C Program to Display Armstrong Number Between Two Intervals

A positive integer with digits a, b, c, d… is called an Armstrong number of order n if the following condition is satisfied:

abcd... = a^n + b^n + c^n + d^n +...

Example:

153 = 1*1*1 + 5*5*5 + 3*3*3  

       =   1 + 125 + 27

       =     153        

Therefore, 153 is an Armstrong number.

To display the Armstrong number between two intervals we can use 2 different Methods with 4 approaches:

1. Without using the pow() function
2. Using pow() function

We will keep the same input in all the mentioned approaches and get an output accordingly.

Input: start = 1, end = 500
Output: 1, 153, 370, 371, 407
Explanation: The Armstrong number is the sum of the cubes of its digits. i.e.
1=1³ 
153= 1³ + 5³ + 3³ 
370= 3³ + 7³ + 0³ etc.

1. Without using the pow() function

C

// C program to demonstrate an armstrong number
// between the given intervals
#include <stdio.h>

int main()
{
    int s = 1, e = 500, num1, n, arm = 0,
          i, num2, c;

    // Iterating the for loop using
    // the given intervals
    for (i = s; i <= e; i++) {
        num1 = i;
        num2 = i;

        // Finding the number of digits
        while (num1 != 0) {
            num1 = num1 / 10;
            ++c;
        }

        // Finding the armstrong number
        while (num2 != 0) {
            n = num2 % 10;
            int pow=1;
            for(int i=1;i<=c;i++)
            pow= pow*n;
            arm = arm + (pow);
            num2 = num2 / 10;
        }

        // If number is equal to the arm
        // then it is a armstrong number
        if (arm == i) {
            printf("%d\n", i);
        }
        arm = 0;
        c = 0;
    }
    return 0;
}

Output

1
153
370
371
407

2. Using pow() function

C

// C program to demonstrate an
// armstrong number between the
// given intervals using pow()
#include <math.h>
#include <stdio.h>

// Driver code
int main()
{
    int s = 1, e = 500, num1, n,
          arm = 0, i, num2, c;

    // Iterating the for loop using the
    // given intervals
    for (i = s; i <= e; i++) {
        num1 = i;
        num2 = i;

        // Finding the number of digits
        while (num1 != 0) {
            num1 = num1 / 10;
            ++c;
        }

        // Finding the armstrong number
        while (num2 != 0) {
            n = num2 % 10;
            arm = arm + pow(n, c);
            num2 = num2 / 10;
        }

        // If number is equal to the arm then
        // it is a armstrong number
        if (arm == i) {
            printf("%d\n", i);
        }

        arm = 0;
        c = 0;
    }

    return 0;
}

Output

1
153
370
371
407