# 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: Please solve it on “PRACTICE” first, before moving on to the solution.

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 implementation of above idea.

## C++

 `// C++ program to check whether a number is Desoriam ` `// 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 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. `

Output:

```Disarium Number
```

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