Problem 5 solution - Naive algorithm (Improved over problem_5/sol1.c) More...

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

Include dependency graph for sol2.c:

Functions
static int	check_number (unsigned long long n)
	Checks if a given number is devisable by every number between 1 and 20.

int	main (void)
	Main function.

Variables
static unsigned int	divisors []
	Hack to store divisors between 1 & 20.

Detailed Description

Problem 5 solution - Naive algorithm (Improved over problem_5/sol1.c)

Little bit improved version of the naive problem_5/sol1.c. Since the number has to be divisable by 20, we can start at 20 and go in 20 steps. Also we don't have to check against any number, since most of them are implied by other divisions (i.e. if a number is divisable by 20, it's also divisable by 2, 5, and 10). This all gives a 97% perfomance increase on my machine (9.562 vs 0.257)

See also: Slower: problem_5/sol1.c; Faster: problem_5/sol3.c

Function Documentation

◆ check_number()

static int check_number ( unsigned long long n )

static

Checks if a given number is devisable by every number between 1 and 20.

Parameters

n	number to check

Returns: 0 if not divisible; 1 if divisible

{
    for (size_t i = 0; i < 7; ++i)
    {
        if (n % divisors[i] != 0)
        {
            return 0;
        }
    }
 
    return 1;
}

◆ main()

int main ( void )

Main function.

Returns: 0 on exit

{
    for (unsigned long long n = 20;; n += 20)
    {
        if (check_number(n))
        {
            printf("Result: %llu\n", n);
            break;
        }
    }
    return 0;
}

Here is the call graph for this function:

Variable Documentation

◆ divisors

unsigned int divisors[]

static

Initial value:

= {
    11, 13, 14, 16, 17, 18, 19, 20,
}

Hack to store divisors between 1 & 20.

                                 {
    11, 13, 14, 16, 17, 18, 19, 20,
};