C program to find Armstrong numbers between 1 to n

Example

Input

Enter lower limit: 1
Enter upper limit: 1000

Output

Armstrong number between 1 to 1000 are:
1, 2, 3, 4, 5, 6, 7, 8, 9, 370, 371, 407

An Armstrong number is a n-digit number that is equal to the sum of n^th power of its digits. For example,
6 = 6¹ = 6
371 = 3³ + 7³ + 1³ = 371

/**
 * C program to print Armstrong numbers from 1 to n
 */
#include <stdio.h>
#include <math.h>

int main()
{
    int num, lastDigit, digits, sum, i, end;

    /* Input upper limit from user */
    printf("Enter upper limit: ");
    scanf("%d", &end);

    printf("Armstrong number between 1 to %d are: \n", end);

    for(i=1; i<=end; i++)
    {
        sum = 0;

        /* Copy the value of num for processing */
        num = i;

        /* Find total digits in num */
        digits = (int) log10(num) + 1;

        /* Calculate sum of power of digits */
        while(num > 0)
        {
            /* Extract last digit */
            lastDigit = num % 10;

            // Find sum of power of digits
            // Use ceil() function to overcome any rounding errors by pow()
            sum = sum + ceil(pow(lastDigit, digits));

            /* Remove the last digit */
            num = num / 10;
        }

        /* Check for Armstrong number */
        if(i == sum)
        {
            printf("%d, ", i);
        }

    }

    return 0;
}

Program to find Armstrong numbers in given range

/**
 * C program to generate Armstrong numbers in a given range
 */
#include <stdio.h>
#include <math.h>

int main()
{
    int num, lastDigit, digits, sum, i;
    int start, end;

    /* Input lower and upper limit from user */
    printf("Enter lower limit: ");
    scanf("%d", &start);
    printf("Enter upper limit: ");
    scanf("%d", &end);

    printf("Armstrong number between %d to %d are: \n", start, end);

    for(i=start; i<=end; i++)
    {
        sum = 0;

        /* Copy the value of num for processing */
        num = i;

        /* Find total digits in num */
        digits = (int) log10(num) + 1;

        /* Calculate sum of power of digits */
        while(num > 0)
        {
            /* Extract the last digit */
            lastDigit = num % 10;

            // Find sum of power of digits
            // Use ceil() function to overcome any rounding errors by pow()
            sum = sum + ceil(pow(lastDigit, digits));

            /* Remove the last digit */
            num = num / 10;
        }

        /* Check for Armstrong number */
        if(i == sum)
        {
            printf("%d, ", i);
        }
    }

    return 0;
}

/*
Output :

Enter lower limit: 1
Enter upper limit: 10000
Armstrong number between 1 to 10000 are:
1, 2, 3, 4, 5, 6, 7, 8, 9, 153, 370, 371, 407, 1634, 8208, 9474,
*/