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939811371e
Eclipse integration for simpler use with Windows. Still have to perform some good work on prime numbers! Signed-off-by: Julien Lengrand-Lambert <julien@lengrand.fr>
64 lines
1.9 KiB
Python
64 lines
1.9 KiB
Python
'''
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Created on 7 feb. 2012
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@author: Julien Lengrand-Lambert
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DESCRIPTION : Solves problem 52 of Project Euler
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It can be seen that the number, 125874, and its double, 251748, contain exactly the same digits, but in a different order.
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Find the smallest positive integer, x, such that 2x, 3x, 4x, 5x, and 6x, contain the same digits.
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'''
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def smallest_match(max_mul):
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"""
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Find the smallest positive integer, x, such that ..., till max_mul*x contain the same digits.
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"""
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cpt = 2
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nlist = [(i + 1) * cpt for i in range(max_mul)]
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while(1):
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if cpt % 10000 == 0:
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print cpt
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if same_size(nlist):
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if same_digits(nlist):
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print cpt, nlist
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return cpt
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cpt += 1
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nlist = [(i + 1) * cpt for i in range(max_mul)]
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def all_permutations(seq):
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"""permutate a sequence and return a list of the permutations"""
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if not seq:
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return [seq] # is an empty sequence
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else:
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temp = []
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for k in range(len(seq)):
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part = seq[:k] + seq[k+1:]
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for m in all_permutations(part):
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temp.append(seq[k:k+1] + m)
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return temp
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def same_digits(nlist):
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"""
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Returns True if all numbers in nlist contain the same digits.
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Which means they are all permutations of the first value
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"""
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perms = all_permutations(str(nlist[0])) # finds all permutations
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for el in nlist:
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if not(str(el) in perms):
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return False
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return True
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def same_size(nlist):
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"""
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Returns True if all numbers in nlist have the same number of digits
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"""
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nlen = len(str(nlist[0]))
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for el in nlist :
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if len(str(el)) != nlen:
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return False
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return True
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if __name__ == '__main__':
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print "Answer : %d " % (smallest_match(6)) |