mirror of
https://github.com/TheAlgorithms/C
synced 2024-11-29 08:43:09 +03:00
bb6c62aa62
* rename existing code as sol3
* Added naive implementation for Problem 5
* Added a solution for Euler Problem 5 with easy improvements
* rename new files
* code formatting
* update documentations
* fix docs
* updating DIRECTORY.md
Co-authored-by: buffet <niclas@countingsort.com>
Co-authored-by: github-actions <${GITHUB_ACTOR}@users.noreply.github.com>
60 lines
1.4 KiB
C
60 lines
1.4 KiB
C
/**
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* \file
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* \brief [Problem 5](https://projecteuler.net/problem=5) solution - Naive
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* algorithm (Improved over problem_5/sol1.c)
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* @details Little bit improved version of the naive `problem_5/sol1.c`. Since
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* the number has to be divisable by 20, we can start at 20 and go in 20 steps.
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* Also we don't have to check against any number, since most of them are
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* implied by other divisions (i.e. if a number is divisable by 20, it's also
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* divisable by 2, 5, and 10). This all gives a 97% perfomance increase on my
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* machine (9.562 vs 0.257)
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*
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* \see Slower: problem_5/sol1.c
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* \see Faster: problem_5/sol3.c
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*/
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#include <stdio.h>
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#include <stdlib.h>
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/**
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* @brief Hack to store divisors between 1 & 20
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*/
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static unsigned int divisors[] = {
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11, 13, 14, 16, 17, 18, 19, 20,
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};
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/** Checks if a given number is devisable by every number between 1 and 20
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* @param n number to check
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* @returns 0 if not divisible
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* @returns 1 if divisible
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*/
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static int check_number(unsigned long long n)
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{
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for (size_t i = 0; i < 7; ++i)
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{
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if (n % divisors[i] != 0)
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{
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return 0;
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}
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}
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return 1;
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}
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/**
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* @brief Main function
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*
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* @return 0 on exit
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*/
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int main(void)
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{
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for (unsigned long long n = 20;; n += 20)
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{
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if (check_number(n))
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{
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printf("Result: %llu\n", n);
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break;
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}
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}
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return 0;
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}
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