本文整理匯總了Python中permutation.Permutation.get_all_permutations方法的典型用法代碼示例。如果您正苦於以下問題：Python Permutation.get_all_permutations方法的具體用法？Python Permutation.get_all_permutations怎麽用？Python Permutation.get_all_permutations使用的例子？那麽, 這裏精選的方法代碼示例或許可以為您提供幫助。您也可以進一步了解該方法所在類permutation.Permutation的用法示例。
在下文中一共展示了Permutation.get_all_permutations方法的1個代碼示例，這些例子默認根據受歡迎程度排序。您可以為喜歡或者感覺有用的代碼點讚，您的評價將有助於係統推薦出更棒的Python代碼示例。
示例1: __init__
# 需要導入模塊: from permutation import Permutation [as 別名]
# 或者: from permutation.Permutation import get_all_permutations [as 別名]
class Divisor:
def __init__(self,):
self.__mod_prime = Prime()
self.__mod_permutation = Permutation()
def get_count_of_divisors(self, n):
"""
>>> D = Divisor()
>>> D.get_count_of_divisors(2)
2
>>> D.get_count_of_divisors(100)
9
>>> D.get_count_of_divisors(-1)
Traceback (most recent call last):
ValueError: input has to be a positive int or long: -1
"""
if isinstance(n, (int, long)) == False or n <= 0:
raise ValueError("input has to be a positive int or long: %s" %n)
if n == 1:
return 1
# n = (a ** p) * (b ** q) * (c ** r) のとき、
# n の約數は (p + 1) * (q + 1) * (r + 1) で求められる
factors = self.__mod_prime.get_prime_factors(n)
powers = [v + 1 for v in factors.values()] # [p+1, q+1, r+1, ...]
return reduce(lambda x, y: x * y, powers)
def get_proper_divisors(self, n, n_large = 2000):
"""
>>> D = Divisor()
>>> D.get_proper_divisors(28)
[1, 2, 4, 7, 14]
>>> D.get_proper_divisors(200)
[1, 2, 4, 5, 8, 10, 20, 25, 40, 50, 100]
"""
if n >= n_large:
return self._get_pds1(n)
else:
return self._get_pds2(n)
def _get_pds1(self, n):
prime_factors = self.__mod_prime.get_prime_factors(n)
# prime_all_powers
prime_all_powers = {}
# for n=600 primes = [2, 3, 5]
for prime,power in prime_factors.items(): # for n=200 prime=2,5
prime_all_powers[prime] = [prime * pw for pw in range(power)]
# for n=600 prime_all_powers = {2:[1,2,4,8], 3:[1,3], 5:[1,5,25]}
res = []
for pair in self.__mod_permutation.get_all_permutations(prime_all_powers.values()):
# pairs: [1,1,1], [1,1,5], [1,1,25], [1,3,1], [1,3,5], [1,3,25], .... [8,3,25]
res.append( reduce(lambda x,y: int(x)*int(y), pair) )
return res
def _get_pds2(self, n):
pds = []
for i in range(1, int(n / 2) + 1):
if n % i == 0:
pds.append( i )
return pds