Initial import

ProjectEuler/001/desc.yml

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+title: Add all the natural numbers below one thousand that are multiples of 3 or 5.
+url: http://projecteuler.net/problem=1
+
+desc: |
+  If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23.
+  Find the sum of all the multiples of 3 or 5 below 1000.
+  
+solution: |
+  Loop numbers from 1-1000 - Use mod to find the multiples
+
+solutions:
+  solve.php:
+    desc: Basic solution
+    language: php
+  solve.rb:
+    desc: Basic solution in Ruby
+    language: ruby
+  solve.c:
+    desc: ANSI C solution (Tested with TCC)
+    language: c

ProjectEuler/001/solve.c

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+#include <stdio.h>
+
+int main( )
+{
+	int sum = 0;
+	int i = 0;
+	for(i = 1; i < 1000; i++)
+	{
+		if(i % 3 == 0 || i % 5 == 0)
+		{
+			sum += i;
+		}
+	}
+	printf( "%i", sum );
+}

ProjectEuler/001/solve.php

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+<?php
+$sum = 0;
+$i = 0;
+for($i=1; $i<1000;$i++) {
+	if($i % 3 == 0 OR $i % 5 == 0) {
+		$sum += $i;
+	}
+}
+
+echo $sum;

ProjectEuler/001/solve.rb

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+sum = 0
+(1..1000).each do |i| 
+	if(i % 3 == 0 or i % 5 == 0) 
+		sum += i
+	end
+end
+puts sum

ProjectEuler/002/desc.yml

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+title: By considering the terms in the Fibonacci sequence whose values do not exceed four million, find the sum of the even-valued terms.
+url: http://projecteuler.net/problem=2
+
+desc: |
+  Each new term in the Fibonacci sequence is generated by adding the previous two terms. By starting with 1 and 2, the first 10 terms will be:
+  1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ...
+  By considering the terms in the Fibonacci sequence whose values do not exceed four million, find the sum of the even-valued terms.
+  
+solution: |
+  Step through the fibonacci sequence while you sum each even entry. Quit as you reach 4million
+
+solutions:
+  solve.php:
+    desc: Basic solution
+    language: php
+  solve.rb:
+    desc: Basic solution in Ruby
+    language: ruby
+  solve.c:
+    desc: ANSI C solution (Tested with TCC)
+    language: c

ProjectEuler/002/solve.c

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+#include <stdio.h>
+
+int main( )
+{
+	int sum = 2;
+	int fib[3] = { 1, 2, 3 };
+
+	while(fib[2] < 4000000)
+	{
+		fib[2] = fib[0] + fib[1];
+		if(fib[2] % 2 == 0)
+			sum += fib[2];
+
+		fib[0] = fib[1];
+		fib[1] = fib[2];
+	}
+	printf( "%i", sum );
+}

ProjectEuler/002/solve.php

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+<?php
+$fib = array(1,2,3);
+$sum = 2;
+while($fib[2] < 4000000) {
+	$fib[2] = $fib[0] + $fib[1];
+	if($fib[2] % 2 == 0) {
+		$sum += $fib[2];
+	} 
+	$fib[0] = $fib[1];
+	$fib[1] = $fib[2];
+}
+echo $sum;

ProjectEuler/002/solve.rb

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+fib = [1, 2, 3];
+sum = 2;
+while(fib[2] < 4000000) 
+	fib[2] = fib[0] + fib[1];
+	if(fib[2] % 2 == 0) 
+		sum += fib[2];
+	end
+	fib[0] = fib[1];
+	fib[1] = fib[2];
+end
+puts sum;