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57 lines
2.0 KiB
Matlab
57 lines
2.0 KiB
Matlab
function [] = Problem1()
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%ProjectEuler/Octave/Problem1.m
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%Matthew Ellison
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% Created: 03-28-19
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%Modified: 10-26-20
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%What is the sum of all the multiples of 3 or 5 that are less than 1000
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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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%Setup your variables
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fullSum = 0; %To hold the sum of all the numbers
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numbers = 0; %To hold all of the numbers
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counter = 0; %The number. It must stay below 1000
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%Start the timer
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startTime = clock();
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%Get the sum of the progressions of 3 and 5 and remove the sum of progressions of the overlap
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sumOfMultiples = sumOfProgression(3) + sumOfProgression(5) - sumOfProgression(3 * 5);
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%Stop the timer
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endTime = clock();
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%Print the results
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printf("The sum of all the numbers less than 1000 that is divisibly by 3 or 5 is: %d\n", sumOfMultiples)
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printf("It took %f seconds to run this algorithm\n", etime(endTime, startTime))
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end %End of Problem1()
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%Gets the sum of the progression of the multiple
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function [sum] = sumOfProgression(multiple)
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numTerms = floor(999 / multiple); %This gets the number of multiples of a particular number that is < MAX_NUMBER
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%The sum of progression formula is (n / 2)(a + l). n = number of terms, a = multiple, l = last term
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sum = ((numTerms / 2) * (multiple + (numTerms * multiple)));
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end %End of sumOfProgression
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%{
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Results:
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The sum of all the numbers less than 1000 that is divisibly by 3 or 5 is: 233168
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It took 0.014717 seconds to run this algorithm
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%}
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