Apply embed.ly

solutions

problem 1:
>>> fc = lambda n: 1 if n <= 1 else n * fc(n - 1)
>>> def r(n):
...     s = str(n)
...     d = 0
...     for i in s:
...             d += int(i)
...     return d
>>> R = lambda n: r(fc(n))
>>> i = 0
>>> while True:
...     i += 1
...     x = R(i)
...     if x == 8001:
...             break
>>> print(i)
787

problem 2:
>>> import xml.parsers.expat
>>> u = 'http://apply.embed.ly/static/data/2.html'
>>> from urllib.request import urlopen
>>> html = urlopen(u)
>>> h = html.read()
>>> h = h.encode('utf-8')
>>> d = 0
>>> from collections import defaultdict
>>> t = defaultdict(list)
>>> def start_h(n, a):
...     global d, t
...     d += 1
...     t[n] = d - 1
... 
>>> def end_h(n):
...     global d
...     d -= 1
... 
>>> p = xml.parsers.expat.ParserCreate()
>>> p.StartElementHandler = start_h
>>> p.EndElementHandler = end_h
>>> p.Parse('<?xml version="1.0">' + h)
1
>>> def sd(l):
...     m = sum(l)/len(l)
...     d = 0
...     for i in l:
...             d += (m - i) ** 2
...     d /= len(l) - 1
...     return sqrt(d)
... 
>>> from math import sqrt
>>> sd(t['p'])
1.4303439938073574

problem 3:
>>> fs = 0
>>> for i in range(1, 901):
...     fs += 900/i
... 
>>> fs
6642.149292810678
>>> fc = 0
>>> for i in range(1, 901):
...     fc += 900/i
...     if fc > fs / 2:
...             break
... 
>>> i
22