sol1.c File Reference

Problem 23 solution More...

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

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Functions
char	get_perfect_number (unsigned long N)
	Returns: -1 if N is deficient 1 if N is abundant 0 if N is perfect.
unsigned long	is_abundant (unsigned long N)
	Is the given number an abundant number (1) or not (0)
unsigned long	get_next_abundant (unsigned long N)
	Find the next abundant number after N and not including N.
char	is_sum_of_abundant (unsigned long N)
	check if a given number can be represented as a sum of two abundant numbers. More...
int	main (int argc, char **argv)
	Main function.

Variables
unsigned long	MAX_N = 28123
	upper limit of numbers to check

Detailed Description

Problem 23 solution

Author

Krishna Vedala

Function Documentation

is_sum_of_abundant()

char is_sum_of_abundant

(

unsigned long

N

)

check if a given number can be represented as a sum of two abundant numbers.

Returns

1 - if yes

0 - if not

{
    /* optimized logic:
     * i + j = N   where both i and j should be abundant
     * hence we can simply check for j = N - i as we loop through i
     */
    for (unsigned long i = get_next_abundant(1); i <= (N >> 1);
         i = get_next_abundant(i))
        if (is_abundant(N - i))
        {
#ifdef DEBUG
            printf("\t%4lu + %4lu = %4lu\n", i, N - i, N);
#endif
            return 1;
        }
    return 0;
}

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