C Program to check if the given number is a Disarium Number

Disarium Number

A number is said to be the Disarium number when the sum of its digit raised to the power of their respective positions becomes equal to the number itself.

For example, 175 is a Disarium number as follows:

1¹+ 7² + 5³ = 1+ 49 + 125 = 175

ALGORITHM:

main()

STEP 1: START
STEP 2: DEFINE n
STEP 3: SET num =175, sum =0, rem =0
STEP 4: len = calculateLength(num)
STEP 5: n =num
STEP 6: REPEAT STEP 7 to 10 UNTIL (num>0)
STEP 7: rem = num%10
STEP 8: sum =sum + (rem)^len
STEP 9: num = num/10
STEP 10: len--
STEP 11: if(sum==n) then PRINT "Yes" else PRINT "No"
STEP 12: RETURN 0
STEP 13: END

calculateLength(n)

STEP 1: START
STEP 2: SET length =0
STEP 3: REPEAT STEP 4 and STEP 5 UNTIL (n!=0)
STEP 4: length = length+1
STEP 5: n =n/10
STEP 6: RETURN length
STEP 7: END

PROGRAM:

#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;  
}  
   
int main()  
{  
    int num = 175, sum = 0, rem = 0, n;  
    int len = calculateLength(num);  
      
    //Makes a copy of the original number num  
    n = num;  
      
    //Calculates the sum of digits powered with their respective position  
    while(num > 0){  
        rem = num%10;  
        sum = sum + (int)pow(rem,len);  
        num = num/10;  
        len--;  
    }  
      
    //Checks whether the sum is equal to the number itself  
    if(sum == n)  
        printf("%d is a disarium number", n);  
    else  
        printf("%d is not a disarium number", n);  
   
    return 0;  
}  

Output: