def solve():
max_factors = defaultdict(int)
for i in xrange(1, 21):
factors = defaultdict(int)
for j in factor(i):
factors[j] += 1
for k in factors:
max_factors[k] = max(max_factors[k], factors[k])
return product(k ** v for k, v in max_factors.iteritems())
def findRepeatedSequenceOffsetsAndCounts(text):
# Remove all spaces
text = text.replace(' ', '')
allSeq = []
for i in range(0, len(text) - 3):
allSeq.append([i, text[i:i+3]])
allSeq.append([i, text[i:i+4]])
lastStart = len(text) - 3
allSeq.append([lastStart, text[lastStart:]])
# Now all possible sequences of 3 and 4 chars are stored in allSeq
def keyFunc(item):
return item[1]
# Sort it based on text contents, so that those that match are back to back
sortedSeq = sorted(allSeq, key=keyFunc)
matchDiffs = {}
lastItem = None
for item in sortedSeq:
if lastItem is None:
lastItem = item
continue
if lastItem[1] == item[1]:
diff = item[0] - lastItem[0]
if diff in matchDiffs:
matchDiffs[diff] += 1
else:
matchDiffs[diff] = 1
lastItem = item
# END FOR
matchDiffsAndFactors = dict(matchDiffs)
for key, value in matchDiffs.items():
# print(str(key) + ':'+ str(value))
factors = util.factor(key)
for factor in factors:
if factor in matchDiffsAndFactors:
matchDiffsAndFactors[factor] += value
else:
matchDiffsAndFactors[factor] = value
return sorted(matchDiffsAndFactors.items(), key=operator.itemgetter(1), reverse=True)
def smallestDivisibleByRange(start, end):
factor_count = defaultdict(lambda: 1)
# Analyze the maximum number of times a factor appears.
for n in xrange(max(start, 2), end+1):
local_count = defaultdict(lambda: 0)
factors = factor(n)
for f in factors:
local_count[f] += 1
for f, count in local_count.iteritems():
factor_count[f] = max(factor_count[f], count)
n = 1
for f, count in factor_count.iteritems():
n *= (f**count)
for d in range(max(start, 2), end+1):
assert not n % d, (n, d)
return n
def main():
return max(factor(600851475143))
"""
This file contains a solution for the fifth Project Euler problem.
https://projecteuler.net/problem=5
Author: Clinton Morrison
File: 005.py
"""
import util
lcm_factors = {}
for a in xrange(1, 21):
for (factor, count) in util.factor(a).iteritems():
lcm_factors[factor] = max(count, lcm_factors.get(factor, 0))
result = 1
for (factor, count) in lcm_factors.iteritems():
result *= factor**count
print result
#! /usr/bin/python
from util import factor
factors = factor(28124)
for f in range(len(factors)):
factors[f].remove(f)
abundant = [i for i in range(1, 28124) if i < sum(factors[i])]
result = 0
for i in range(1, 28124):
print i,
can_be = False
for a in abundant:
if a > (i / 2):
break
if i - a in abundant:
can_be = True
break
if not can_be:
result += i
print result
#! /usr/bin/python
from util import factor
factors = factor(28124)
for f in range(len(factors)):
factors[f].remove(f)
abundant = [i for i in range(1,28124) if i < sum(factors[i])]
result = 0
for i in range(1,28124):
print i,
can_be = False
for a in abundant:
if a > (i/2):
break
if i - a in abundant:
can_be = True
break
if not can_be:
result += i
print result
"""What is the value of the first triangle number to have over five hundred
divisors?
"""
from operator import mul
from util import factor, triangle_sequence
for n in triangle_sequence():
old = 1
m = 0
div = []
for f in factor(n):
if f != old:
div.append(m)
m = 1
else: m += 1
old = f
if div and reduce(mul, map(lambda n: n + 1, div)) - 1 > 500:
print n
break
def numfactors_alt(n):
return 0 if n<1 else 1 if n<2 else reduce(mul,[b+1 for (a,b) in factor(n)])
def solve():
return max(factor(600851475143))
"""What is the largest prime factor of the number 600851475143 ? """ from util import factor print[f for f in factor(600851475143)][-1]
""" This file contains a solution for the third Project Euler problem. https://projecteuler.net/problem=3 Author: Clinton Morrison File: 003.py """ import util print max(util.factor(600851475143).keys())
def divisorcount(n):
factorcount = defaultdict(int)
for i in factor(n):
factorcount[i] += 1
return product(count + 1 for count in factorcount.itervalues())
