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ProjectEulerLua/Problem1.lua
2020-10-26 14:44:44 -04:00

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Lua

--ProjectEuler/lua/Problem1.lua
--Matthew Ellison
-- Created: 02-01-19
--Modified: 10-26-20
--What is the sum of all the multiples of 3 or 5 that are less than 1000
--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"
local TOP_NUMBER = 999; --This is the largest number that you are going to check
local sumOfMultiples = 0;
local timer = Stopwatch:create();
timer:start();
--Gets the sum of the progression of the multiple
local function sumOfProgression(multiple)
local numTerms = TOP_NUMBER // multiple; --This gets the number of multiples of a particular number that is < MAX_NUMBER
--The sum of progression formula is (n / 2)(a + l). n = number of terms, a = multiple, l = last term
return ((numTerms / 2) * (multiple + (numTerms * multiple)));
end
--Get the sum of the progressions of 3 and 5 and remove the sum of progressions of the overlap
sumOfMultiples = math.floor(sumOfProgression(3)) + math.floor(sumOfProgression(5)) - math.floor(sumOfProgression(3 * 5));
timer:stop()
--Print the results
print("The sum of all numbers < 1000 is " .. sumOfMultiples);
print("It took " .. timer:getMicroseconds() .. " microseconds to run this algorithm");
--[[Results:
The sum of all numbers < 1000 is 233168
It took 2.0 microseconds to run this algorithm
]]