%ProjectEuler/Octave/Problem5.m
%Matthew Ellison
% Created: 03-16-19
%Modified: 03-28-19
%What is the smallest positive number that is evenly divisible by all of the numbers from 1 to 20?
%{
	Copyright (C) 2019  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/>.
%}


%Create your variables
nums = [1:20];
factors = [1];	%The factors that are already in the number
list = [];	%For a temperary list of the factors of the current number
counter = 1;

%Start the timer
startTime = clock();

%You need to find the factors of all the numbers from 1->20
while(counter <= size(nums)(2))
	list = factor(nums(counter));

	%Search factors and try to match all elements in list
	listCnt = 1;
	factorCnt = 1;
	while(listCnt <= size(list)(2))
		if((factorCnt > size(factors)(2)) || (factors(factorCnt) > list(listCnt)))
			%If it was not found add the factor to the list for the number and reset the counters
			factors(end + 1) = list(listCnt);
			factors = sort(factors);
			factorCnt = 1;
			listCnt = 1;
		elseif(factors(factorCnt) == list(listCnt))
			++listCnt;
			++factorCnt;
		else
			++factorCnt;
		end
	end
	++counter;
end

%Stop the timer
endTime = clock();

%Print the results
printf("The smallest positive number that is evenly divisible by all numbers 1-20 is %d\n", prod(factors))
printf("It took %f seconds to run this algorithm\n", etime(endTime, startTime))

%Cleanup your variables
clear counter;
clear factorCnt;
clear listCnt;
clear list;
clear nums;
clear factors;
clear startTime;
clear endTime;
clear ans;

%{
Results:
The smallest positive number that is evenly divisible by all numbers 1-20 is 232792560
It took 0.010742 seconds to run this algorithm
%}