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87 lines
2.9 KiB
Lua
87 lines
2.9 KiB
Lua
--ProjectEuler/lua/Problem26.lua
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--Matthew Ellison
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-- Created: 08-02-19
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--Modified: 06-19-20
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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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--All of my requires, unless otherwise listed, can be found at https://bitbucket.org/Mattrixwv/luaClasses
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--[[
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Copyright (C) 2020 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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require "Stopwatch"
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local TOP_NUMBER = 999 --The largest denominator to tbe checked
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--Setup the variables
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local timer = Stopwatch:create();
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local longestCycle = 0;
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local longestNumber = 1;
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--Start the timer
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timer:start();
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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 do
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local remainderList = {}; --Holds the list of remainders
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local endFound = false; --Holds whether we have found an end to the number (either a cycle or a 0 for remainder)
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local cycleFound = false; --Holds whether a cycle was detected
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local numerator = 1; --The numerator that will be divided
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while(not endFound) do
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--Get the remainder after the division
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local remainder = 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) then
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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(remainderList[remainder]) then
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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[remainder] = true;
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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) then
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--If it is larger than the largest, set it as the new largest
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if(#remainderList > longestCycle) then
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longestCycle = #remainderList;
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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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timer:stop();
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--Print the results
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io.write("The longest cycle is " .. longestCycle .. " digits long\n");
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io.write("It is started with the number " .. longestNumber .. '\n');
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io.write("It took " .. timer:getString() .. " to run this algorithm\n");
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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 28.222 milliseconds to run this algorithm
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]]
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