Python の宿題ここで答えます Part 2

[24 デフォルトの名無しさん 2009/11/19(木) 19:00:14]

from itertools import count

def divisor(n):
  return set(reduce(lambda x, y: x + y, [[i, n / i] for i in range(2, int(n**.5) + 1) if not (n % i)], []))

def perfectnumber(limit):
  g = count(2)
  while True:
    p = g.next()
    if sum(divisor(p)) + 1 == p: yield p
    if p >= limit: break

if __name__ == '__main__':
  for n in perfectnumber(1000):
    print n