04 42 16 73 38 25 39 11 24 94 72 18 08 46 29 32 40 62 76 36
20 69 36 41 72 30 23 88 34 62 99 69 82 67 59 85 74 04 36 16
20 73 35 29 78 31 90 01 74 31 49 71 48 86 81 16 23 57 05 54
01 70 54 71 83 51 54 69 16 92 33 48 61 43 52 01 89 19 67 48
"""
ns = [[int(i) for i in line.split()] for line in grid.split("\n") if line]
def number_tuples(ns, size):
maxrow = len(ns)
maxcol = len(ns[0])
# for each starting corner
for row in xrange(maxrow):
for col in xrange(maxcol):
if col <= maxcol - size: # -
yield ns[row][col : (col + size)]
if row <= maxrow - size: # |
yield [nsr[col] for nsr in ns[row : (row + size)]]
if row <= maxrow - size and col <= maxcol - size: # \
yield [ns[row + i][col + i] for i in xrange(size)]
if row >= size - 1 and col <= maxcol - size: # /
yield [ns[row - i][col + i] for i in xrange(size)]
from pe5 import prod
s = max(prod(nt) for nt in number_tuples(ns, 4))
assert s == 70600674
"""
http://projecteuler.net/index.php?section=problems&id=9
A Pythagorean triplet is a set of three natural numbers, a < b < c, for which,
a2 + b2 = c2
For example, 32 + 42 = 9 + 16 = 25 = 52.
There exists exactly one Pythagorean triplet for which a + b + c = 1000.
Find the product abc.
"""
from pe5 import prod
def find( k ) :
# a + b > c; c = k - a - b
# b > k/2 - a
for a in xrange(1, k/3) :
for b in xrange( k/2-a, k/2 ) :
c = k - a - b
if a**2 + b**2 == c**2 :
return a,b,c
assert prod(find(1000)) == 31875000
73167176531330624919225119674426574742355349194934
96983520312774506326239578318016984801869478851843
85861560789112949495459501737958331952853208805511
12540698747158523863050715693290963295227443043557
66896648950445244523161731856403098711121722383113
62229893423380308135336276614282806444486645238749
30358907296290491560440772390713810515859307960866
70172427121883998797908792274921901699720888093776
65727333001053367881220235421809751254540594752243
52584907711670556013604839586446706324415722155397
53697817977846174064955149290862569321978468622482
83972241375657056057490261407972968652414535100474
82166370484403199890008895243450658541227588666881
16427171479924442928230863465674813919123162824586
17866458359124566529476545682848912883142607690042
24219022671055626321111109370544217506941658960408
07198403850962455444362981230987879927244284909188
84580156166097919133875499200524063689912560717606
05886116467109405077541002256983155200055935729725
71636269561882670428252483600823257530420752963450
""".replace(
"\n", ""
)
from pe5 import prod
ns = [int(i) for i in s]
mp = max(prod(ns[i : i + 5]) for i in xrange(4, len(ns) - 5))
assert mp == 40824
def proper_divisors(factors) :
multiplicants = [ [p**i for i in xrange(e+1)] for p,e in factors.items() ]
orig_n = prod( m[-1] for m in multiplicants )
return filter( lambda pd : pd < orig_n, (prod(ns) for ns in product(*multiplicants)) )
def num_divisors2( factors ) :
return prod( e+1 for e in factors.values() )
def num_divisors(n) :
f = factorize(n)
return prod( e+1 for e in f.values() )