Print 1 to 100 without loop using Goto and Recursive-main

Last Updated : 06 Apr, 2023

Our task is to print all numbers from 1 to 100 without using a loop. There are many ways to print numbers from 1 to 100 without using a loop. Two of them are the goto statement and the recursive main.

Print numbers from 1 to 100 Using Goto statement

Follow the steps mentioned below to implement the goto statement:

declare variable i of value 0
declare the jump statement named begin, which increases the value of i by 1 and prints it.
write an if statement with condition i < 100, inside which call the jump statement begin

Below is the implementation of the above approach:

C++

   #include <iostream>
using namespace std;
 
int main()
{
    int i = 0;
 
begin:
    i = i + 1;
    cout << i << " ";
 
    if (i < 100) {
        goto begin;
    }
    return 0;
}
 
// This code is contributed by ShubhamCoder
 
  

C

   #include <stdio.h>
 
int main()
{
    int i = 0;
begin:
    i = i + 1;
    printf("%d ", i);
 
    if (i < 100)
        goto begin;
    return 0;
}
 
  

Java

   public class Main {
    public static void main(String[] args) {
        int i = 0;
 
        while (i < 100) {
            i = i + 1;
            System.out.print(i + " ");
        }
    }
}
 
  

C#

   using System;
 
class GFG {
 
    static public void Main()
    {
        int i = 0;
    begin:
        i = i + 1;
        Console.Write(" " + i + " ");
 
        if (i < 100) {
            goto begin;
        }
    }
}
 
// This code is contributed by ShubhamCoder
 
  

Python3

   i = 0
 
while i < 100:
    i = i + 1
    print(i, end=' ')

Javascript

   // Javascript
let i = 0;
 
while (i < 100) {  // loop will iterate 100 times
    i += 1;
    console.log(i, end=' '); // print i and add a space
}
 
  

Output

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100

Time complexity: O(1)
Auxiliary Space: O(1)

Print numbers from 1 to 100 Using recursive-main

Follow the steps mentioned below to implement the recursive main:

declare variable i of value 1.
keep calling the main function till i < 100.

Below is the implementation of the above approach:

C

   // C program to count all pairs from both the
// linked lists whose product is equal to
// a given value
#include <stdio.h>
 
int main()
{
    static int i = 1;
    if (i <= 100) {
        printf("%d ", i++);
        main();
    }
    return 0;
}
 
  

C++

   // C++ program to count all pairs from both the
// linked lists whose product is equal to
// a given value
#include <iostream>
using namespace std;
 
int main()
{
    static int i = 1;
 
    if (i <= 100) {
        cout << i++ << " ";
        main();
    }
    return 0;
}
 
// This code is contributed by ShubhamCoder
 
  

Java

   // Java program to count all pairs from both the
// linked lists whose product is equal to
// a given value
class GFG {
    static int i = 1;
 
    public static void main(String[] args)
    {
 
        if (i <= 100) {
            System.out.printf("%d ", i++);
            main(null);
        }
    }
}
 
// This code is contributed by Rajput-Ji
 
  

Python3

   # Python3 program to count all pairs from both
# the linked lists whose product is equal to
# a given value
 
def main(i):
 
    if (i <= 100):
        print(i, end=" ")
        i = i + 1
        main(i)
 
i = 1
main(i)
 
# This code is contributed by SoumikMondal

C#

   // C# program to count all pairs from both the
// linked lists whose product is equal to
// a given value
using System;
 
class GFG {
    static int i = 1;
 
    public static void Main(String[] args)
    {
        if (i <= 100) {
            Console.Write("{0} ", i++);
            Main(null);
        }
    }
}
 
// This code is contributed by Rajput-Ji
 
  

Javascript

   // Program to count all pairs from both
//the linked lists whose product is equal to
// a given value
function main(i) {
    if (i <= 100) {
        console.log(i + " ");
        i = i + 1;
        main(i);
    }
}
 
let i = 1;
main(i);
 
  

Output

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100

Time complexity: O(1)
Auxiliary Space: O(1)

How to print a number 100 times without using loop and recursion in C?

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Print 1 to 100 without loop using Goto and Recursive-main

Print numbers from 1 to 100 Using Goto statement

C++

C

Java

C#

Python3

Javascript

Print numbers from 1 to 100 Using recursive-main

C

C++

Java

Python3

C#

Javascript

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