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88 lines
2.6 KiB
Python
88 lines
2.6 KiB
Python
#ProjectEuler/Python/Problem23.py
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#Matthew Ellison
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# Created: 03-22-19
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#Modified: 03-28-19
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#Find the sum of all the positive integers which cannot be written as the sum of two abundant numbers
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#All of my imports can be gotten from my pyClasses repository at https://bitbucket.org/Mattrixwv/pyClasses
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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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from Stopwatch import Stopwatch
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import Algorithms
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__maxNum = 28123
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#A function that returns true if num can be created by adding two elements from abund and false if it cannot
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def isSum(abund: list, num: int) -> bool:
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sumOfNums = 0
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#Pick a number for the first part of the sum
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for firstNum in range(0, len(abund)):
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#Pick a number for the second part of the sum
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for secondNum in range(0, len(abund)):
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sumOfNums = abund[firstNum] + abund[secondNum]
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if(sumOfNums == num):
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return True
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elif(sumOfNums > num):
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break
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#If you have run through the entire list and did not find a sum then it is false
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return False
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def Problem23():
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#Setup the variables
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divisorSums = []
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#Make sure every element has a 0 in it's location
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for cnt in range(0, __maxNum):
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divisorSums.append(0)
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#Get the sum of the divisors of all numbers < __maxNum
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for cnt in range(1, __maxNum):
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div = Algorithms.getDivisors(cnt)
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if(len(div) > 1):
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div.remove(div[len(div) - 1])
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divisorSums[cnt] = sum(div)
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#Get the abundant numbers
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abund = []
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for cnt in range(0, len(divisorSums)):
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if(divisorSums[cnt] > cnt):
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abund.append(cnt)
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#Check if each number can be the sum of 2 abundant numbers and add to the sum if no
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sumOfNums = 0
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for cnt in range(1, __maxNum):
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if(not isSum(abund, cnt)):
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sumOfNums += cnt
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#Print the results
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print("The answer is " + str(sumOfNums))
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if __name__ == "__main__":
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timer = Stopwatch()
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timer.start()
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Problem23()
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timer.stop()
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print("It took " + timer.getString() + " to run this algorithm")
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""" Results:
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The answer is 4179871
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It took 27.738 minutes to run this algorithm
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"""
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