feat: Project Euler Problem 5 - #162 (#599)

* 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 <[email protected]>
Co-authored-by: github-actions <${GITHUB_ACTOR}@users.noreply.github.com>

DIRECTORY.md

@@ -319,7 +319,9 @@
   * Problem 401
     * [Sol1](https://github.com/TheAlgorithms/C/blob/master/project_euler/problem_401/sol1.c)
   * Problem 5
-    * [Sol](https://github.com/TheAlgorithms/C/blob/master/project_euler/problem_5/sol.c)
+    * [Sol1](https://github.com/TheAlgorithms/C/blob/master/project_euler/problem_5/sol1.c)
+    * [Sol2](https://github.com/TheAlgorithms/C/blob/master/project_euler/problem_5/sol2.c)
+    * [Sol3](https://github.com/TheAlgorithms/C/blob/master/project_euler/problem_5/sol3.c)
   * Problem 6
     * [Sol](https://github.com/TheAlgorithms/C/blob/master/project_euler/problem_6/sol.c)
   * Problem 7

project_euler/problem_5/sol1.c

@@ -0,0 +1,48 @@
+/**
+ * \file
+ * \brief [Problem 5](https://projecteuler.net/problem=5) solution - Naive
+ * algorithm (slowest)
+ *
+ * \see Faster: problem_5/sol2.c
+ * \see Fastest: problem_5/sol3.c
+ */
+#include <stdio.h>
+#include <stdlib.h>
+
+/** Pretty naive implementation. Just checks every number if it's devisable by 1
+ * through 20
+ * @param n number to check
+ * @returns 0 if not divisible
+ * @returns 1 if divisible
+ */
+static char check_number(unsigned long long n)
+{
+    for (unsigned long long i = 1; i <= 20; ++i)
+    {
+        if (n % i != 0)
+        {
+            return 0;
+        }
+    }
+
+    return 1;
+}
+
+/**
+ * @brief Main function
+ *
+ * @return 0 on exit
+ */
+int main(void)
+{
+    for (unsigned long long n = 1;; ++n)
+    {
+        if (check_number(n))
+        {
+            printf("Result: %llu\n", n);
+            break;
+        }
+    }
+
+    return 0;
+}

project_euler/problem_5/sol2.c

@@ -0,0 +1,59 @@
+/**
+ * \file
+ * \brief [Problem 5](https://projecteuler.net/problem=5) solution - Naive
+ * algorithm (Improved over problem_5/sol1.c)
+ * @details 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 Slower: problem_5/sol1.c
+ * \see Faster: problem_5/sol3.c
+ */
+#include <stdio.h>
+#include <stdlib.h>
+
+/**
+ * @brief Hack to store divisors between 1 & 20
+ */
+static unsigned int divisors[] = {
+    11, 13, 14, 16, 17, 18, 19, 20,
+};
+
+/** Checks if a given number is devisable by every number between 1 and 20
+ * @param n number to check
+ * @returns 0 if not divisible
+ * @returns 1 if divisible
+ */
+static int check_number(unsigned long long n)
+{
+    for (size_t i = 0; i < 7; ++i)
+    {
+        if (n % divisors[i] != 0)
+        {
+            return 0;
+        }
+    }
+
+    return 1;
+}
+
+/**
+ * @brief Main function
+ *
+ * @return 0 on exit
+ */
+int main(void)
+{
+    for (unsigned long long n = 20;; n += 20)
+    {
+        if (check_number(n))
+        {
+            printf("Result: %llu\n", n);
+            break;
+        }
+    }
+    return 0;
+}

project_euler/problem_5/sol.c → project_euler/problem_5/sol3.c

@@ -1,12 +1,19 @@
 /**
  * \file
- * \brief [Problem 5](https://projecteuler.net/problem=5) solution
+ * \brief [Problem 5](https://projecteuler.net/problem=5) solution (Fastest).
+ * @details Solution is the LCM of all numbers between 1 and 20.
+ *
+ * \see Slowest: problem_5/sol1.c
+ * \see Slower: problem_5/sol2.c
  */
 #include <stdio.h>
 
 /** Compute [Greatest Common Divisor
  * (GCD)](https://en.wikipedia.org/wiki/Greatest_common_divisor) of two numbers
  * using Euclids algorithm
+ * @param a first number
+ * @param b second number
+ * @return GCD of `a` and `b`
  */
 unsigned long gcd(unsigned long a, unsigned long b)
 {
@@ -27,14 +34,19 @@ unsigned long gcd(unsigned long a, unsigned long b)
 
 /** Compute [Least Common Multiple
  * (LCM)](https://en.wikipedia.org/wiki/Least_common_multiple) of two numbers
+ * @param a first number
+ * @param b second number
+ * @return LCM of `a` and `b`
  */
 unsigned long lcm(unsigned long a, unsigned long b)
 {
     unsigned long long p = (unsigned long long)a * b;
     return p / gcd(a, b);
 }
 
-/** Main function */
+/** Main function
+ * @returns 0 on exit
+ */
 int main(void)
 {
     unsigned long ans = 1;