C++ Program to Display Armstrong Number Between Two Intervals

An armstrong number is a positive integer that is equal to the sum of its digits, each raised to the power of the total number of digits in the number. Our goal here is to display the armstrong numbers between two intervals in C++. In simple terms, for a number with n digits:

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

If this condition is satisfied, we can say the number is an armstrong number.

Let's understand with examples:

153 is an Armstrong number because:
1^3 + 5^3 + 3^3 = 1 + 125 + 27 = 153

1634 is also an Armstrong number because:
1^4 + 6^4 + 3^4 + 4^4 = 1 + 1296 + 81 + 256 = 1634

But 123 is not an Armstrong number because:
1^3 + 2^3 + 3^3 = 1 + 8 + 27 = 36 ? 123

Display Armstrong Numbers Between Intervals

To display armstrong numbers between two intervals in C++, we need to check each number in the given range to see if it meets the armstrong condition.

Here are the steps we follow:

• We define a function that checks whether a number is an Armstrong number.
• In the main function, we loop through the numbers from the start to the end of the interval, calling this function for each number.
• Then, for each number, we count its digits and calculate the sum of each digit raised to the power of the total number of digits.
• If this sum equals the original number, it is an armstrong number.
• Finally, we display all the armstrong numbers found within the given interval.

C++ Program to Display Armstrong Numbers Between Intervals

Here's a complete C++ program where we display Armstrong numbers between two intervals:

#include <iostream>
#include <cmath>
using namespace std;

// Function to check if a number is Armstrong
bool isArmstrong(int num) {
    int sum = 0, temp, remainder, n = 0;

    // Store the number in a temporary variable
    temp = num;
    // Find the number of digits
    while (temp != 0) {
        temp /= 10;
        ++n;
    }
    temp = num;
    // Calculate the sum of the powers of digits
    while (temp != 0) {
        remainder = temp % 10;
        sum += pow(remainder, n);
        temp /= 10;
    }
    // If sum equals the original number, it's an Armstrong number
    return (sum == num);
}

int main() {
    int start = 100, end = 999;

    // Loop through the range and check for Armstrong numbers
    cout << "Armstrong numbers between " << start << " and " << end << " are: " << endl;
    for (int i = start; i <= end; ++i) {
        if (isArmstrong(i)) {
            cout << i << " ";
        }
    }
    return 0;
}

The output below displays the armstrong numbers between the two intervals:

Armstrong numbers between 100 and 999 are: 
153 370 371 407 

Time Complexity: O(n * d * log d) because we check each number with d digits in the range n.

Space Complexity: O(1) because we only use a constant amount of extra space.