How can i print the count number in first?

Harry Potter is worried about the recent outbreak of Dementor from the Prison of Azkaban. Every Dementor is marked with a specific number. Write a Program to Assist Harry identify all of the Dementor Numbers between two ranges. Dementor numbers are those numbers which have following

CONDITIONS:
a. Number of Digits is greater than 2 and less than 6
b. Sum of the digits in EVEN Positions is greater than 8 and divided by 4

For Example: 1319 is a Dementor Number since
· Number of Digits 4 is greater than 2 and less than 6
· Sum of the digits in EVEN Positions: 3 + 9 = 12 is greater than 8 and divided by 4

Sample Input
1300 1350

Sample Output
5: 1309, 1319, 1329, 1339, 1349

Sample Input
15450 15475

Sample Output
6: 15470, 15471, 15472, 15473, 15474, 15475

#include <stdio.h>
int main()
{
    int n1, n2;
    scanf("%d %d", &n1, &n2);

    int count = 0;
    for (int i = n1; i <= n2; i++)
    {
        int digit = 0;
        int temp = i;
        while (temp != 0)
        {
            temp = temp / 10;
            digit++;
        }
        if (digit > 2 && digit < 6)
        {
            int a[digit];
            int num = i;
            int index = digit - 1;
            while (num != 0)
            {
                a[index] = num % 10;
                num = num / 10;
                index--;
            }
            int sum = 0;
            for (int j = 0; j < digit; j++)
            {
                if (j % 2 == 1)
                {
                    sum += a[j];
                }
            }

            if (sum > 8 && (sum % 4 == 0))
            {
                for (int j = 0; j < digit; j++)
                {
                    printf("%d", a[j]);
                }
                printf(" ");
                count++;
            }
        }
    }
    printf("%d ", count);

    return 0;
}

I have solved the problem.
But the output is printing this way:
15470 15471 15472 15473 15474 15475 6

I can't figure out how to print this way:
6 15470, 15471, 15472, 15473, 15474, 15475
I tried to take another array of count size to keep the output value but didn't work

how to print count number in first?

Note: using only if-else,loop,array
and it is not programming contest problem.

The way you execute your algorithm imposes you the order of printing. There's a reason for that. You don't know a priori the number of integers you will find in the range, so you cannot print it before you have got all the numbers.

To cope with this, you need to store (and never to print) the numbers (e.g. in an array, like list is suggested below) while they are all being computed, until you end the loop, so you can then print the count, and make a second loop to navigate the array list, and print all the numbers then.

Printers have the infamous custom of printing top to bottom, and left to right for each line, so the printer imposes you the order you have to use to print the numbers. If you print first the list of numbers (in the loop), you will print the count at the end (where the printer head is)

Let me apologize for not showing complete code to do this, but it seems to be an academic homework example, so you have to work it out, but basically the idea is to have one more variable:

• int list[A_BIG_NUMBER_HERE]; to store the individual list elements. The A_BIG_NUMBER_HERE constant should be large enough to allow you to manage a big list of numbers (there is a possibility of making the list dimension variable so you are not limited by the size of the array, but escapes to the subject, so I'll not use that approach here). Although, if you wan to be conservative, as the problem states that candidates will never be of equal or more than 6 digits, you can set an upper bound of 1,000,000 elements in the array, so you can

#define A_BIG_NUMBER_HERE 1000000

and that will make never truncate a list.

Then change the printf() calls you do in the first loop to print the list elements into something like:

    if (list_cnt < A_BIG_NUMBER_HERE) /* if there's still room in the array */
        list[count++] = copy_of_num;  /* you should have made a copy of num as you destroyed it on the computation of the digits. */

Then, after you print the count, just use another loop to print, one by one, each number in list array. Remember that you need to save the num value at the beginning of the loop, because you destroy it (after the first loop to get all the digits, you get num == 0) so, a declaration like:

    const int copy_of_num = num;

just after getting the candidate number stored in num, would be interesting to make it work correctly.

One more thing, if you pass a range that makes you find more numbers than can be stored in list it should be nice to print a warning message indicating that the list will be truncated, as no more than A_BIG_NUMBER_HERE elements are supported by your program.

After analizing your code a bit, I've seen sevaral places wher you can optimize your computation, by reusing the data you have acquired about the considered number.

...
        int digit = 0;
        int temp = i;
        while (temp != 0)
        {
            temp = temp / 10;
            digit++;
        }

this step can be optimized by considering that a number that is 2 digit or less is a number that is less than 100, and a number that is 6 digits or more can be tested to be larger or equal to 1000000. So a test like:

        if (i < 100 || i >= 1000000) {
            continue; /* continue to the next number */
        }

        if (digit > 2 && digit < 6)
        {
            int a[digit];

try not to use a variable length array. This requires the compiler to do far more calculations to accomodate local variables (if any of the local variables has to be adjusted in length, tracking all the other variables that afe located after it, becomes a mesh, as their positions have to be calculated at runtime, making your code slower) and an integer will never require more than 9 digits (your problem doesn't allow you to process numbers bigger than 6 digits) a good alternative is to give a a fixed size of, let's say, 10 digits:

            int a[10]; /* far better */

next:

            int num = i;
            int index = digit - 1;
            while (num != 0)
            {
                a[index] = num % 10;
                num = num / 10;
                index--;
            }

your solution here was clever, I was trying to see how did you solve the requirement of getting the digits from right to left, while having to process them left ro right (odds and evens change positions depending on the total number of digits, when considered left to right). Another alternative that doesn't require to save the digits in an array is to carry the odds and evens sums in parallel, and exchange them at the end, which makes the algorithm independent of the actual number of digits you consider.

            int sum = 0;
            for (int j = 0; j < digit; j++)
            {
                if (j % 2 == 1)
                {
                    sum += a[j];
                }
            }

correct! good job.

and here

        if (sum > 8 && (sum % 4 == 0))
        {
            for (int j = 0; j < digit; j++)
            {
                printf("%d", a[j]);
            }
            printf(" ");
            count++;
        }
    }
}

you need to change the computation you made by appending the number
to `list` above, instead of printing it.

and this piece of code:

    printf("%d ", count);

will need to be changed into:

    printf("%d:", count);
    for (int i = 0; i < count; i++) {
        printf(" %d", list[i]);
    }
    printf("\n");

Other considerations:

Your code looks good, but I'd not implement the modification you ask for, but only if it has been included as an output requirement. It requires extra processing, considering storing temporarily a list of the computed numbers and the code is a bit more complex than, e.g. you just print the list in one line, and the number of found elements in the next line (or not to print the number of found elements at all if it is not required) As an academic exercise the problem is good, but it is also a good things to show compact and readable code, than a complex thing that requires to do a lot of coding just to write the output in a format that is neither required nor needed.