp[][] = new int[n + 1][k + 1];
        
        // Base cases
        dp[0][0] = 1;
        for (int i = 1; i <= n; i++)
            dp[i][0] = 0;   
      
        // Finding dp[i][j]
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= Math.min(i, k);
                                          j++)
                dp[i][j] = dp[i][j - 1] +
                           dp[i - 1][i - j];
            }
      
        return dp[n][k];
    }
     
    /* Driver program to test above function */
    public static void main(String[] args)
    {
        int n = 4, k = 3;
        System.out.println(zigzag(n, k));
    }
}
     
// This code is contributed by Arnav Kr. Mandal.   

Python3

# Python3 Program to find Entringer
# Number E(n, k)
 
# Return Entringer Number E(n, k)
def zigzag(n, k):
    dp = [[0 for x in range(k+1)]
             for y in range(n+1)]
 
    # Base cases
    dp[0][0] = 1
    for i in range(1, n+1):
        dp[i][0] = 0
 
    # Finding dp[i][j]
    for i in range(1, n+1):
        for j in range(1, k+1):
            dp[i][j] = (dp[i][j - 1]
                 + dp[i - 1][i - j])
                         
    return dp[n][k]
 
# Driven Program
n = 4
k = 3
print(zigzag(n, k))
 
# This code is contributed by
# Prasad Kshirsagar

C＃

// C# Code For Entringer Number
using System;
 
class GFG {
 
    // Return Entringer Number E(n, k)
    static int zigzag(int n, int k)
    {
        int[, ] dp = new int[n + 1, k + 1];
 
        // Base cases
        dp[0, 0] = 1;
        for (int i = 1; i <= n; i++)
            dp[i, 0] = 0;
 
        // Finding dp[i][j]
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= Math.Min(i, k);
                j++)
                dp[i, j] = dp[i, j - 1] + dp[i - 1, i - j];
        }
 
        return dp[n, k];
    }
 
    /* Driver program to test above function */
    public static void Main()
    {
        int n = 4, k = 3;
        Console.WriteLine(zigzag(n, k));
    }
}
 
// This code is contributed by vt_m.

Java脚本

 
// JavaScript program For Entringer Number
 
    // Return Entringer Number E(n, k)
    function zigzag(n, k)
    {
        let dp = new Array(n+1);
         
        // Loop to create 2D array using 1D array
        for (var i = 0; i < dp.length; i++) {
            dp[i] = new Array(2);
        }
          
        // Base cases
        dp[0][0] = 1;
        for (let i = 1; i <= n; i++)
            dp[i][0] = 0;   
        
        // Finding dp[i][j]
        for (let i = 1; i <= n; i++) {
            for (let j = 1; j <= Math.min(i, k);
                                          j++)
                dp[i][j] = dp[i][j - 1] +
                           dp[i - 1][i - j];
            }
        
        return dp[n][k];
    }
 
// Driver code
 
        let n = 4, k = 3;
        document.write(zigzag(n, k));
                             

输出 ：

5