pset 1 credit 2022

Question

At the outset, I want to say that I tried to complete this task with arrays and don't know how to make it witout them.

Here is my code:

#include <cs50.h>
#include <stdio.h>

int main(void)
{

    long n = get_long("Card number: ");
    long nc = n; // to define how many digits in a card number; after first loop gets to 0
    int x = 0;
    int c = 0; // to store the answer after first step of  the algorithm. *MAIN NUMBER*
    int c1 = 0; // rest. *MAIN NUMBER*
    int a; // *MAIN NUMBER*

    //defines how many digits in a card number
    for (int y = 0; nc > 0; y++)
    {
        nc /= 10;
        x = y + 1;
    }
    // stores every number in the array
    int nm [x]; // main array
    int v = 0; // how many in each part
    int half1 [0]; // to be multiplied by 2
    int half2 [0]; // rest


    for (int z = 0; z < x; z++)
    {
        nm [z] =  n % 10;
        n /= 10;                  //   я не понимал, почему значение *n*  не обновляется. оказалось, что я просто забыл *пересобрать эту переменную*
    }

    if (x % 2 == 0)
    {
         half1 [x / 2];
         half2 [x / 2];
         half1 [0] = nm [0];
    }

    else
    {
          half1 [x / 2 + 1];
          half2 [x / 2];
          half1 [0] = nm [0];
    }



    for (int b = 2, g = 1; b < x; b += 2, g += 1, v++) // when solve this problem try to change variable name b to z
    {
         half1 [v] = nm [b];
         half2 [v] = nm [g];
    }

    for (int h = 0; h < v; h++)
    {
        half1 [h] *= 2;
        c1 += half2 [h];
    }

    for (int h = 0; h < v; h++)
    {
        if (half1 [h] >= 10)
        {
            half1 [v + 1] =  half1 [h] / 10;
            half1 [v + 2] =  half1 [h] % 10;
        }
        else
        {
            c += half1 [h];
        }
    }


    if(a % 10 != 0)
    {
        printf("INVALID CARD NUMBER\n");
    }
    else if ((nm [0] == 3 && nm [1] == 4) || (nm [0] == 3 && nm [1] == 7))
    {
        printf("AMEX\n");
    }
    else if ((nm [0] == 5 && nm [1] == 1) || (nm [0] == 5 && nm [1] == 2) || (nm [0] == 5 && nm [1] == 3) || (nm [0] == 5 && nm [1] == 4) || (nm [0] == 5 && nm [1] == 5))
    {
        printf("MASTERCARD");
    }
    else if (nm [0] == 4)
    {
        printf("VISA\n");
    }

}

Now I got problelms there, from line 33 to 45:

    if (x % 2 == 0)
    {
         half1 [x / 2];
         half2 [x / 2];
         half1 [0] = nm [0];
    }

    else
    {
          half1 [x / 2 + 1];
          half2 [x / 2];
          half1 [0] = nm [0];
    }

And have got such an output at the terminal. I thought this part of my code would define size of an array, because it isn't strict and depends on how many is a card number.