Armstrong number c program

Armstrong number c program: c programming code to check whether a number is Armstrong or not. Armstrong number is a number which is equal to sum of digits raise to the power total number of digits in the number. Some Armstrong numbers are: 0, 1, 2, 3, 153, 370, 407, 1634, 8208 etc. Read more about Armstrong numbers at Wikipedia. We will consider base 10 numbers in our program. Algorithm to check Armstrong is: First we calculate number of digits in our program and then compute sum of individual digits raise to the power number of digits. If this sum equals input number then number is Armstrong otherwise not an Armstrong number.

Examples:
7 = 7^1
371 = 3^3 + 7^3 + 1^3 (27 + 343 +1)
8208 = 8^4 + 2^4 +0^4 + 8^4 (4096 + 16 + 0 + 4096).
1741725 = 1^7 + 7^7 + 4^7 + 1^7 + 7^7 + 2^7 +5^7 (1 + 823543 + 16384 + 1 + 823543 +128 + 78125)

.

#include <stdio.h>
 
int power(int, int);
 
int main()
{
   int n, sum = 0, temp, remainder, digits = 0;
 
   printf("Input an integer\n");
   scanf("%d", &n);
 
   temp = n;
   // Count number of digits
   while (temp != 0) {
      digits++;
      temp = temp/10;
   }
 
   temp = n;
 
   while (temp != 0) {
      remainder = temp%10;
      sum = sum + power(remainder, digits);
      temp = temp/10;
   }
 
   if (n == sum)
      printf("%d is an Armstrong number.\n", n);
   else
      printf("%d is not an Armstrong number.\n", n);
 
   return 0;
}
 
int power(int n, int r) {
   int c, p = 1;
 
   for (c = 1; c <= r; c++) 
      p = p*n;
 
   return p;   
}

=================================================================

C program to check Armstrong number using function


#include <stdio.h>
 
int check_armstrong(long long);
long long power(int, int);
 
int main () {
   long long n;
 
   printf("Input a number\n");
   scanf("%lld", &n);
 
   if (check_armstrong(n) == 1)
      printf("%lld is an armstrong number.\n", n); 
   else
      printf("%lld is not an armstrong number.\n", n);   
 
   return 0;
}
 
int check_armstrong(long long n) {
   long long sum = 0, temp;
   int remainder, digits = 0;
 
   temp = n;
 
   while (temp != 0) {
      digits++;
      temp = temp/10;
   }
 
   temp = n;
 
   while (temp != 0) {
      remainder = temp%10;
      sum = sum + power(remainder, digits);
      temp = temp/10;
   }
 
   if (n == sum)
      return 1;
   else
      return 0;
}
 
long long power(int n, int r) {
   int c;
   long long p = 1;
 
   for (c = 1; c <= r; c++) 
      p = p*n;
 
   return p;   
}