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

• Difficulty Level : Basic
• Last Updated : 16 Aug, 2022

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 ``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 ` `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 ``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 ` `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)

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