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94 lines
2.7 KiB
Matlab
94 lines
2.7 KiB
Matlab
function [] = Problem12()
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%ProjectEuler/Octave/Problem12.m
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%Matthew Ellison
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% Created:
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%Modified: 03-28-19
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%What is the value of the first triangle number to have over five hundred divisors?
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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 your variables
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number = 1; %To hold the current triangle number
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nextNumber = 2; %To hold the current number you have counted up to
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found = false; %To tell when the answer has been found
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numDivisors = 0;
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%Start the timer
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startTime = clock();
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%Keep checking to find the correct numbers until you
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while((~found) && (number > 0))
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%See how many divisors the number has
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numDivisors = size(getDivisors(number))(2);
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%If it is enough set the flag to stop the loop
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if(numDivisors > 500)
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found = true;
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else
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number += nextNumber;
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nextNumber += 1;
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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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%Print your result
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printf("The first triangular number with more than 500 divisors is %d\n", number)
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printf("It took %f seconds to run this algorithm\n", etime(endTime, startTime))
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end %End of Problem12()
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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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The first triangular number with more than 500 divisors is 76576500
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It took 344.606964 seconds to run this algorithm
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%}
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