Finding whether a number is armstrong number of not

The need:

     This one is trivial program at this point of time. Still putting on demand of Rahul Devra.For those who dont know what is Armstrong number (Narcissistic number) first have a look at Narcissistic number on wikipedia.

The code:

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/*Find whether entered number is armstrong number or not*/
#include<stdio.h>
#include<math.h>
int digit_count(int x);
int isArmstrong(int x);

int main()
{
    int flg=0,num;
    printf("This is a program to find whether a number is armstrong number\n");
    printf("enter a natural number\n");
    scanf("%d",&num);
    if(isArmstrong(num))
    printf("%d is an armstrong number\n",num);
    else
    printf("%d is not an armstrong number\n",num);
    getchar();
    return 0;
}

int digit_count(int x)
{
    int count=0;
    if(x==0)
    return 0;
    while(x/=10)
    count++;
    return count+1;
}

int isArmstrong(int x)
{
    int i,num_digit,sum=0,current_digit,holder;
    num_digit = digit_count(x);
    holder=x;
    for(i=0;i<num_digit;i++)
    {
        current_digit = x%10;
        sum+=pow(current_digit,num_digit);
        x/=10;
    }
    if(sum==holder)
    return 1;
    else
    return 0;
}

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Approach:
   The approach is simple. Go by the definition, take digit by digit check for the condition for Armstrong number and that's it.