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42 lines
1001 B
Python
42 lines
1001 B
Python
#-.- coding: latin-1 -.-
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from __future__ import print_function
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from math import sqrt
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'''
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Amicable Numbers
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Problem 21
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Let d(n) be defined as the sum of proper divisors of n (numbers less than n which divide evenly into n).
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If d(a) = b and d(b) = a, where a ≠ b, then a and b are an amicable pair and each of a and b are called amicable numbers.
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For example, the proper divisors of 220 are 1, 2, 4, 5, 10, 11, 20, 22, 44, 55 and 110; therefore d(220) = 284. The proper divisors of 284 are 1, 2, 4, 71 and 142; so d(284) = 220.
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Evaluate the sum of all the amicable numbers under 10000.
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'''
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try:
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xrange #Python 2
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except NameError:
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xrange = range #Python 3
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def sum_of_divisors(n):
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total = 0
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for i in xrange(1, int(sqrt(n)+1)):
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if n%i == 0 and i != sqrt(n):
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total += i + n//i
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elif i == sqrt(n):
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total += i
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return total-n
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sums = []
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total = 0
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for i in xrange(1, 10000):
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n = sum_of_divisors(i)
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if n < len(sums):
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if sums[n-1] == i:
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total += n + i
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sums.append(n)
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print(total) |