How will you print numbers from 1 to 100 without using a loop?
Last Updated :
23 Jul, 2025
If we take a look at this problem carefully, we can see that the idea of "loop" is to track some counter value, e.g., "i = 0" till "i <= 100". So, if we aren't allowed to use loops, how can we track something in the C language?
Well, one possibility is the use of 'recursion', provided we use the terminating condition carefully. Here is a solution that prints numbers using recursion.
Method-1:
C++
// C++ program to How will you print
// numbers from 1 to 100 without using a loop?
#include <iostream>
using namespace std;
class gfg
{
// It prints numbers from 1 to n.
public:
void printNos(unsigned int n)
{
if(n > 0)
{
printNos(n - 1);
cout << n << " ";
}
return;
}
};
// Driver code
int main()
{
gfg g;
g.printNos(100);
return 0;
}
// This code is contributed by SoM15242
#include <stdio.h>
// Prints numbers from 1 to n
void printNos(unsigned int n)
{
if(n > 0)
{
printNos(n - 1);
printf("%d ", n);
}
return;
}
// Driver code
int main()
{
printNos(100);
getchar();
return 0;
}
import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;
class GFG
{
// Prints numbers from 1 to n
static void printNos(int n)
{
if(n > 0)
{
printNos(n - 1);
System.out.print(n + " ");
}
return;
}
// Driver Code
public static void main(String[] args)
{
printNos(100);
}
}
// This code is contributed by Manish_100
# Python3 program to Print
# numbers from 1 to n
def printNos(n):
if n > 0:
printNos(n - 1)
print(n, end = ' ')
# Driver code
printNos(100)
# This code is contributed by Smitha Dinesh Semwal
// C# code for print numbers from
// 1 to 100 without using loop
using System;
class GFG
{
// Prints numbers from 1 to n
static void printNos(int n)
{
if(n > 0)
{
printNos(n - 1);
Console.Write(n + " ");
}
return;
}
// Driver Code
public static void Main()
{
printNos(100);
}
}
// This code is contributed by Ajit
<script>
// Javascript code for print numbers from
// 1 to 100 without using loop
// Prints numbers from 1 to n
function printNos(n)
{
if(n > 0)
{
printNos(n - 1);
document.write(n + " ");
}
return;
}
printNos(100);
// This code is contributed by rameshtravel07.
</script>
<?php
// PHP program print numbers
// from 1 to 100 without
// using loop
// Prints numbers from 1 to n
function printNos($n)
{
if($n > 0)
{
printNos($n - 1);
echo $n, " ";
}
return;
}
// Driver code
printNos(100);
// This code is contributed by vt_m
?>
Output1 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...
Time Complexity: O(n)
Auxiliary Space: O(n)
Method 2:
C++
// C++ program
#include <bits/stdc++.h>
using namespace std;
void printNos(int initial, int last)
{
if (initial <= last) {
cout << initial << " ";
printNos(initial + 1, last);
}
}
int main()
{
printNos(1, 100);
return 0;
}
// This code is contributed by ukasp.
/*package whatever //do not write package name here */
import java.io.*;
class GFG {
public static void main(String[] args)
{
printNos(1, 100);
}
public static void printNos(int initial, int last)
{
if (initial <= last) {
System.out.print(initial + " ");
printNos(initial + 1, last);
}
}
}
def printNos(initial, last):
if(initial<=last):
print(initial)
printNos(initial+1,last)
printNos(1,10)
/*package whatever //do not write package name here */
using System;
public class GFG {
public static void Main(String[] args)
{
printNos(1, 100);
}
public static void printNos(int initial, int last)
{
if (initial <= last) {
Console.Write(initial + " ");
printNos(initial + 1, last);
}
}
}
// This code contributed by gauravrajput1
/*package whatever //do not write package name here */
printNos(1, 100);
function printNos(initial, last)
{
if (initial <= last) {
document.write(initial + " ");
printNos(initial + 1, last);
}
}
// This code is contributed by shivanisinghss2110
Output1 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...
Time Complexity : O(n)
Auxiliary Space: O(n)
Method 3: Using a MACRO
C++
#include <iostream>
using namespace std;
#define COUT(i) cout << i++ << " ";
#define LEVEL(N) N N N N N N N N N N
#define PRINT(i) LEVEL(LEVEL(COUT(i))); // 100 = 10×10
int main()
{
int i = 1;
// prints numbers from 1 to 100
PRINT(i);
return 0;
}
public class Main {
static int i = 1;
static void OUT() {
System.out.print(i++ + " ");
}
static void LEVEL() {
OUT(); OUT(); OUT(); OUT(); OUT(); OUT(); OUT(); OUT(); OUT(); OUT();
}
static void PRINT() {
// 100 = 10×10
LEVEL(); LEVEL(); LEVEL(); LEVEL(); LEVEL(); LEVEL(); LEVEL(); LEVEL(); LEVEL(); LEVEL();
}
public static void main(String[] args) {
// prints numbers from 1 to 100
PRINT();
}
}
class Main:
i = 1
@staticmethod
def OUT():
print(Main.i, end=' ')
Main.i += 1
@staticmethod
def LEVEL():
Main.OUT(); Main.OUT(); Main.OUT(); Main.OUT(); Main.OUT(); Main.OUT(); Main.OUT(); Main.OUT(); Main.OUT(); Main.OUT()
@staticmethod
def PRINT():
# 100 = 10×10
Main.LEVEL(); Main.LEVEL(); Main.LEVEL(); Main.LEVEL(); Main.LEVEL(); Main.LEVEL(); Main.LEVEL(); Main.LEVEL(); Main.LEVEL(); Main.LEVEL()
@staticmethod
def main(args):
# prints numbers from 1 to 100
Main.PRINT()
Main.main(None)
using System;
public class MainClass
{
static int i = 1;
static void OUT()
{
Console.Write(i++ + " ");
}
static void LEVEL()
{
OUT(); OUT(); OUT(); OUT(); OUT(); OUT(); OUT(); OUT(); OUT(); OUT();
}
static void PRINT()
{
// 100 = 10×10
LEVEL(); LEVEL(); LEVEL(); LEVEL(); LEVEL(); LEVEL(); LEVEL(); LEVEL(); LEVEL(); LEVEL();
}
public static void Main(string[] args)
{
// prints numbers from 1 to 100
PRINT();
}
}
// Define a variable i and initialize it to 1
let i = 1;
// Define a function OUT that prints the current value of i and increments it by 1
function OUT() {
console.log(i++ + " ");
}
// Define a function LEVEL that calls OUT() 10 times
function LEVEL() {
OUT(); OUT(); OUT(); OUT(); OUT(); OUT(); OUT(); OUT(); OUT(); OUT();
}
// Define a function PRINT that calls LEVEL() 10 times to print numbers from 1 to 100
function PRINT() {
// 100 = 10×10
LEVEL(); LEVEL(); LEVEL(); LEVEL(); LEVEL(); LEVEL(); LEVEL(); LEVEL(); LEVEL(); LEVEL();
}
// Main function
function main() {
// Prints numbers from 1 to 100
PRINT();
}
// Call the main function
main();
Output1 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...
Now try writing a program that does the same but without any "if" condition.
Hint : use some operator which can be used instead of "if".
Please note that the recursion technique is good, but every call to the function creates one "stack-frame" in the program stack. So if there's a constraint to the limited memory and we need to print a large set of numbers, "recursion" might not be a good idea. So what could be the other alternative?
Another alternative is the "goto" statement. Though use of "goto" is not suggested as a general programming practice as the "goto" statement changes the normal program execution sequence, in some cases, use of "goto" is the best working solution.
So please give a try to printing numbers from 1 to 100 with the "goto" statement.
Python range:
The given code uses the range() function to generate a sequence of numbers from 1 to 100, and then uses the map() function to apply the print() function to each number in the sequence. This approach allows the program to print the numbers from 1 to 100 without using a loop.
C++
#include <iostream>
int main() {
// Use a for loop to generate a sequence of numbers from 1 to 100
for (int i = 1; i <= 100; ++i) {
// Print each number in the sequence
std::cout << i << std::endl;
}
return 0;
}
//Thos code is contrbiuted by Utkarsh.
import java.util.stream.IntStream;
public class Main {
public static void main(String[] args) {
// Use IntStream.rangeClosed to generate a sequence of numbers from 1 to 100
IntStream numbers = IntStream.rangeClosed(1, 100);
// Use forEach to print each number in the sequence
numbers.forEach(System.out::println);
}
}
# Use the range function to generate a sequence of numbers from 1 to 100
numbers = range(1, 101)
# Use the map function to print each number in the sequence
list(map(print, numbers))
// Generate a sequence of numbers from 1 to 100
const numbers = Array.from({ length: 100 }, (_, index) => index + 1);
// Use forEach to print each number in the sequence
numbers.forEach(number => console.log(number));
Output1
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...
Note that the map function returns a map object, which needs to be converted to a list in order to print the values.
The time complexity of this approach is O(n), where n is the number of elements in the range generated by range(1, 101). In this case, n is 100, so the time complexity is constant. The space complexity of this approach is also O(n), since it creates a sequence of n integers.
This approach is justified because it is simple and efficient. The range() function generates a sequence of integers without having to create a list in memory, which can save memory and improve performance. The map() function applies a function to each element of a sequence, and using print() as the function allows for the numbers to be printed without needing a loop. Overall, this approach allows for a concise and efficient solution to the problem of printing numbers from 1 to 100 without using a loop.
Print 1 to 100 in C++, without loop and recursion
Another approach to print numbers from 1 to 100 without using a loop is to use recursion with a lambda function.
Here's an example implementation in Python:
C++
#include <iostream>
using namespace std;
// Define a recursive function to emulate the lambda function
// The function prints the current value of x, followed by a space,
// then recursively calls itself with x+1 until x reaches 100.
// Once x reaches 100, the function stops recursion.
int f(int x) {
// Print the current value of x followed by a space
cout << x << " ";
// If x is less than 100, recursively call f with x+1
// Otherwise, stop recursion (return 0)
return x < 100 ? f(x + 1) : 0;
}
int main() {
// Call the function f with an initial value of 1
f(1);
return 0;
}
//This code ios contribited by monu.
f = lambda x: print(x, end=' ') or f(x+1) if x < 100 else None
f(1)
function f(x) {
// Print the current value of x followed by a space
process.stdout.write(x + " ");
// If x is less than 100, recursively call f with x+1
// Otherwise, stop recursion
if (x < 100) {
return f(x + 1);
}
}
function main() {
// Call the function f with an initial value of 1
f(1);
}
// Calling the main function
main();
Output1 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...
In this code, we define a lambda function f that takes an argument x. The lambda function first prints x, and then calls itself with x+1 as the argument if x is less than 100. This recursive call continues until x reaches 101, at which point the lambda function returns None and the program terminates.
We call the lambda function f with x=1 to print the numbers from 1 to 100.
Note: The or operator is used in the lambda function to concatenate two expressions, so that the lambda function is still valid even if the print() function returns None (which it does by default).
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