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100 lines
3.0 KiB
Matlab
100 lines
3.0 KiB
Matlab
function [] = Problem33()
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%ProjectEuler/Octave/Problem33.m
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%Matthew Ellison
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% Created: 02-07-21
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%Modified: 02-07-21
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%{
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The fraction 49/98 is a curious fraction, as an inexperienced mathematician in attempting to simplify it may incorrectly believe that 49/98 = 4/8, which is correct, is obtained by cancelling the 9s
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We shall consider fractions like, 30/50 = 3/5, to be trivial examples
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There are exactly four non-trivial examples of this type of fraction, less than one in value, and containing two digits in the numerator and denominator
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If the product of these four fractions is given in its lowest common terms, find the value of the denominator
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%}
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%{
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Copyright (C) 2021 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 variables
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MIN_NUMERATOR = 10;
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MAX_NUMERATOR = 98;
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MIN_DENOMINATOR = 11;
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MAX_DENOMINATOR = 99;
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NUMERATORS = [];
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DENOMINATORS = [];
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prodDenominator = 1;
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%Start the timer
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startTime = clock();
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%Search every possible numerator/denominator pair
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for denominator = MIN_DENOMINATOR : MAX_DENOMINATOR
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for numerator = MIN_NUMERATOR : (denominator - 1)
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num = num2str(numerator);
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denom = num2str(denominator);
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tempNum = 0;
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tempDenom = 0;
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%Check that this isn't a trivial example
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if((num(2) == '0') && (denom(2) == '0'))
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continue;
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elseif(num(1) == denom(1))
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tempNum = str2num(num(2));
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tempDenom = str2num(denom(2));
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elseif(num(1) == denom(2))
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tempNum = str2num(num(2));
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tempDenom = str2num(denom(1));
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elseif(num(2) == denom(1))
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tempNum = str2num(num(1));
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tempDenom = str2num(denom(2));
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elseif(num(2) == denom(2))
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tempNum = str2num(num(1));
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tempDenom = str2num(denom(1));
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end
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if(tempDenom == 0)
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continue;
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end
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%Test if the new fraction is the same as the old one
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if((tempNum / tempDenom) == (numerator / denominator))
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numerators(end + 1) = numerator;
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denominators(end + 1) = denominator;
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end
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end
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end
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%Get the product of the numbers
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numProd = prod(numerators);
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denomProd = prod(denominators);
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%Get the gcd to reduce to lowest terms
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GCD = gcd(numProd, denomProd);
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%Save the denominator
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prodDenominator = denomProd / GCD;
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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 denominator of the product is %d\n", prodDenominator)
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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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%{
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Results:
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The denominator of the product is 100
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It took 4.728714 seconds to run this algorithm
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%}
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