How to convert the digits of a number into elements of array?

Question

For example in pset1 hacker edition program asks user to input the number of credit card. Imagine that a digits are: 1234567890 So can anyone tell me does it possible to convert this number into elements of array {1,2,3,4,5,6,7,8,9,0} and if it possible than how?

in my code it looks like this:

//askin the user to input number

long long cc_num; //here is another question do i need to declare cc_num equals zero?
do{
printf("number:);
cc_num = GetLongLong();
}
while (cc_num < 10 && cc_num > 10); //we assume that number always be 10 digits

//so here i want to declare an array using the digits from cc_num like elements of an array

     int array[11];
        for (int i = 0, n = strlen(cc_num); i < n; i++)
        {
            array[i] = ???

            for (i = 9; i > 0; i -= 2) //here i want to start counting from second-to-last element
            {
                //multiply each second element by 2 
                //sum multiplyed elements = a
            }
            for (i = 10; i >= 0; i -= 2) //here i want to start counting from last element to get result of sum non-multiplyed elements
            {
                //sum non-multiplyed elements = b
            }
            //sum multiplyed and non-multiplyed  (a + b) 

        }

i have already spent 2 days and 2 nights solving this problem...heeeelp me

Answer 1

If you want to store each digit of a number as an element of an array you can do the following.

long long number = 1234567890;
int digits[10];

for (int i = 0; i < 10; i++)
{
    digits[i] = number % 10;
    number = number / 10;
}
// At the end digits = {0, 9, 8, 7, 6, 5, 4, 3, 2, 1}

But if you want the elemnts to keep the same order than the original number you can do this instead:

for (int i = 9; i >= 0; i--)
{
    digits[i] = number % 10;
    number = number / 10;
}

// At the end digits = {1, 2, 3, 4, 5, 6, 7, 8, 9, 0}

Answer 2

Why not input the number as a string to begin with? (This would also make it easier to check the length is right - you seem to be comparing the size of the number, rather than its length, with 10). Then it is already an array.

Of course then you will store the ascii codes of the digits initially, rather than their values, but the conversion is easy.