Find the sum of sums of digits of a given number, as per the method mentioned below?

Question

Alex has been asked by his teacher to do an assignment on sums of digits of a number. The assignment requires Alex to find the sum of sums of digits of a given number, as per the method mentioned below.
If the given number is 582109, the Sum of Sums of Digits will be calculated as =
= (5 + 8 + 2 + 1 + 0 + 9) + (8 + 2 + 1 + 0 + 9) + (2 + 1 + 0 + 9) + (1 + 0 + 9) + (0 + 9) + (9)
= 25 + 20 + 12 + 10 + 9 + 9 = 85

Alex contacts you to help him write a program for finding the Sum of Sums of Digits for any given number, using the above method.

Help Alex by completing the login in the given function sumOfSumsOfDigits which takes as input an integer input1 representing the given number.
The function is expected to return the "Sum of Sums of Digits" of input1.

Assumptions: For this assignment, let us assume that the given number will always contain more than 1 digit, i.e. the given number will always be >9.

Answer 1

include

include

int main()
{

int a = 582109;
int sum = 0;
int count = 0;
int master_sum = 0;
int res = 0;
while(a > 0)
{
res = a %10; //gives remainder, lets ay 1991 % 10 => 1
count++;
a = a / 10;//quotient
}

int flag = count;
while(count >0)
{
a = 582109;
while(a > 0 && flag>0)
{
res = a %10; //gives remainder, lets ay 1991 % 10 => 1
sum = sum+res;
a = a / 10;//quotient
flag = flag-1;
}
printf("Sum is %d\n",sum);
master_sum = master_sum + sum;
sum = 0;
count = count-1;
flag = count;
}
printf("master sum is %d\n",master_sum);
}

Comments

please update if this code can be optimized, feedback will be appreciated