How will you print numbers from 1 to 100 without using loop?
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 loop, how else can be track something in 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.
C++
// C++ program to How will you print // numbers from 1 to 100 without using loop? #include <iostream> using namespace std; class gfg { // 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 |
C
#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; } |
Java
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
# 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#
// 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 |
PHP
<?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 ?> |
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)
Now try writing a program that does the same but without any “if” construct.
Hint — use some operator which can be used instead of “if”.
Please note that recursion technique is good but every call to the function creates one “stack-frame” in program stack. So if there’s constraint to the limited memory and we need to print large set of numbers, “recursion” might not be a good idea. So what could be the other alternative?
Another alternative is “goto” statement. Though use of “goto” is not suggestible as a general programming practice as “goto” statement changes the normal program execution sequence yet in some cases, use of “goto” is the best working solution.
So please give a try printing numbers from 1 to 100 with “goto” statement. You can use GfG IDE!
Print 1 to 100 in C++, without loop and recursion
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