# Disarium Number

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```
Recommended Practice

The idea is to first 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 Disarium` `// or not` `#include` `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 disarium` `// 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"``);` `    ``}` `}`

## Python3

 `# Python program to check whether a number is Disarium` `# 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 Disarium 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`

## Javascript

 ``

Output

`Disarium Number`

Time complexity : O(logn)
Auxiliary Space : O(1)

If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

Feeling lost in the world of random DSA topics, wasting time without progress? It's time for a change! Join our DSA course, where we'll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 geeks!