打印数字系列，不使用任何循环

问题–给定两个数N和K，我们的任务是从N中减去数K，直到number(N)大于零，一旦N变为负数或零，我们就开始加K直到该数成为原始数(N )。
注意：不允许使用任何循环。

例子 ：

Input : N = 15 K = 5  
Output : 15 10 5 0 1 5 10 15

Input : N = 20 K = 6
Output : 20 14 8 2 -4 2 8 14 20 

解释–我们可以使用递归的想法来做到这一点，那就是我们一次又一次地调用该函数，直到N大于零(在每个函数调用中，我们将N减去K)。一旦数字变为负数或零，我们就开始在每个函数调用中添加K，直到数字变为原始数字为止。这里我们使用一个单一的函数加法和减法，而是加法或减法函数之间，我们使用了布尔标志开关。

C++

// C++ program to Print Number
// series without using loop
#include 
  
using namespace std;
  
// function print series
// using recursion
void PrintNumber(int N, int Original, int K, bool flag)
{
  
    // print the number
    cout << N << " ";
  
    // change flag if number
    // become negative
    if (N <= 0)
        flag = !flag;
  
    // base condition for
    // second_case (Adding K)
    if (N == Original && !flag)
  
        return;
  
    // if flag is true
    // we subtract value until
    // number is greater then zero
    if (flag == true) {
  
        PrintNumber(N - K, Original, K, flag);
  
        return;
    }
  
    // second case (Addition )
    if (!flag) {
  
        PrintNumber(N + K, Original, K, flag);
  
        return;
    }
}
  
// driver program
int main()
{
  
    int N = 20, K = 6;
  
    PrintNumber(N, N, K, true);
  
    return 0;
}

Java

// Java program to Print Number
// series without using loop
  
import java.io.*;
import java.util.*;
  
class GFG
{
    public static void PrintNumber(int N, int Original, int K, boolean flag)
    {
  
        // print the number
        System.out.print(N + " ");
  
        // change flag if number
        // become negative
        if (N <= 0)
            flag = !flag;
  
        // base condition for
        // second_case (Adding K)
        if (N == Original && !flag)
  
            return;
  
        // if flag is true
        // we subtract value until
        // number is greater then zero
        if (flag == true)
        {
            PrintNumber(N - K, Original, K, flag);
            return;
        }
  
        // second case (Addition )
        if (!flag)
        {
            PrintNumber(N + K, Original, K, flag);
            return;
        }
    }
  
    public static void main (String[] args)
    {
        int N = 20, K = 6;
        PrintNumber(N, N, K, true);
    }
}
// This code is contributed by Mohit Gupta_OMG

Python3

# Python program to Print Number
# series without using loop
  
def PrintNumber(N, Original, K, flag):
    #print the number
    print(N, end = " ")
      
    # change flag if number
    # become negative
      
    if (N <= 0):
        if(flag==0):
            flag = 1
        else:
            flag = 0
          
    # base condition for
    # second_case (Adding K)
      
    if (N == Original and (not(flag))):
        return
      
    # if flag is true
    # we subtract value until
    # number is greater then zero
      
    if (flag == True):
        PrintNumber(N - K, Original, K, flag)
        return
      
    # second case (Addition )
    if (not(flag)):
        PrintNumber(N + K, Original, K, flag);
        return
      
N = 20
K = 6
PrintNumber(N, N, K, True)
  
# This code is contributed by Mohit Gupta_OMG

C#

// C# program to Print Number
// series without using loop
using System;
  
public class GFG {
          
    // function print series
    // using recursion
    static void PrintNumber(int N,
             int Original, int K, bool flag)
    {
      
        // print the number
        Console.Write(N + " ");
      
        // change flag if number
        // become negative
        if (N <= 0)
            flag = !flag;
      
        // base condition for
        // second_case (Adding K)
        if (N == Original && !flag)
            return;
      
        // if flag is true
        // we subtract value until
        // number is greater then zero
        if (flag == true)
        {
            PrintNumber(N - K, Original, K,
                                      flag);
            return;
        }
      
        // second case (Addition )
        if (!flag)
        {
            PrintNumber(N + K, Original, K,
                                      flag);
            return;
        }
    }
      
    // driver program
    static public void Main ()
    {
        int N = 20, K = 6;
  
        PrintNumber(N, N, K, true);
    }
}
  
// This code is contributed by vt_m.

输出 ：

20 14 8 2 -4 2 8 14 20