Armstrong Numbers

Input An integer and number of digits in the integer
Output	Number is Armstrong or not.
Complexity O(d) where d is the number of digits in the integer

Armstrong(num,digits)
1    n=num
2    while(num>0) do
3        digit =n%10;
4        sum=sum+pow(digit,digits)
5        n= n/10;
6   if (sum =num)
7        print it is an Armstrong number
8    Else 
9        print it is not an Armstrong number

#include<stdio.h>
int main()
{
   int n,i,d,sum=0,m;
   printf("\n enter any integer number  :");
   scanf("%d",&n);
   printf("\n enter number of digits in the number  :");
   scanf("%d",&n);
   m=n;
   while(n>0)
   {
        d=n%10;
        sum+=d*d*d;
        n=n/10;
   }
   if(m==sum)
   {
    	  printf("\ its armstrong number");
   }
   else
   {
	  printf("\ its not armstrong number");
   }
   return 0;     
}