Author: | ElasticDog |
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Mode: | factor |

Date: | Thu, 20 Nov 2008 06:47:56 |

! Copyright (c) 2008 Aaron Schaefer. ! See http://factorcode.org/license.txt for BSD license. USING: arrays kernel locals math math.primes sequences ; IN: project-euler.050 ! http://projecteuler.net/index.php?section=problems&id=50 ! DESCRIPTION ! ----------- ! The prime 41, can be written as the sum of six consecutive primes: ! 41 = 2 + 3 + 5 + 7 + 11 + 13 ! This is the longest sum of consecutive primes that adds to a prime below ! one-hundred. ! The longest sum of consecutive primes below one-thousand that adds to a ! prime, contains 21 terms, and is equal to 953. ! Which prime, below one-million, can be written as the sum of the most ! consecutive primes? ! SOLUTION ! -------- ! 1) Create an sequence of all primes under 1000000. ! 2) Start summing elements in the sequence until the next number would put you ! over 1000000. ! 3) Check if that sum is prime, if not, subtract the last number added. ! 4) Repeat step 3 until you get a prime number, and store it along with the ! how many consecutive numbers from the original sequence it took to get there. ! 5) Drop the first number from the sequence of primes, and do steps 2-4 again ! 6) Compare the longest chain from the first run with the second run, and store ! the longer of the two. ! 7) If the sequence of primes is still longer than the longest chain, then ! repeat steps 5-7...otherwise, you've found the longest sum of consecutive ! primes! <PRIVATE :: sum-upto ( seq limit -- length sum ) 0 seq [ + dup limit > ] find [ swapd - ] [ drop seq length swap ] if* ; : pop-until-prime ( seq sum -- seq prime ) over length 0 > [ [ unclip-last-slice ] dip swap - dup prime? [ pop-until-prime ] unless ] [ 2drop { } 0 ] if ; ! a pair is { length of chain, prime the chain sums to } : longest-prime ( seq limit -- pair ) dupd sum-upto dup prime? [ 2array nip ] [ [ head-slice ] dip pop-until-prime [ length ] dip 2array ] if ; : longest ( pair pair -- longest ) 2dup [ first ] bi@ > [ drop ] [ nip ] if ; : continue? ( pair seq -- ? ) [ first ] [ length 1- ] bi* < ; : (find-longest) ( best seq limit -- best ) [ longest-prime longest ] 2keep 2over continue? [ [ rest-slice ] dip (find-longest) ] [ 2drop ] if ; : find-longest ( seq limit -- best ) { 1 2 } -rot (find-longest) ; : solve ( n -- answer ) [ primes-upto ] keep find-longest second ; PRIVATE> : euler050 ( -- answer ) 1000000 solve ; ! [ euler050 ] 100 ave-time ! 291 ms run / 20.6 ms GC ave time - 100 trials MAIN: euler050

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