How to count the non-null digits of a number and the digits divisible by 3

Question

if (c % 3 == 0)
{
    d=d+1;
}
else
{
    cout << "The number has no digits divisible with 3" << endl;
}

But the thing is, c was used before in a while structure because I had to use it to calculate the sum of the number's digits and other things. If I try to write this outside of the while, I believe c will get the value only of the first digit of the number because of the loop. I tried giving the value o c to another variable but it's still the same.

Answer 1

Maybe you should try putting your logic in another while loop, similar to the one you used before.

You can also try calculating the number of digits divisible by 3 in the same loop that you calculate the sum in

while(...)
{
    ...
    sum = sum + c;
    if(c%3 == 0)
    {
        d = d+1;
    }
    ...
}