Skip to content
Related Articles

Related Articles

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

View Discussion
Improve Article
Save Article
  • Difficulty Level : Basic
  • Last Updated : 16 Aug, 2022
View Discussion
Improve Article
Save Article

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;
}


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


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


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;
}


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


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(N)
Auxiliary Space: O(1)


My Personal Notes arrow_drop_up
Recommended Articles
Page :

Start Your Coding Journey Now!