#ProjectEuler/Python/Problem21.py
#Matthew Ellison
# Created: 03-18-19
#Modified: 03-28-19
#Evaluate the sum of all the amicable numbers under 10000
#Unless otherwise listed, all of my non-standard imports can be gotten from my pyClasses repository at https://bitbucket.org/Mattrixwv/pyClasses
"""
	Copyright (C) 2019  Matthew Ellison

	This program is free software: you can redistribute it and/or modify
	it under the terms of the GNU Lesser General Public License as published by
	the Free Software Foundation, either version 3 of the License, or
	(at your option) any later version.

	This program is distributed in the hope that it will be useful,
	but WITHOUT ANY WARRANTY; without even the implied warranty of
	MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the
	GNU Lesser General Public License for more details.

	You should have received a copy of the GNU Lesser General Public License
	along with this program.  If not, see <https://www.gnu.org/licenses/>.
"""


from Stopwatch import Stopwatch
import Algorithms

__limit = 10000	#The top number that will be evaluated


def Problem21():
	#Setup the variables
	divisorSum = []	#Holds the sum of the divisors of the subscript number
	divisorSum.append(0)	#Start with a 0 in the [0] location

	#Generate the divisors of all the numbers < 10000, get their sum, and add it to the list
	for cnt in range(1, __limit):
		divisors = Algorithms.getDivisors(cnt)	#Get all the divisors of a number
		if(len(divisors) > 1):
			divisors.pop()	#Remove the last entry because it will be the number itself
		divisorSum.append(int(sum(divisors)))
	#Check every sum of divisors in the list for a matching sum
	amicable = []
	for cnt in range(1, len(divisorSum)):
		currentSum = divisorSum[cnt]
		#If the sum is greater than the number of divisors then it is impossible to be amicable. Skip the number and continue
		if(currentSum >= len(divisorSum)):
			continue
		#We know that divisorSum[cnt] == currentSum, so if divisorSum[currentSum] == cnt we found an amicable number
		if(divisorSum[currentSum] == cnt):
			#A number can't be amicable with itself
			if(currentSum == cnt):
				continue
			#Add the number to the amicable vector
			amicable.append(cnt)

	#Print the results
	print("All amicable numbers less than 10000 are")
	for num in amicable:
		print(str(num))
	print("The sum of all of these amicable numbers is " + str(sum(amicable)))

#Run the correct function if this script is called stand along
if __name__ == "__main__":
	timer = Stopwatch()
	timer.start()
	Problem21()
	timer.stop()
	print("It took " + timer.getString() + " to run this algorithm")

""" Results:
All amicable numbers less than 10000 are
220
284
1184
1210
2620
2924
5020
5564
6232
6368
The sum of all of these amicable numbers is 31626
It took 59.496 milliseconds to run this algorithm
"""