Given a number N, we need to print numbers from 1 to N with out direct recursion, loops, labels. Basically we need to insert in above code snippet so that it can be able to print numbers from 1 to N?
#include <stdio.h> #define N 20; int main() { // Your code goes Here. } |
Examples :
Input : 10 Output : 1 2 3 4 5 6 7 8 9 10 Input : 5 Output : 1 2 3 4 5
We have already discussed solutions in below posts :
Print 1 to 100 in C++, without loop and recursion
How will you print numbers from 1 to 100 without using loop?
Here’s the code that can print the numbers from 1 to 100 with out direct recursion, loops and labels. The code uses indirect recursion.
C
// C program to print from 1 to N using // indirect recursion/ #include<stdio.h> // We can avoid use of these using references #define N 20; int n = 1; // Prints n, increments n and calls fun1() void fun1() { if (n <= N) { printf ( "%d" , n); n++; fun2(); } else return ; } // Prints n, increments n and calls fun2() void fun2() { if (n <= N) { printf ( "%d" , n); n++; fun1(); } else return ; } // Driver Program int main( void ) { fun1(); return 0; } |
Java
// Java program to print from 1 to N using // indirect recursion class GFG { // We can avoid use of these using references static final int N = 20 ; static int n = 1 ; // Prints n, increments n and calls fun1() static void fun1() { if (n <= N) { System.out.printf( "%d " , n); n++; fun2(); } else { return ; } } // Prints n, increments n and calls fun2() static void fun2() { if (n <= N) { System.out.printf( "%d " , n); n++; fun1(); } else { return ; } } // Driver Program public static void main(String[] args) { fun1(); } } // This code is contributed by Rajput-Ji |
Python3
# Python program to prfrom 1 to N using # indirect recursion # We can avoid use of these using references N = 20 ; n = 1 ; # Prints n, increments n and calls fun1() def fun1(): global N, n; if (n < = N): print (n, end = " " ); n + = 1 ; fun2(); else : return ; # Prints n, increments n and calls fun2() def fun2(): global N, n; if (n < = N): print (n, end = " " ); n + = 1 ; fun1(); else : return ; # Driver Program if __name__ = = '__main__' : fun1(); # This code is contributed by 29AjayKumar |
C#
// C# program to print from 1 to N using // indirect recursion using System; class GFG { // We can avoid use of these using references static readonly int N = 20; static int n = 1; // Prints n, increments n and calls fun1() static void fun1() { if (n <= N) { Console.Write( "{0} " , n); n++; fun2(); } else { return ; } } // Prints n, increments n and calls fun2() static void fun2() { if (n <= N) { Console.Write( "{0} " , n); n++; fun1(); } else { return ; } } // Driver Code public static void Main(String[] args) { fun1(); } } // This code is contributed by Rajput-Ji |
PHP
<?php // PHP program to print // from 1 to N using // indirect recursion // We can avoid use of // these using references $N = 20; $n = 1; // Prints n, increments // n and calls fun1() function fun1() { global $N ; global $n ; if ( $n <= $N ) { echo $n , " " ; $n ++; fun2(); } else return ; } // Prints n, increments // n and calls fun2() function fun2() { global $N ; global $n ; if ( $n <= $N ) { echo $n , " " ; $n ++; fun1(); } else return ; } // Driver Code fun1(); // This code is contributed // by m_kit ?> |
Output :
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
How This works?:
In above Above program, we just used two functions. One calls others and other one calls previous one, therefore indirect recursion.
Exercise :
Modify the above program to use N as a parameter instead of making it global.
This article is contributed by Umamaheswararao Tumma of Jntuh college of Engineering Hyderabad. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
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