http://projecteuler.net/problem=3

最开始自己的方法很麻烦，每次运行，都会卡。经过xw_y_am童鞋的点拨：http://tieba.baidu.com/p/1558545720 自己也做出来啦 在此对xw_y_am表示感谢。

#! /usr/bin/env python
# Filename: po3.py

from math import floor
from math import sqrt

def isprimer(num):
    if num < 2 or num % 2 == 0 and num != 2:
        return False
    end = floor(sqrt(num)) + 1
    fac  = 3
    while fac < end:
        if num % fac == 0: 
            return False
        fac += 2
    return True

def main():
    num = 600851475143L
    fac_small = 1
    fac_big = 0
    step = 1 if num % 2 == 0 else 2
    end = floor(sqrt(num)) + 1
    max_primer_fac = 0

    while fac_small < end:
        if num % fac_small == 0:
            fac_big = num / fac_small
            if isprimer(fac_big):
                max_primer_fac = fac_big
                break
            elif isprimer(fac_small):
                max_primer_fac = fac_small    
        fac_small += step

    if max_primer_fac != 0:
        print("%d" % (max_primer_fac))

 if __name__ == '__main__':
    main()

第一版：

#! /usr/bin/env python
# -*- coding: utf-8 -*-
# Filename: _3.py
from math import sqrt
from math import floor

def isprimer(num):
    fac_max = floor(sqrt(num)) + 1
    for fac in range(2, fac_max):
        if num % fac == 0: return False
    return True

def primer(lim):
    for n in range(3, lim, 2):
        if isprimer(n):
            yield n

def main():
    num = 600851475143
    primer_fac = list(primer(num))
    primer_fac_rev = primer_fac[-1::-1]
    for pf in primer_fac_rev:
        if num % pf == 0:
            print("the largest prime factor of %d is %d" % (num, pf))
            break

if __name__ == '__main__':
    main()