sum_abundant_nums.c

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

#define MAX 28123

int get_divisors(int, int **);

int main(void) {
	int *divisors;
	
    int n, i;
    
    int abundant_nums[MAX] = {0};
    int abundant_nums_count = 0;
    
    // Get abundant numbers
    for (n = 1; n < MAX; n++) {
        int sum = 0;
        
        // Get sum of divisors
        int num_divisors = get_divisors(n, &divisors);
        for (i = 0; i < num_divisors; i++) {
            sum += divisors[i];
        }

        free(divisors);
        
        if (sum > n) {
            abundant_nums[abundant_nums_count++] = n;
        }
    }
    
    // Make an array for the possible values
    int all_sums[MAX + 1];
    for (i = 0; i < MAX + 1; i++)
        all_sums[i] = 1;
    
    // Remove all the bad values
    int add_a;
    int add_b;
    for (add_a = 0; add_a < abundant_nums_count; add_a++) {
        for (add_b = add_a; add_b < abundant_nums_count; add_b++) {
            int sum = abundant_nums[add_a] + abundant_nums[add_b];
            if (sum < MAX + 1)
                all_sums[sum] = 0;
        }
    }
    
    // Now find the total
    long int solution  = 0;
    for (i = 0; i < MAX + 1; i++) {
        if (all_sums[i])
            solution += i;
    }
    
    printf("Solution: %ld\n", solution);

	return 0;
}

/* get_divisors: Need to pass in a pointer to an array of integers */
int get_divisors(int n, int **result_ptr) {

	int num_divisors = 0;
	
	// Just picking a large number
	int *result = malloc(sizeof(int) * 1000);
	
	// 1 is always in the result
	result[num_divisors++] = 1;
	
	// Proper divisors only
	/*
	if (n != 1) {
	    result[num_divisors++] = n;
	}
	//*/
	
	int max = floor(sqrt(n));
	int divisor = 2;
	while (divisor <= max) {
		if (n % divisor == 0) {
			result[num_divisors++] = divisor;
			
			int other_divisor = n / divisor;
			if (other_divisor > divisor)
				result[num_divisors++] = other_divisor;
		}
		
		divisor++;
	}
	
	*result_ptr = result;

	// Return number of divisors
	return num_divisors;
}