示例#1
0
"""
Let d(n) be defined as the sum of proper divisors of n (numbers less than n which divide evenly into n).
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.

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.

Evaluate the sum of all the amicable numbers under 10000.
"""

import divisors

divisorSums = {}
sumOfAmicableNumbers = 0

for i in xrange(1, 10000):
    divisorSums[i] = sum(divisors.get_proper_divisors(i))
    if divisorSums[i] != i and divisorSums[i] in divisorSums:
        if divisorSums[divisorSums[i]] == i:
            sumOfAmicableNumbers += i
            sumOfAmicableNumbers += divisorSums[i]

print str(sumOfAmicableNumbers)
def is_abundant_number(number):
    return sum(divisors.get_proper_divisors(number)) > number