89 lines
2.5 KiB
Factor
89 lines
2.5 KiB
Factor
! Copyright (c) 2008 Aaron Schaefer.
|
|
! See http://factorcode.org/license.txt for BSD license.
|
|
USING: arrays kernel locals math math.order math.primes
|
|
project-euler.common 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 ;
|
|
|
|
: continue? ( pair seq -- ? )
|
|
[ first ] [ length 1 - ] bi* < ;
|
|
|
|
: (find-longest) ( best seq limit -- best )
|
|
[ longest-prime max ] 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
|
|
|
|
SOLUTION: euler050
|