Finish coin piles problem

coin-stack.cpp

@@ -0,0 +1,49 @@
+#include <iostream>
+#include <utility>
+
+int
+main(void)
+{
+	long t;
+	std::cin >> t;
+
+	long a, b;
+	for (long i = 0; i < t; i++)
+	{
+		std::cin >> a >> b;
+
+		// A pattern emerges where, if the sum of the coin stacks is
+		// even, there is no way to empty them following this scheme
+		// unless the sum is divisible by 3. Also, the size of
+		// one stack cannot be more than double the size of the other.
+
+		// First solution was failing on some numbers, but since it's
+		// too difficult to go through the hundred thousand outputs that
+		// were tested, let's just try doing this manually.
+		if (a < b)
+		{
+			std::swap(a, b);
+		}
+
+		// No way to empty stacks without having one with leftover coins
+		if (a > 2 * b)
+		{
+			std::cout << "NO" << std::endl;
+		}
+		else
+		{
+			// Stacks must be divisible by three to be valid, so
+			// let's reduce the values to the bare minimum.
+			a %= 3;
+			b %= 3;
+			if (a < b)
+				std::swap(a, b);
+			if ((a == 2 && b == 1) || (a == b && b == 0))
+				std::cout << "YES" << std::endl;
+			else
+				std::cout << "NO" << std::endl;
+
+		}
+	}
+	return 0;
+}