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. More...
unsigned long	is_abundant (unsigned long N)
	Is the given number an abundant number (1) or not (0) More...
unsigned long	get_next_abundant (unsigned long N)
	Find the next abundant number after N and not including N. More...
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. More...

Detailed Description

Problem 23 solution

Author

Krishna Vedala

Function Documentation

get_next_abundant()

unsigned long get_next_abundant

(

unsigned long

N

)

Find the next abundant number after N and not including N.

{
    unsigned long i;
    for (i = N + 1; !is_abundant(i); i++)
    {
        ;
    }
    return i;
}

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get_perfect_number()

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 sum = 1;
    char ret = 0;
 
    for (unsigned long i = 2; i * i <= N; i++)
    {
        if (N % i == 0)
        {
            sum += i;
            unsigned long tmp = N / i;
            if (tmp != i)
            {
                sum += tmp;
            }
        }
    }
 
    ret = sum == N ? 0 : (sum > N ? 1 : -1);
    // #ifdef DEBUG
    //     printf("%5lu: %5lu : %d\n", N, sum, ret);
    // #endif
    return ret;
}

is_abundant()

unsigned long is_abundant

(

unsigned long

N

)

Is the given number an abundant number (1) or not (0)

{
    return get_perfect_number(N) == 1 ? 1 : 0;
}

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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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main()

int main	(	int	argc,
		char **	argv
	)

Main function.

{
    unsigned long MAX_N = 28123; /* upper limit of numbers to check */
 
    unsigned long sum = 0;
    if (argc == 2)
    {
        MAX_N = strtoul(argv[1], NULL, 10);
    }
 
#ifdef _OPENMP
    printf("Using OpenMP parallleization with %d threads\n",
           omp_get_max_threads());
#else
    printf("Not using parallleization!\n");
#endif
 
    double total_duration = 0.f;
    long i;
#ifdef _OPENMP
#pragma omp parallel for reduction(+ : sum) schedule(runtime)
#endif
    for (i = 1; i <= MAX_N; i++)
    {
        clock_t start_time = clock();
        if (!is_sum_of_abundant(i))
        {
            sum += i;
        }
        clock_t end_time = clock();
        total_duration += (double)(end_time - start_time) / CLOCKS_PER_SEC;
 
        printf("... %5lu: %8lu\r", i, sum);
        if (i % 100 == 0)
        {
            fflush(stdout);
        }
    }
 
    printf("Time taken: %.4g s\n", total_duration);
    printf(
        "Sum of numbers that cannot be represented as sum of two abundant "
        "numbers : %lu\n",
        sum);
 
    return 0;
}

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