#
# For example, when the list is sorted into alphabetical order, COLIN,
# which is worth 3 + 15 + 12 + 9 + 14 = 53, is the 938th name in the
# list. So, COLIN would obtain a score of 938 × 53 = 49714.
#
# What is the total of all the name scores in the file?
import eutils.strings as strings
import eutils.vectorize as vec
import re
import time
start = time.time()
val = dict(zip(list('ABCDEFGHIJKLMNOPQRSTUVWXYZ'), list(range(1, 27))))
f = open('../data/names.txt', 'r')
n = ""
for line in f:
n += line.rstrip()
f.close()
names = [re.sub("[^A-Z]", "", x) for x in n.split(',')]
names.sort()
ans = vec.v_sum([vec.v_sum(strings.map_vals(x, val)) * (i + 1)
for i, x in enumerate(names)])
end = time.time()
print("The answer to Problem 22 is: %s" % ans)
print("<< Returned in %s seconds >>" % (end - start))
# Problem 16
#
# 2^15 = 32768 and the sum of its digits is 3 + 2 + 7 + 6 + 8 = 26.
#
# What is the sum of the digits of the number 2^1000?
import eutils.vectorize as vec
import time
start = time.time()
ans = vec.v_sum([int(x) for x in list(str(2 ** 1000))])
end = time.time()
print("The answer to Problem 16 is: %s" % ans)
print("<< Returned in %s seconds >>" % (end - start))
# Problem 20
#
# n! means n × (n − 1) × ... × 3 × 2 × 1
#
# For example, 10! = 10 × 9 × ... × 3 × 2 × 1 = 3628800, and the sum of
# the digits in the number 10! is 3 + 6 + 2 + 8 + 8 + 0 + 0 = 27.
#
# Find the sum of the digits in the number 100!
import eutils.lists as lists
import eutils.vectorize as vec
import math
import time
start = time.time()
ans = vec.v_sum(lists.l_digits(math.factorial(100)))
end = time.time()
print("The answer to Problem 20 is: %s" % ans)
print("<< Returned in %s seconds >>" % (end - start))
# Problem 13
#
# Work out the first ten digits of the sum of the following one-hundred
# 50-digit numbers.
#
# [See 0013.txt]
import eutils.vectorize as vec
import time
start = time.time()
f = open('../data/0013.txt', 'r')
mat = [int(x) for x in f]
f.close()
ans = int(str(vec.v_sum(mat))[0:10])
end = time.time()
print("The answer to Problem 13 is: %s" % ans)
print("<< Returned in %s seconds >>" % (end - start))
#
# Using words.txt (right click and 'Save Link/Target As...'), a 16K text
# file containing nearly two-thousand common English words, how many are
# triangle words?
import eutils.strings as strings
import eutils.vectorize as vec
import re
import time
start = time.time()
tri = dict(zip([int(0.5 * n * (n + 1)) for n in list(range(1, 25))],
list(range(1, 25))))
val = dict(zip(list('ABCDEFGHIJKLMNOPQRSTUVWXYZ'), list(range(1, 27))))
f = open('../data/words.txt', 'r')
n = ""
for line in f:
n += line.rstrip()
f.close()
words = [re.sub("[^A-Z]", "", x) for x in n.split(",")]
wvals = [vec.v_sum(strings.map_vals(x, val)) for x in words]
ans = len([x for x in wvals if x in tri])
end = time.time()
print("The answer to Problem 42 is: %s" % ans)
print("<< Returned in %s seconds >>" % (end - start))
# a prime, contains 21 terms, and is equal to 953.
#
# Which prime, below one-million, can be written as the sum of the most
# consecutive primes?
import eutils.primes as primes
import eutils.vectorize as vec
import time
start = time.time()
ceil = 1000000
p = [2] + [n for n in list(range(3, ceil, 2)) if primes.is_prime(n)]
max = [0, 0]
for i in range(0, len(p)):
j = i + 1 + max[1]
ssum = vec.v_sum(p[i:j])
while j <= len(p) and ssum < ceil:
if primes.is_prime(ssum):
max = [ssum, len(p[i:j])]
j += 1
ssum = vec.v_sum(p[i:j])
ans = max[0]
end = time.time()
print("The answer to Problem 50 is: %s" % ans)
print("<< Returned in %s seconds >>" % (end - start))
#
# As 12 is the smallest abundant number, 1 + 2 + 3 + 4 + 6 = 16, the
# smallest number that can be written as the sum of two abundant numbers
# is 24. By mathematical analysis, it can be shown that all integers
# greater than 28123 can be written as the sum of two abundant numbers.
# However, this upper limit cannot be reduced any further by analysis
# even though it is known that the greatest number that cannot be
# expressed as the sum of two abundant numbers is less than this limit.
#
# Find the sum of all the positive integers which cannot be written as
# the sum of two abundant numbers.
import eutils.factors as factors
import eutils.vectorize as vec
import time
start = time.time()
abun = []
for n in range(1, 28124):
if n < vec.v_sum(factors.l_factors(n)[:-1]):
abun.append(n)
n = list(range(1,28124))
able = [x + y for i, x in enumerate(abun) for y in abun[i:]]
able = dict(zip(able, [0] * len(able)))
ans = vec.v_sum([y for y in n if y not in able])
end = time.time()
print("The answer to Problem 23 is: %s" % ans)
print("<< Returned in %s seconds >>" % (end - start))
# * d4d5d6=635 is divisible by 5
# * d5d6d7=357 is divisible by 7
# * d6d7d8=572 is divisible by 11
# * d7d8d9=728 is divisible by 13
# * d8d9d10=289 is divisible by 17
#
# Find the sum of all 0 to 9 pandigital numbers with this property.
import eutils.strings as strings
import eutils.vectorize as vec
import math
import time
start = time.time()
n = 10
digits = list(range(0, n))
check = [1,2,3,5,7,11,13,17]
perms = strings.l_permute(digits)[int(math.factorial(n) / n):]
perms = [''.join([str(y) for y in x]) for x in perms]
ans = 0
for s in perms:
pan = vec.v_sum([int(s[i:i + 3]) % d for i, d in enumerate(check)])
if pan == 0:
ans += int(s)
end = time.time()
print("The answer to Problem 43 is: %s" % ans)
print("<< Returned in %s seconds >>" % (end - start))
