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97 lines
2.9 KiB
Matlab
97 lines
2.9 KiB
Matlab
function [] = Problem26()
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%ProjectEuler/Octave/Problem26.m
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%Matthew Ellison
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% Created: 08-02-19
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%Modified: 08-02-19
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%Find the value of d < 1000 for which 1/d contains the longest recurring cycle in its decimal fraction part.
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%{
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Copyright (C) 2019 Matthew Ellison
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This program is free software: you can redistribute it and/or modify
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it under the terms of the GNU Lesser General Public License as published by
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the Free Software Foundation, either version 3 of the License, or
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(at your option) any later version.
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This program is distributed in the hope that it will be useful,
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but WITHOUT ANY WARRANTY; without even the implied warranty of
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MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the
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GNU Lesser General Public License for more details.
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You should have received a copy of the GNU Lesser General Public License
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along with this program. If not, see <https://www.gnu.org/licenses/>.
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%}
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TOP_NUMBER = 999; %The largest denominator to the checked
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%Setup variables
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longestCycle = 0; %
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longestNumber = 1;
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%Start the timer
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startTime = clock();
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%Start with 1/2 and find out how long the longest cycle is by checking the remainders
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%Loop through every number from 2-999 and use it for the denominator
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for denominator = 2 : TOP_NUMBER
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remainderList = []; %Holds the list of remainders
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endFound = false; %Holds whether we have found an end ot the number (either a cycle or a 0 for a remainder)
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cycleFound = false; %Holds whether a cycle was detected
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numerator = 1; %The numerator that will be divided
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while(~endFound)
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%Get the remainder after the division
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remainder = mod(numerator, denominator);
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%Check if the remainder is 0
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%If it is, set the flag
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if(remainder == 0)
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endFound = true;
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%Check if the remainder is in the list
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%If it is in the list, set the appropriate flags
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elseif(isFound(remainderList, remainder))
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endFound = true;
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cycleFound = true;
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%Else add it to the list
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else
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remainderList(end + 1) = remainder;
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end
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%Multiply the remainder by 10 to continue finding the next remainder
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numerator = remainder * 10;
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end
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%If a cycle was found check the size of the list against the largest cycle
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if(cycleFound)
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%If it is larger than the largest, set it as the new largest
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if(size(remainderList)(2) > longestCycle)
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longestCycle = size(remainderList)(2);
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longestNumber = denominator;
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end
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end
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end
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%End the timer
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endTime = clock();
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%Print the results
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printf("The longest cycle is %d digits long\n", longestCycle);
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printf("It is started with the number %d\n", longestNumber);
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printf("It took %f seconds to run this algorithm\n", etime(endTime, startTime));
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end
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function [found] = isFound(array, key)
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found = false; %Start with a false. It only turns true if you find key in array
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for location = 1 : size(array)(2)
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if(key == array(location))
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found = true;
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return;
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end
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end
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end
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%{
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Results:
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The longest cycle is 982 digits long
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It is started with the number 983
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It took 49.173325 seconds to run this algorithm
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%}
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