Solution to http://100.70.19.162:8000/contest/32/9

compliance.py

import sys
from math import factorial
from collections import Counter
from functools import reduce

"""
  Solution to http://100.70.19.162:8000/contest/32/9
"""


""" Counts number of permutations of a given multiset
  let m be the number of items in a multiset with n unique items
  let each ith item of the multiset occur m_i times (multiplicity of m_i)
    such that \sum_{i=1}^{n} m_i = m
  then the number of unique permutations of the multiset is
    \frac{m!}{\Pi_{i=1}^{n} m_i!}
  
  The multiset will be given as a list of multiplicities
  numer = m!
  denom = \Pi_{i=1}^{n} m_i!
"""
def multiset_perms(multiset):
  numer = factorial(sum(multiset))

  # product of all permutations of each multiplicity
  multiplicity_perms = map(lambda v: factorial(v), multiset)
  denom = reduce(lambda x, y: x * y, multiplicity_perms)

  return numer // denom


"""  Counts number of possible unique palindromes from a Counter representing distribution of chars """
def count_unique_pals(counter):
  if len([v for v in counter.values() if v % 2 == 1]) > 1:
    # more than 1 character with odd occurrences
    return 0

  # partition into one side of the pivot
  side = list(map(lambda v: v // 2, counter.values()))
  return multiset_perms(side)


chars = sys.stdin.read().splitlines()[0]
print(count_unique_pals(Counter(chars)))