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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>
75 lines
1.9 KiB
Python
75 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 45 of Project Euler
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Triangle, pentagonal, and hexagonal numbers are generated by the following formulae:
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Triangle Tn=n(n+1)/2 1, 3, 6, 10, 15, ...
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Pentagonal Pn=n(3(n-1))/2 1, 5, 12, 22, 35, ...
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Hexagonal Hn=n(2(n-1)) 1, 6, 15, 28, 45, ...
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It can be verified that T285 = P165 = H143 = 40755.
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Find the next triangle number that is also pentagonal and hexagonal.
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'''
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def tn(value):
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"""
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Returns the triangle number of the given value
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"""
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return value * (value + 1) / 2
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def pn(value):
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"""
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Returns the Pentagonal number of the given value
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"""
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return value * ((3 * value) - 1) / 2
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def hn(value):
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"""
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Returns the Hexagonal number of the given value
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"""
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return value * ((2 * value) - 1)
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def tripenhex(value):
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"""
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Returns True of the triangle, pentagonal and hexagonal number of value are the same
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"""
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return (tn(value) == pn(value) == hn(value))
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def big_while():
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"""
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Returns the value for which tm = pn = ho
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"""
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cpt_t = 2
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cpt_p = 2
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cpt_h = 2
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val_t = tn(cpt_t)
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val_p = pn(cpt_p)
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val_h = hn(cpt_h)
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res = []
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while len(res) < 2: # to get the two first values
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while not (val_t == val_p == val_h):
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cpt_t += 1
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val_t = tn(cpt_t)
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# updating val_p
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while val_p < val_t:
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cpt_p += 1
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val_p = pn(cpt_p)
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# updating val_h
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while val_h < val_t:
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cpt_h += 1
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val_h = hn(cpt_h)
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res.append(val_t)
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cpt_t += 1
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val_t = tn(cpt_t)
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return res[1] # outputs the second value
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if __name__ == '__main__':
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print "Answer : %d " % (big_while()) |