Disarium Number

Difficulty Level : Easy
Last Updated : 09 Nov, 2020

Given a number “n”, find if it is Disarium or not. A number is called Disarium if sum of its digits powered with their respective positions is equal to the number itself.

Examples:

Input   : n = 135
Output  : Yes 
1^1 + 3^2 + 5^3 = 135
Therefore, 135 is a Disarium number

Input   : n = 89
Output  : Yes 
8^1+9^2 = 89
Therefore, 89 is a Disarium number

Input   : n = 80
Output  : No
8^1 + 0^2 = 8

The idea is to fist count digits in given numbers. Once we have count, we traverse all digits from right most (using % operator), raise its power to digit count and decrement the digit count.

Below is the implementation of above idea.

C++

// C++ program to check whether a number is Desoriam
// or not
#include<bits/stdc++.h>
using namespace std;
  
// Finds count of digits in n
int countDigits(int n)
{
    int count_digits = 0;
  
    // Count number of digits in n
    int x = n;
    while (x)
    {
        x = x/10;
  
        // Count the no. of digits
        count_digits++;
    }
    return count_digits;
}
  
// Function to check whether a number is disarium or not
bool check(int n)
{
    // Count digits in n.
    int count_digits = countDigits(n);
  
    // Compute sum of terms like digit multiplied by
    // power of position
    int sum = 0; // Initialize sum of terms
    int x = n;
    while (x)
    {
        // Get the rightmost digit
        int r = x%10;
  
        // Sum the digits by powering according to
        // the positions
        sum = sum + pow(r, count_digits--);
        x = x/10;
    }
  
    // If sum is same as number, then number is
    return (sum == n);
}
  
//Driver code to check if number is disarium or not
int main()
{
    int n = 135;
    if( check(n))
        cout << "Disarium Number";
    else
        cout << "Not a Disarium Number";
    return 0;
}

Java

// Java program to check whether a number is Desoriam
// or not
  
class Test
{
    // Method to check whether a number is disarium or not
    static boolean check(int n)
    {
        // Count digits in n.
        int count_digits = Integer.toString(n).length();
       
        // Compute sum of terms like digit multiplied by
        // power of position
        int sum = 0; // Initialize sum of terms
        int x = n;
        while (x!=0)
        {
            // Get the rightmost digit
            int r = x%10;
       
            // Sum the digits by powering according to
            // the positions
            sum = (int) (sum + Math.pow(r, count_digits--));
            x = x/10;
        }
       
        // If sum is same as number, then number is
        return (sum == n);
    }
      
    // Driver method
    public static void main(String[] args) 
    {
        int n = 135;
          
        System.out.println(check(n) ? "Disarium Number" : "Not a Disarium Number");
    }
}

Python

# Python program to check whether a number is Desarium
# or not
import math 
  
# Method to check whether a number is disarium or not
def check(n) :
  
    # Count digits in n.
    count_digits = len(str(n))
       
    # Compute sum of terms like digit multiplied by
    # power of position
    sum = 0  # Initialize sum of terms
    x = n
    while (x!=0) :
  
        # Get the rightmost digit
        r = x % 10
           
        # Sum the digits by powering according to
        # the positions
        sum = (int) (sum + math.pow(r, count_digits))
        count_digits = count_digits - 1
        x = x/10
         
    # If sum is same as number, then number is
    if sum == n :
        return 1
    else :
        return 0
        
# Driver method
n = 135
if (check(n) == 1) :
    print "Disarium Number"
else :
    print "Not a Disarium Number"
   
# This code is contributed by Nikita Tiwari.

C#

// C# program to check whether a number
// is Desoriam or not
using System;
  
class GFG{
      
// Method to check whether a number
// is disarium or not
static bool check(int n)
{
      
    // Count digits in n.
    int count_digits = n.ToString().Length;
   
    // Compute sum of terms like digit 
    // multiplied by power of position
    // Initialize sum of terms
    int sum = 0; 
    int x = n;
      
    while (x != 0)
    {
          
        // Get the rightmost digit
        int r = x % 10;
   
        // Sum the digits by powering according
        // to the positions
        sum = (int)(sum + Math.Pow(
              r, count_digits--));
        x = x / 10;
    }
   
    // If sum is same as number, 
    // then number is
    return (sum == n);
}
  
// Driver code
public static void Main(string[] args) 
{
    int n = 135;
      
    Console.Write(check(n) ? "Disarium Number" :
                       "Not a Disarium Number");
}
}
  
// This code is contributed by rutvik_56

Output:

Disarium Number

This article is contributed by Sahil Chhabra(KILLER). If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
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