USACO FEB07LEVEL1 Problem 'dsums' Analysis

by Rob Kolstad

The code of interest for this task is the code that figures out the 'digitsum' of an integer n. As with most digits-of-a-number problems, we use the 'mod' operator ('%' in C and so many languages) to find the last digit and then integer divide to remove the last digit from an integer. Some folks might consider changing the number to a string, but that is just so very slow when you have a lot of numbers to process!

Here's one way to code the digit sum routine:

dsum(n) {
    int sum;
    do {
        for (sum = 0 ; n; n /= 10)      /* peel one digit each iter */
            sum += n % 10;              /* bottom digit */
        n = sum;
    } while (sum > 9);
    return sum;
}

#include <stdio.h>
#include <stdlib.h>

dsum(n) {
    int sum;
    do {
        for (sum = 0 ; n; n /= 10)
            sum += n % 10;
        n = sum;
    } while (sum > 9);
    return sum;
}

int main() {
    FILE *fin = fopen ("dsums.in", "r");
    FILE *fout = fopen ("dsums.out", "w");
    int a, b, d, i, count;
    fscanf (fin, "%d %d %d", &a, &b, &d);
    count = 0;
    for (i = a; i <= b; i++)
        if (dsum (i) == d) count++;
    fprintf (fout, "%d\n", count);
    exit (0);
}