CS50 (2018) | pset1 Credit | Solution by Normantas.

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

void outputCard(int sum, int digit[]);
int sumCal(int digit[], int howLong);
int devisionIntCal(long long credit);


int main() //Main Code;
{
    long long credit = get_long_long("Credit: "); // Test Code: 378282246310005 // Declares Needed/To-be-Used Data;
    int digit[16], sum = 0;

    int devisionInt = devisionIntCal(credit);
    for(int i = 0; i < devisionInt; i++) //splits ints
    {
        long long division = pow(10, devisionInt - i - 1);
        digit[i] = credit / division % 10;
    }
   sum = sumCal(digit, devisionInt );
   outputCard(sum, digit);

}

void outputCard(int sum, int digit[])
{
    if(sum % 10 == 0){ //Checks Card;
        if(digit[0] == 4){
            printf("Visa \n");
        }
        if(digit[0] * 10 + digit[1] == 37 || digit[0] * 10 + digit[1] == 34){
            printf("AMEX \n");
        }
        if(digit[0] * 10 + digit[1] == 51 || digit[0] * 10 + digit[1] == 52 || digit[0] * 10 + digit[1] == 53 || digit[0] * 10 + digit[1] == 54 || digit[0] * 10 + digit[1] == 55)
        {
            printf("MASTERCARD \n");
        }
        }
    else
    {
        printf("Invalid! \n");
    }
}

int sumCal(int digit[], int howLong)  //Calculates sum
{
    int sum;
       for(int i = 0; i < howLong + 1; i = i + 2)
    {
        sum = sum + digit[i];
    }
    for(int i = 1; i < howLong + 1; i = i + 2)
    {
        sum = sum + (digit[i] * 2) / 10 + (digit[i] * 2) % 10;
    }
    return sum;
}

int devisionIntCal(long long credit) //checks how long is the card, cause Amex can be 15digit, Visa can Be 13 or 15digits long, or Mastercard 16;
{
    int i = 16, checker;
    do
    {
        i--;
        long long dev = pow(10, i);
        checker = credit / dev % 10;
    }
    while(checker == 0);
    i++;
    return i;
}