def fund(n):
lower, upper = bounds(n)
p = 0
for b in range(lower, upper+1):
r = n - b
for seq in mtools.comb(range(1,n+1), r):
a = mtools.mul(seq)
p += a
return mtools.fac(n+1)/p
def facs(lim):
l = []
i = 1
k = 1
while k < lim:
l.append(k)
i = i + 1
k = mtools.fac(i)
l.append(k)
return l
def c(n, r):
a = mtools.fac(r)
b = 1
for i in range(n-r+1,n+1):
b = b*i
return b/a
import math, time, mtools, sys
fac_tab = [mtools.fac(x) for x in range(10)]
chain_tab = {}
def solve():
cnt = 0
for i in range(1, 1000000):
n = chain(i)
chain_tab[i] = n
if n == 60:
cnt = cnt + 1
return cnt
def sum_fac(n):
l = mtools.separate_digits(n)
l2 = [fac_tab[x] for x in l]
return sum(l2)
def chain(i):
c = 1
d = {i:True}
while True:
v = sum_fac(i)
if chain_tab.has_key(v):
return c + chain_tab[v]
if d.has_key(v):
return c
d[v] = True
c = c + 1
i = v
'''
145 is a curious number, as 1! + 4! + 5! = 1 + 24 + 120 = 145.
Find the sum of all numbers which are equal to the sum of the factorial of their digits.
Note: as 1! = 1 and 2! = 2 are not sums they are not included.
'''
import time, mtools
tab = [mtools.fac(x) for x in range(0, 10)]
def is_curious(n):
global tab
l = get_digits(n)
l2 = [tab[x] for x in l]
return n == sum(l2)
def get_digits(n):
l = []
while n > 0:
l.append(n%10)
n = n /10
l.reverse()
return l
def test():
assert [1, 2, 3] == get_digits(123)
assert is_curious(145)
def solve():