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++
#include <iostream>
using namespace std;
class gfg
{
public:
void printNos(unsigned int n)
{
if(n > 0)
{
printNos(n - 1);
cout << n << " ";
}
return;
}
};
int main()
{
gfg g;
g.printNos(100);
return 0;
}
|
C
#include <stdio.h>
void printNos(unsigned int n)
{
if(n > 0)
{
printNos(n - 1);
printf("%d ", n);
}
return;
}
int main()
{
printNos(100);
getchar();
return 0;
}
|
Java
import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;
class GFG
{
static void printNos(int n)
{
if(n > 0)
{
printNos(n - 1);
System.out.print(n + " ");
}
return;
}
public static void main(String[] args)
{
printNos(100);
}
}
|
Python3
def printNos(n):
if n > 0:
printNos(n - 1)
print(n, end = ' ')
printNos(100)
|
C#
using System;
class GFG
{
static void printNos(int n)
{
if(n > 0)
{
printNos(n - 1);
Console.Write(n + " ");
}
return;
}
public static void Main()
{
printNos(100);
}
}
|
Javascript
<script>
function printNos(n)
{
if(n > 0)
{
printNos(n - 1);
document.write(n + " ");
}
return;
}
printNos(100);
</script>
|
PHP
<?php
function printNos($n)
{
if($n > 0)
{
printNos($n - 1);
echo $n, " ";
}
return;
}
printNos(100);
?>
|
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 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(n)
Method 2:
C++
#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;
}
|
Java
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);
}
}
}
|
Python3
def printNos(initial, last):
if(initial<=last):
print(initial)
printNos(initial+1,last)
printNos(1,10)
|
C#
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);
}
}
}
|
Javascript
printNos(1, 100);
function printNos(initial, last)
{
if (initial <= last) {
document.write(initial + " ");
printNos(initial + 1, last);
}
}
|
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 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(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;
PRINT(i);
return 0;
}
|
Java
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() {
LEVEL(); LEVEL(); LEVEL(); LEVEL(); LEVEL(); LEVEL(); LEVEL(); LEVEL(); LEVEL(); LEVEL();
}
public static void main(String[] args) {
PRINT();
}
}
|
Python3
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():
Main.LEVEL(); Main.LEVEL(); Main.LEVEL(); Main.LEVEL(); Main.LEVEL(); Main.LEVEL(); Main.LEVEL(); Main.LEVEL(); Main.LEVEL(); Main.LEVEL()
@staticmethod
def main(args):
Main.PRINT()
Main.main(None)
|
C#
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()
{
LEVEL(); LEVEL(); LEVEL(); LEVEL(); LEVEL(); LEVEL(); LEVEL(); LEVEL(); LEVEL(); LEVEL();
}
public static void Main(string[] args)
{
PRINT();
}
}
|
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 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
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. You can use the GFG IDE!
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.
Python3
numbers = range(1, 101)
list(map(print, numbers))
|
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
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
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:
Python3
f = lambda x: print(x, end=' ') or f(x+1) if x < 100 else None
f(1)
|
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 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
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).