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First solution came good, but slow. Could be optimized easily Signed-off-by: Julien Lengrand-Lambert <julien@lengrand.fr>
76 lines
2.5 KiB
Python
76 lines
2.5 KiB
Python
#!/usr/bin/env python
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"""
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##---
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# Julien Lengrand-Lambert
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#Created on : Thu Jan 19 10:12:06 CET 2012
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#
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# DESCRIPTION : Solves problem 23 of Project Euler
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A perfect number is a number for which the sum of its proper divisors is
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exactly equal to the number. For example, the sum of the proper divisors of 28
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would be 1 + 2 + 4 + 7 + 14 = 28, which means that 28 is a perfect number.
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A number n is called deficient if the sum of its proper divisors is less than
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n and it is called abundant if this sum exceeds n.
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As 12 is the smallest abundant number, 1 + 2 + 3 + 4 + 6 = 16, the smallest
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number that can be written as the sum of two abundant numbers is 24.
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By mathematical analysis, it can be shown that all integers greater than 28123
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can be written as the sum of two abundant numbers. However, this upper limit
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cannot be reduced any further by analysis even though it is known that the
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greatest number that cannot be expressed as the sum of two abundant numbers
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is less than this limit.
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Find the sum of all the positive integers which cannot be written as the sum
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of two abundant numbers.
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Hum : Once again, solution is wayyy to long ! (about 20 minuts)
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Starts slowly, but gets faster as soon as abundants numbers are big.
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##---
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"""
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max_sum_abundant = 28123
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def sum_abundant_numbers():
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"""
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Returns the sum of all the positive integers that cannot be written as the sum of
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two abundant numbers.
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"""
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global max_sum_abundant
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numbers = range(max_sum_abundant + 1)
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abundants = [i for i in numbers if is_abundant(i)]
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print "starting to count"
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for ii in abundants:
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print "%d /%d" %(ii, max_sum_abundant)
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for jj in abundants:
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curr_sum = ii + jj
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if curr_sum <= max_sum_abundant: #We don't want to go over the max
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try: # ok because all numbers are unique
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numbers.remove(curr_sum)
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except ValueError: # if not exists, already removed
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pass
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return sum(numbers)
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def get_divisors(value):
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"""
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Returns the list of proper divisors of value
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"""
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div = []
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for val in range(1, value ):
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if (value % val == 0):
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div.append(val)
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return div
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def is_abundant(value):
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"""
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Returns True or False depending whether a number is abundant or not.
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"""
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return sum(get_divisors(value)) > value
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if __name__ == '__main__' :
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print "Answer is : %d" %( sum_abundant_numbers())
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raw_input() # USed to keep Windows terminal open
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