43 lines
1.4 KiB
Factor
43 lines
1.4 KiB
Factor
! Copyright (c) 2007 Samuel Tardieu.
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! See http://factorcode.org/license.txt for BSD license.
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USING: arrays kernel lazy-lists math.algebra math math.functions math.primes
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math.ranges sequences ;
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IN: project-euler.134
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! http://projecteuler.net/index.php?section=problems&id=134
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! DESCRIPTION
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! -----------
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! Consider the consecutive primes p1 = 19 and p2 = 23. It can be
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! verified that 1219 is the smallest number such that the last digits
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! are formed by p1 whilst also being divisible by p2.
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! In fact, with the exception of p1 = 3 and p2 = 5, for every pair of
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! consecutive primes, p2 p1, there exist values of n for which the last
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! digits are formed by p1 and n is divisible by p2. Let S be the
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! smallest of these values of n.
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! Find S for every pair of consecutive primes with 5 p1 1000000.
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! SOLUTION
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! --------
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! Compute the smallest power of 10 greater than m or equal to it
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: next-power-of-10 ( m -- n )
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10 swap log 10 log / ceiling >integer ^ ; foldable
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! Compute S for a given pair (p1, p2) -- that is the smallest positive
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! number such that X = p1 [npt] and X = 0 [p2] (npt being the smallest
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! power of 10 above p1)
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: s ( p1 p2 -- s )
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over 0 2array rot next-power-of-10 rot 2array chinese-remainder ;
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: euler134 ( -- answer )
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5 lprimes-from [ 1000000 > ] luntil [ [ s + ] keep ] leach drop ;
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! [ euler134 ] 10 ave-time
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! 6743 ms run / 79 ms GC ave time - 10 trials
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MAIN: euler134
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