**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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