mirror of
https://bitbucket.org/Mattrixwv/projecteuleroctave.git
synced 2025-12-06 17:43:57 -05:00
97 lines
2.9 KiB
Matlab
97 lines
2.9 KiB
Matlab
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
|
|
%}
|