--ProjectEuler/lua/Problem23.lua
--Matthew Ellison
-- Created: 03-22-19
--Modified: 06-19-20
--Find the sum of all the positive integers which cannot be written as the sum of two abundant numbers
--All of my requires, unless otherwise listed, can be found at https://bitbucket.org/Mattrixwv/luaClasses
--[[
	Copyright (C) 2020  Matthew Ellison

	This program is free software: you can redistribute it and/or modify
	it under the terms of the GNU Lesser General Public License as published by
	the Free Software Foundation, either version 3 of the License, or
	(at your option) any later version.

	This program is distributed in the hope that it will be useful,
	but WITHOUT ANY WARRANTY; without even the implied warranty of
	MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the
	GNU Lesser General Public License for more details.

	You should have received a copy of the GNU Lesser General Public License
	along with this program.  If not, see <https://www.gnu.org/licenses/>.
]]


require "Stopwatch"
require "Algorithms"


MAX_NUM = 28123;	--The highest number that will be evaluated


local function isSum(abund, num)
	local sumOfNums = 0;
	--Pick a number for the first part of the sum
	for firstNum = 1, #abund do
		for secondNum = firstNum, #abund do
			sumOfNums = abund[firstNum] + abund[secondNum];
			if(sumOfNums == num) then
				return true;
			elseif(sumOfNums > num) then
				break;
			end
		end
	end
	return false;
end


--Setup the variables
local timer = Stopwatch:create();
local divisorSums = {};

--Start the timer
timer:start();

--Get the sum of the divisors of all numbers < MAX_NUM
for cnt = 1, MAX_NUM do
	local div = getDivisors(cnt);
	if(#div > 1) then
		table.remove(div, #div);
	end
	divisorSums[cnt] = getSum(div);
end

--Get the abundant numbers
local abund = {};
for cnt = 1, #divisorSums do
	if(divisorSums[cnt] > cnt) then
		abund[#abund + 1] = cnt;
	end
end

--Check if each number can be the sum of 2 abundant numbers and add to the sum if no
local sumOfNums = 0;
for cnt = 1, MAX_NUM do
	if(not isSum(abund, cnt)) then
		sumOfNums = sumOfNums + cnt;
	end
end

--Stop the timer
timer:stop();

--Print the results
io.write("The answer is " .. sumOfNums .. '\n');
io.write("It took " .. timer:getSeconds() .. " seconds to run this algorithm\n");

--[[Results:
The answer is 4179871
It took 393.139 seconds to run this algorithm
]]