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122 lines
4.3 KiB
Python
122 lines
4.3 KiB
Python
#ProjectEuler/Python/Problem29.py
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#Matthew Ellison
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# Created: 10-10-19
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#Modified: 07-19-20
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#How many distinct terms are in the sequence generated by a^b for 2 <= a <= 100 and 2 <= b <= 100?
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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 Stopwatch import Stopwatch
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from Unsolved import Unsolved
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class Problem29(Problem):
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#Variables
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__bottomA = 2 #The lowest possible value for A
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__topA = 100 #The highest possible value for A
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__bottomB = 2 #The lowest possible value for B
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__topB = 100 #The highest possible value for B
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#Functions
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#Constructor
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def __init__(self):
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super().__init__("How many distinct terms are in the sequence generated by a^b for 2 <= a <= 100 and 2 <= b <= 100?")
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self.unique = []
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#Operational functions
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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 the first A and move towards the top
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for currentA in range(self.__bottomA, self.__topA + 1):
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#Start with the first B and move towards the top
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for currentB in range(self.__bottomB, self.__topB + 1):
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#Get the new number
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currentNum = currentA ** currentB
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#If the new number isn't in the list add it
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if currentNum not in self.unique:
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self.unique.append(currentNum)
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#Stop the timer
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self.timer.stop()
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#Print the results
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self.result = "The number of unique values generated by a^b for " + str(self.__bottomA) + " <= a < = " + str(self.__topA) + " and " + str(self.__bottomB) + " <= b <= " + str(self.__topB) + " is " + str(len(self.unique))
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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.unique.clear()
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#Gets
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#Returns the lowest possible value for a
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def getBottomA(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 can you see the lowest possible A")
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return self.__bottomA
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#Returns the lowest possible value for a
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def getTopA(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 can you see the highest possible A")
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return self.__topA
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#Returns the lowest possible value for a
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def getBottomB(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 can you see the lowest possible B")
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return self.__bottomB
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#Returns the lowest possible value for a
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def getTopB(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 can you see the highest possible B")
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return self.__topB
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#Returns a list of all unique values for a^b
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def getUnique(self) -> list:
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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 list of unique values")
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return self.unique
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#This calls the appropriate functions if the script is called stand alone
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if __name__ == "__main__":
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problem = Problem29()
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print(problem.getDescription()) #Print the description
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problem.solve() #Call the function that answers the problem
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#Print the results
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print(problem.getResult())
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print("It took " + problem.getTime() + " to solve this algorithm")
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""" Results:
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The number of unique values generated by a^b for 2 <= a < = 100 and 2 <= b <= 100 is 9183
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It took 304.630 milliseconds to run this algorithm
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"""
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