I am writing a recursive function to print all the divisor's of a number including 1 and itself.
I am using a functionand inside that I am usingCode:print_recur(int number)how can i get it to print out iff a number is divisible and just skip the ones that arent.Code:number % (number - 1)
It will end up printing for 10: 10, 5, 2, 1
When I do this it prints only the number 1, can anyone help w/ that?Code:void print_all_divisor(int number) { int divis; if (number <=0) return 0; if ((number%(number-1)) == 0) { divis = number-1; printf("%d", divis); return print_all_divisor(number-1); } if((number%(number-1)) != 0) return print_all_divisor(number-1);
May be this will help you out..
Code:#include <stdio.h> void divisor(int n,int i) { if(i<=n) { if(n%i==0) printf("%d ",n/i); divisor(n,i+1); } } void print_recur(int n) { divisor(n,1); } int main() { int n=10; print_recur(n); return 0; }
Last edited by chottachatri; 10-27-2008 at 12:34 AM.
Or even this one?
Code:#include <stdio.h> void print_recur(int n) { static int i=1; if(i<=n) { if(n%i==0) printf("%d ",n/i); i++; print_recur(n); } else i=1; } int main() { int n=10; print_recur(n); return 0; }
Last edited by chottachatri; 10-27-2008 at 01:02 AM.
No number will will ever be divided by its one-lesser other than 2 (1).
Without using loops or statics, I don't think this can be done with only a single parameter recursion function. If you're implementing this for pedantic reasons, you may want to take a look at the gcd algo (Greatest Common Divisor) as its generally a good starting point for learning recursion.
There you go Chotta man i owe you a big one thanks a lot
