Restructuring

solutions/ProjectEuler/009/desc.yml

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+title: Find the only Pythagorean triplet, {a, b, c}, for which a + b + c = 1000.
+url: http://projecteuler.net/problem=9
+
+desc: |
+  A Pythagorean triplet is a set of three natural numbers, a  b  c, for which,
+  a^2 + b^2 = c^2
+  For example, 3^2 + 4^2 = 9 + 16 = 25 = 5^2.
+  There exists exactly one Pythagorean triplet for which a + b + c = 1000.
+  Find the product abc.
+  
+solution: |
+  Make a nested forloop for a and b in the range 1-1000 - Then c = 1000-a-b - Test if solution is valid.
+
+solutions:
+  solve.php:
+    desc: Basic solution
+    language: php
+  solve.rb:
+    desc: Basic solution
+    language: ruby
+  solve.c:
+    desc: ANSI C solution compiled with TCC
+    language: c

solutions/ProjectEuler/009/solve.c

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+#include "stdio.h"
+#include "math.h"
+
+int main( )
+{
+	int a;
+	int b;
+	int c;
+	int cmp;
+	for(a = 1; a < 1000; a++)
+	{
+		for(b = 1; b < 1000; b++)
+		{
+			//Calculate the only valid value for c
+			c = 1000 - a - b;
+			if( pow(c,2) == (pow(a, 2 ) + pow( b, 2 )))
+			{
+				int result = a * b * c;
+				printf("%i", result);
+				return 0;
+			}
+		}
+	}
+}

solutions/ProjectEuler/009/solve.php

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+<?php
+for ($a = 1; $a < 1000; $a++) {
+	for ($b = 1; $b < 1000; $b++) {
+		//Make it run reverse in order to find solution quickly
+		$c = 1000 - $a - $b;
+		if (pow($a, 2) + pow($b, 2) == pow($c, 2)) {
+			echo $a * $b * $c;
+			die;
+		}
+	}
+}

solutions/ProjectEuler/009/solve.rb

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+(1..1000).each do |a|
+	(1..1000).each do |b|
+		c = 1000 - a - b
+		if( a ** 2 + b ** 2 == c ** 2) 
+			puts a * b *c
+			exit
+		end
+	end
+end