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121 lines
4.2 KiB
Python
121 lines
4.2 KiB
Python
#ProjectEuler/Python/Problem26.py
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#Matthew Ellison
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# Created: 07-29-19
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#Modified: 10-30-20
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#Find the value of d < 1000 for which 1/d contains the longest recurring cycle in its decimal fraction part.
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#Unless otherwise listed, all of my non-standard imports can be gotten from my pyClasses repository at https://bitbucket.org/Mattrixwv/pyClasses
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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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from Problems.Problem import Problem
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from Unsolved import Unsolved
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class Problem26(Problem):
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#Variables
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__topNumber = 999 #The largest denominator to be checked
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#Function
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#Constructor
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def __init__(self):
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super().__init__("Find the value of d < 1000 for which 1/d contains the longest recurring cycle in its decimal fraction part.")
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self.longestCycle = 0
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self.longestNumber = 0
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#Operational function
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#Solve the problem
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def solve(self):
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#If the problem has already been solved do nothing and end the function
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if(self.solved):
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return
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#Start the timer
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self.timer.start()
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#Start with 1/2 and find out how long the longest cycle is by checking the remainders
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#Loop through every number from 2-999 and use it for the denominator
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for denominator in range(2, self.__topNumber):
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remainderList = [] #Holds the list of remainders
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endFound = False #Holds whether we have found an end to the number (either a cycle or a 0 for remainder)
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cycleFound = False #Holds whether a cycle was detected
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numerator = 1 #The numerator that will be divided
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while(not endFound):
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#Get the remainder after the division
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remainder = numerator % denominator
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#Check if the remainder is 0
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#If it is set the flag
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if(remainder == 0):
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endFound = True
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#Check if the remainder is in the list
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#If it is in the list, set the appropriate flags
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elif remainder in remainderList:
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endFound = True
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cycleFound = True
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#Else add it to the list
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else:
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remainderList.append(remainder)
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#Multiply the remainder by 10 to continue finding the next remainder
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numerator = remainder * 10
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#If a cycle was found check the size of the list against the largest cycle
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if(cycleFound):
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#If it is larger than the largest, set it as the new largest
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if(len(remainderList) > self.longestCycle):
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self.longestCycle = len(remainderList)
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self.longestNumber = denominator
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#Stop the timer
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self.timer.stop()
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#Throw a flag to show the problem is solved
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self.solved = True
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#Reset the problem so it can be run again
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def reset(self):
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super().reset()
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self.longestCycle = 0
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self.longestNumber = 0
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#Gets
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#Returns the result of solving the problem
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def getResult(self):
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#If the problem hasn't been solved throw an exception
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if(not self.solved):
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raise Unsolved("You must solve the problem before you can see the result")
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return f"The longest cycle is {self.longestCycle} digits long\n" \
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f"It is started with the number {self.longestNumber}"
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#Returns the length of the longest cycle
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def getLongestCycle(self) -> int:
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#If the problem hasn't been solved throw an exception
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if(not self.solved):
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raise Unsolved("You must solve the problem before can you see the length of the longest cycle")
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return self.longestCycle
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#Returns the denominator that starts the longest cycle
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def getLongestNumber(self) -> int:
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#If the problem hasn't been solved throw an exception
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if(not self.solved):
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raise Unsolved("You must solve the problem before can you see the denominaotr that started the cycle")
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return self.longestNumber
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""" Results:
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The longest cycle is 982 digits long
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It is started with the number 983
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It took an average of 139.229 milliseconds to run this problem through 100 iterations
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"""
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