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122 lines
3.5 KiB
Matlab
122 lines
3.5 KiB
Matlab
function [] = Problem21()
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%ProjectEuler/Octave/Problem21.m
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%Matthew Ellison
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% Created: 03-19-19
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%Modified: 03-28-19
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%Evaluate the sum of all the amicable numbers under 10000
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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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%Setup the varibles
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LIMIT = 10000; %The top number that will be evaluated
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divisorSum = []; %Holds the sum of the factors of the subscript number
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%Start the timer
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startTime = clock();
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%Generate the factors of all the numbers < 10000, get their sum, and add it to the list
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for cnt = 1 : LIMIT
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divisors = getDivisors(cnt); %Get all the divisors of a number
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if(size(divisors)(2) > 1)
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divisors(end) = []; %Remove the last entry because it will be the number itself
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end
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divisorSum(end + 1) = sum(divisors); %Add the sum of the divisors to the vector
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end
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%Check every sum of divisors in the list for a matching sum
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amicable = [];
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for cnt = 1 : size(divisorSum)(2)
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sumOfDivisors = divisorSum(cnt);
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%If the sum is greater than the number of divisors then it is impossible to be amicable. Skip the number and continue
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if(sumOfDivisors >= size(divisorSum)(2))
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continue;
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%We know that divisorSum[cnt] == sum, do if divisorSum[sum] == cnt we have found an amicable number
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elseif(divisorSum(sumOfDivisors) == cnt)
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%A number can't be amicable with itself, so skip those numbers
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if(sumOfDivisors == cnt)
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continue;
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end
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amicable(end + 1) = cnt;
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end
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end
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%Stop the timer
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endTime = clock();
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%Sort the vector for neatness
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sort(amicable);
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%Print the results
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printf("All amicable numbers less than %d are\n", LIMIT)
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%Print the list of amicable numbers
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for location = 1: size(amicable)(2)
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printf("%d\n", amicable(location))
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end
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printf("The sum of all of these amicable numbers is %d\n", sum(amicable))
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printf("It took %f seconds to run this algorithm\n", etime(endTime, startTime))
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end %Problem21()
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function [divisors] = getDivisors(goalNumber)
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divisors = [];
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%Start by checking that the number is positive
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if(goalNumber <= 0)
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return;
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%If the number is 1 return just itself
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elseif(goalNumber == 1)
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divisors(end + 1) = 1;
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return;
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%Otherwise add 1 and itself to the list
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else
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divisors(end + 1) = 1;
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divisors(end + 1) = goalNumber;
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end
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%Start at 3 and loop through all numbers < sqrt(goalNumber) looking for a number that divides it evenly
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topPossibleDivisor = ceil(sqrt(goalNumber));
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for possibleDivisor = 2 : topPossibleDivisor
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%If you find one add it and the number it creates to the list
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if(mod(goalNumber, possibleDivisor) == 0)
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divisors(end + 1) = possibleDivisor;
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%Account for the possibility sqrt(goalNumber) being a divisor
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if(possibleDivisor != topPossibleDivisor)
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divisors(end + 1) = goalNumber / possibleDivisor;
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end
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end
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end
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%Sort the list before returning for neatness
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divisors = sort(divisors);
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end
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%{
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Results:
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All amicable numbers less than 10000 are
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220
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284
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1184
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1210
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2620
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2924
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5020
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5564
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6232
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6368
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The sum of all of these amicable numbers is 31626
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It took 5.650108 seconds to run this algorithm
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%}
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