sum_of_digits_from_one_to_one_million.c

/*

    sum_of_digits_from_one_to_one_million.c

    Calculate total sum of digits of the numbers from 1 to 1000000.
    Number 1000000 is not included.

    Task from
    https://web.facebook.com/photo/?fbid=292086525221428&set=gm.3162235480538814
    https://web.facebook.com/anusha.sritharan.54

    https://www.geeksforgeeks.org/count-sum-of-digits-in-numbers-from-1-to-n/

    Mathematic solutions:
    https://www.xarg.org/puzzle/misc/sum-of-digit-sums-between-one-and-a-million/
    https://www.glassdoor.com/Interview/What-is-the-sum-of-the-digits-of-all-the-numbers-from-1-to-1000000-This-is-different-from-the-sum-of-the-numbers-For-inst-QTN_262688.htm

    Further elaboration of the task:
    How many positive integers less than 1,000,000 have the sum of their digits equal to 19?
    https://www.slader.com/discussion/question/how-many-positive-integers-less-than-1000000-have-the-sum-of-their-digits-equal-to-19/


    You can find all my C programs at Dragan Milicev's pastebin:

    https://pastebin.com/u/dmilicev

*/

#include <stdio.h>

// adds the digits of one number
int sum_of_the_digits_of_the_number( long number )
{
    int digit, sum_of_digits=0; // the sum of the digits of one number
    long mem=number;

    while( number != 0 )
    {
        digit = number % 10;
        sum_of_digits += digit;
        number = number / 10;
    }

    //printf("\n number = %6ld , sum_of_digits = %2d , ", mem, sum_of_digits );

    return( sum_of_digits );

} // sum_of_the_digits_of_the_number()


int main(void)
{
    long number;            // our number
    long max_number=1000000;// we add the digits of the numbers from 1 to max_number
    long total_sum=0;       // the total sum of the digits of all numbers

    for( number=1; number<max_number; number++ )
    {
        total_sum += sum_of_the_digits_of_the_number(number);
        //printf(" total_sum = %6ld \n", total_sum );
    }

    printf("\n Total sum of digits of the numbers from 1 to %6ld is %ld \n", max_number, total_sum );


    return 0;

} // main()