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Problem26.m

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+function [] = Problem26()
+%ProjectEuler/Octave/Problem26.m
+%Matthew Ellison
+% Created: 08-02-19
+%Modified: 08-02-19
+%Find the value of d < 1000 for which 1/d contains the longest recurring cycle in its decimal fraction part.
+%{
+	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/>.
+%}
+
+
+TOP_NUMBER = 999;	%The largest denominator to the checked
+
+
+%Setup variables
+longestCycle = 0;	%
+longestNumber = 1;
+
+%Start the timer
+startTime = clock();
+
+%Start with 1/2 and find out how long the longest cycle is by checking the remainders
+%Loop through every number from 2-999 and use it for the denominator
+for denominator = 2 : TOP_NUMBER
+	remainderList = [];	%Holds the list of remainders
+	endFound = false;	%Holds whether we have found an end ot the number (either a cycle or a 0 for a remainder)
+	cycleFound = false;	%Holds whether a cycle was detected
+	numerator = 1;	%The numerator that will be divided
+	while(~endFound)
+		%Get the remainder after the division
+		remainder = mod(numerator, denominator);
+		%Check if the remainder is 0
+		%If it is, set the flag
+		if(remainder == 0)
+			endFound = true;
+		%Check if the remainder is in the list
+		%If it is in the list, set the appropriate flags
+		elseif(isFound(remainderList, remainder))
+			endFound = true;
+			cycleFound = true;
+		%Else add it to the list
+		else
+			remainderList(end + 1) = remainder;
+		end
+		%Multiply the remainder by 10 to continue finding the next remainder
+		numerator = remainder * 10;
+	end
+	%If a cycle was found check the size of the list against the largest cycle
+	if(cycleFound)
+		%If it is larger than the largest, set it as the new largest
+		if(size(remainderList)(2) > longestCycle)
+			longestCycle = size(remainderList)(2);
+			longestNumber = denominator;
+		end
+	end
+end
+
+%End the timer
+endTime = clock();
+
+%Print the results
+printf("The longest cycle is %d digits long\n", longestCycle);
+printf("It is started with the number %d\n", longestNumber);
+printf("It took %f seconds to run this algorithm\n", etime(endTime, startTime));
+
+end
+
+function [found] = isFound(array, key)
+	found = false;	%Start with a false. It only turns true if you find key in array
+	for location = 1 : size(array)(2)
+		if(key == array(location))
+			found = true;
+			return;
+		end
+	end
+end
+
+%{
+Results:
+The longest cycle is 982 digits long
+It is started with the number 983
+It took 49.173325 seconds to run this algorithm
+%}