Program to print all Disarium numbers between 1 and 100

Python

#calculateLength() will count the digits present in a number  
def calculateLength(n):  
    length = 0;  
    while(n != 0):  
        length = length + 1;  
        n = n//10;  
    return length;  
   
#sumOfDigits() will calculates the sum of digits powered with their respective position  
def sumOfDigits(num):  
    rem = sum = 0;  
    len = calculateLength(num);  
      
    while(num > 0):  
        rem = num%10;  
        sum = sum + (rem**len);  
        num = num//10;  
        len = len - 1;  
    return sum;  
    
result = 0;  
   
#Displays all disarium numbers between 1 and 100  
print("Disarium numbers between 1 and 100 are");  
for i in range(1, 101):  
    result = sumOfDigits(i);  
      
    if(result == i):  
        print(i),  

Output:

Disarium numbers between 1 and 100 are
1 2 3 4 5 6 7 8 9 89 

C

#include <stdio.h>  
   
//calculateLength() will count the digits present in a number  
int calculateLength(int n){  
    int length = 0;  
    while(n != 0){  
        length = length + 1;  
        n = n/10;  
    }  
    return length;  
}  
   
//sumOfDigits() will calculates the sum of digits powered with their respective position  
int sumOfDigits(int num){  
    int sum = 0, rem = 0;  
    int len = calculateLength(num);  
      
    while(num > 0){  
        rem = num%10;  
        sum = sum + pow(rem,len);  
        num = num/10;  
        len--;  
    }  
    return sum;  
}  
      
int main()  
{  
    int result = 0;  
      
    //Displays all disarium numbers between 1 and 100  
    printf("Disarium numbers between 1 and 100 are\n");  
    for(int i = 1; i <= 100; i++){  
        result = sumOfDigits(i);  
      
        if(result == i)  
            printf("%d ", i);  
    }  
   
    return 0;  
} 

Output:

Disarium numbers between 1 and 100 are
1 2 3 4 5 6 7 8 9 89

JAVA

public class DisariumNumbers  
{  
    //calculateLength() will count the digits present in a number  
    public static int calculateLength(int n){  
        int length = 0;  
        while(n != 0){  
            length = length + 1;  
            n = n/10;  
        }  
        return length;  
    }  
      
    //sumOfDigits() will calculates the sum of digits powered with their respective position  
    public static int sumOfDigits(int num){  
        int sum = 0, rem = 0;  
        int len = calculateLength(num);  
          
        while(num > 0){  
            rem = num%10;  
            sum = sum + (int)Math.pow(rem,len);  
            num = num/10;  
            len--;  
        }  
        return sum;  
    }  
      
    public static void main(String[] args) {  
        int result = 0;  
          
        //Displays all disarium numbers between 1 and 100  
        System.out.println("Disarium numbers between 1 and 100 are");  
        for(int i = 1; i <= 100; i++){  
            result = sumOfDigits(i);  
          
            if(result == i)  
                System.out.print(i + " ");  
        }  
    }  
}  

Output:

Disarium numbers between 1 and 100 are
1 2 3 4 5 6 7 8 9 89 

C#

using System;  
                      
public class DisariumNumbers  
{  
    //calculateLength() will count the digits present in a number  
    public static int calculateLength(int n){  
        int length = 0;  
        while(n != 0){  
            length = length + 1;  
            n = n/10;  
        }  
        return length;  
    }  
      
    //sumOfDigits() will calculates the sum of digits powered with their respective position  
    public static int sumOfDigits(int num){  
        int sum = 0, rem = 0;  
        int len = calculateLength(num);  
          
        while(num > 0){  
            rem = num%10;  
            sum = sum + (int)Math.Pow(rem,len);  
            num = num/10;  
            len--;  
        }  
        return sum;  
    }  
      
    public static void Main()  
    {  
        int result = 0;  
          
        //Displays all disarium numbers between 1 and 100  
        Console.WriteLine("Disarium numbers between 1 and 100 are");  
        for(int i = 1; i <= 100; i++){  
            result = sumOfDigits(i);  
          
            if(result == i)  
                Console.Write(i + " ");  
        }  
    }  
}  

Output:

Disarium numbers between 1 and 100 are
1 2 3 4 5 6 7 8 9 89 

PHP

<!DOCTYPE html>  
<html>  
<body>  
<?php  
//calculateLength() will count the digits present in a number  
function calculateLength($n){  
    $length = 0;  
    while($n != 0){  
        $length = $length + 1;  
        $n = intval($n/10);  
    }  
    return $length;  
}  
   
//sumOfDigits() will calculates the sum of digits powered with their respective position  
function sumOfDigits($num){  
    $rem = $sum = 0;  
    $len = calculateLength($num);  
   
    while($num > 0){  
        $rem = $num%10;  
        $sum = $sum + pow($rem,$len);  
        $num = intval($num/10);  
        $len--;  
    }  
    return $sum;  
}  
      
$result = 0;  
   
//Displays all disarium numbers between 1 and 100  
print("Disarium numbers between 1 and 100 are <br>");  
for($i = 1; $i <= 100; $i++){  
    $result = sumOfDigits($i);  
   
    if($result == $i)  
        print($i . " ");  
}  
?>  
</body>  
</html>  

Output:

Disarium numbers between 1 and 100 are
1 2 3 4 5 6 7 8 9 89