def pe172(N=18):
"""
Split 18 digits to 10 piles, each pile contains at most 3 digits.
Get all permutations of all partitions, find out whether it contains 0.
And then count the possible permutations of those digits.
"""
from copy import deepcopy
def countnum(partition):
ap = pec.MP(partition) # all permutations
c = 0
for n in set(partition):
# count how many permutations containing n 0s
temp = deepcopy(partition)
temp.remove(n)
k = pec.MP([temp.count(x) for x in set(temp)])
if n == 0: # if not containing any 0
c += k * ap
else: # else get rid off numbers starting with 0
temp.append(n-1)
c += k * (ap - pec.MP(temp))
return c
c = 0
for r in pec.allPartitions(10, N, init=0):
if all([n < 4 for n in r]):
c += countnum(r)
# answer: 227485267000992000
return c
def pe176(P=47547):
"""
N = 2**a0 * p1**a1 * p2**a2 * ... * pk**ak
A = (2*a1 + 1) * (2*a2 + 1) * ... * (2*ak + 1)
P(N) = (A - 1) / 2, if a0 = 0;
= ((2*a0 - 1) * A - 1) / 2, if a0 > 0.
"""
ODDPRIMES = [3, 5, 7, 11, 13, 17, 19, 23, 29, 31]
def F(plist):
return pef.cprod([2*n+1 for n in plist])
cmin, info = 10**40, []
for i in range(1, 100):
for j in range(1, 10):
for r in pec.allPartitions(j, i):
r = filter(None, r)[::-1]
a = F(r)
if (a - 1) / 2 == P:
c = 1
for x, y in zip(ODDPRIMES[:len(r)], r):
c *= x**y
if c > cmin:
break
if c < cmin:
cmin, info = c, [0] + r
elif (P*2 + 1) % a == 0:
b = (P*2 + 1) / a
if b % 2 == 1 and b / 2 + 1 <= 40:
c = 2**(b / 2 + 1)
for x, y in zip(ODDPRIMES[:len(r)], r):
c *= x**y
if c > cmin:
break
if c < cmin:
cmin, info = c, [b / 2 + 1] + r
elif (a - 1) / 2 > 47547:
break
# answer: 96818198400000 = 2**10 * 3**6 * 5**5 * 7**3 * 11**2
return cmin, info