Basic Programming C++ Examples

C++ Program to Check If a number is Disarium or not

The Program in C++ Program to Check If a number is Disarium or not is given below:

#include <iostream>
#include <cmath>

using namespace std;

int numberOfDigits(int n) {
    int count = 0;
    while (n > 0) {
        count++;
        n /= 10;
    }
    return count;
}

bool isDisariumNumber(int n) {
    int digits = numberOfDigits(n);
    int sum = 0, originalNumber = n;
    while (n > 0) {
        int digit = n % 10;
        sum += pow(digit, digits);
        n /= 10;
        digits--;
    }
    return sum == originalNumber;
}

int main() {
    int n;
    cout << "Hello Codeauri Family,enter a number here to check whether it is disarium number or not!:\n ";
    cin >> n;

    if (isDisariumNumber(n)) {
        cout << n << " is a disarium number." << endl;
    } else {
        cout << n << " is not a disarium number." << endl;
    }
    return 0;
}

Output:

Hello Codeauri Family,enter a number here to check whether it is disarium number or not!:
89
89 is a disarium number.

Pro-Tips💡

This program implements a function numberOfDigits to find the number of digits in a number. It also implements a function isDisariumNumber to check if a number is a disarium number.

The isDisariumNumber function finds the sum of the digits raised to the power of their position (starting from 1 for the rightmost digit) in the number.

If the sum is equal to the original number, the number is a disarium number, otherwise it’s not.

The main function takes a number from the user and calls the isDisariumNumber function to check if it’s a disarium number.

If it is, the program prints that the number is a disarium number, otherwise it prints that it is not.

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