杂耍人序列 |第 2 组（使用递归）

杂耍者序列是一系列整数，其中第一项以正整数a开头，其余项使用以下递归关系从前一项生成：

$a_{k+1}=\begin{Bmatrix} \lfloor a_{k}^{1/2} \rfloor & for \quad even \quad a_k\\ \lfloor a_{k}^{3/2} \rfloor & for \quad odd \quad a_k \end{Bmatrix}$
Juggler Sequence starting with number 3: 
5, 11, 36, 6, 2, 1
Juggler Sequence starting with number 9: 
9, 27, 140, 11, 36, 6, 2, 1

给定一个数字N ，我们必须打印这个数字的杂耍者序列作为序列的第一项。

例子：

Input: N = 9 
Output: 9, 27, 140, 11, 36, 6, 2, 1 
We start with 9 and use above formula to get next terms.

Input: N = 6 
Output: 6, 2, 1

迭代方法：我们已经在这个问题的Set 1中看到了迭代方法。

递归方法：在这种方法中，我们将从 N 开始递归遍历。对于每个递归步骤，请按照以下步骤操作

• 输出N的值
• 如果 N达到 1则结束递归
• 否则，根据奇数或偶数遵循公式，并在新导出的数字上调用递归函数。

下面是该方法的实现：

C++

// C++ code for the above approach
#include 
using namespace std;
 
// Recursive function to print
// the juggler sequence
void jum_sequence(int N)
{
 
    cout << N << " ";
 
    if (N <= 1)
        return;
    else if (N % 2 == 0)
    {
        N = floor(sqrt(N));
        jum_sequence(N);
    }
    else
    {
        N = floor(N * sqrt(N));
        jum_sequence(N);
    }
}
 
// Driver code
int main()
{
   
    // Juggler sequence starting with 10
    jum_sequence(10);
    return 0;
}
 
// This code is contributed by Potta Lokesh

Java

// Java code for the above approach
class GFG
{
   
    // Recursive function to print
    // the juggler sequence
    public static void jum_sequence(int N) {
 
        System.out.print(N + " ");
 
        if (N <= 1)
            return;
        else if (N % 2 == 0) {
            N = (int) (Math.floor(Math.sqrt(N)));
            jum_sequence(N);
        } else {
            N = (int) Math.floor(N * Math.sqrt(N));
            jum_sequence(N);
        }
    }
 
    // Driver code
    public static void main(String args[]) {
 
        // Juggler sequence starting with 10
        jum_sequence(10);
    }
}
 
// This code is contributed by Saurabh Jaiswal

Python3

# Python code to implement the above approach
 
# Recursive function to print
# the juggler sequence
def jum_sequence(N):
     
    print(N, end =" ")
 
    if (N == 1):
        return
    elif N & 1 == 0:
        N = int(pow(N, 0.5))
        jum_sequence(N)
    else:
        N = int(pow(N, 1.5))
        jum_sequence(N)
 
 
# Juggler sequence starting with 10
jum_sequence(10)

C#

// C# code for the above approach
using System;
 
class GFG{
 
// Recursive function to print
// the juggler sequence
public static void jum_sequence(int N)
{
    Console.Write(N + " ");
 
    if (N <= 1)
        return;
    else if (N % 2 == 0)
    {
        N = (int)(Math.Floor(Math.Sqrt(N)));
        jum_sequence(N);
    }
    else
    {
        N = (int)Math.Floor(N * Math.Sqrt(N));
        jum_sequence(N);
    }
}
 
// Driver code
public static void Main()
{
     
    // Juggler sequence starting with 10
    jum_sequence(10);
}
}
 
// This code is contributed by Saurabh Jaiswal

Javascript

// Javascript code for the above approach
 
// Recursive function to print
// the juggler sequence
function jum_sequence(N){
 
    document.write(N +" ");
 
    if (N <= 1)
        return;
    else if (N % 2 == 0)
    {
        N = Math.floor(Math.sqrt(N));
        jum_sequence(N);
    }
    else
    {
        N = Math.floor(N * Math.sqrt(N));
        jum_sequence(N);
    }
}
 
// Driver code
// Juggler sequence starting with 10
 
jum_sequence(10);
     
// This code is contributed by gfgking

输出：

10 3 5 11 36 6 2 1

时间复杂度： O(N)
辅助空间： O(1)