2
0

Delete article

Deleted articles cannot be recovered.

Draft of this article would be also deleted.

Are you sure you want to delete this article?

More than 3 years have passed since last update.

Ruby と Python で解く AtCoder ARC067 C 素因数分解

Last updated at Posted at 2020-05-31

はじめに

AtCoder Problems の Recommendation を利用して、過去の問題を解いています。
AtCoder さん、AtCoder Problems さん、ありがとうございます。

今回のお題

AtCoder Regular Contest C - Factors of Factorial
Difficulty: 925

今回のテーマ、素因数分解

以前解きました Ruby と Python で解く AtCoder ABC057 C 素因数分解 ビット全探索 の簡易版といった問題です。

階乗と言いながら、実際に階乗をするとC言語系では整数オーバーフローが目に見えますので、その点がDifficulty高目なのかもしれませんね。

Ruby

ruby.rb
require 'prime'

n = gets.to_i
if n == 1
  puts '1'
  exit
end
M = 10 ** 9 + 7
h = Hash.new(0)
1.upto(n) do |i|
  i.prime_division.each do |k, v|
    h[k] += v
  end
end
puts h.values.map{|i| (i + 1)}.inject{|u, v| u * v % M}
prime.rb
1.upto(n) do |i|
  i.prime_division.each do |k, v|
    h[k] += v
  end
end

実際の階乗計算は不要で、階乗する数列の数値を素因数分解してハッシュに挿入します。
追記
コメントをいただきましたので修正いたしました。

perm.rb
puts h.values.map{|i| (i + 1)}.inject{|u, v| u * v % M}

ある素数を選択しない(0個選択する)ケースもありますので(i + 1)して組合せ数を求めます。

Ruby 余談

ちなみに1≦N≦1000の最大値1000の階乗は 2568桁の数値になります。

402387260077093773543702433923003985719374864210714632543799910429938512398629020592044208486969404800479988610197196058631666872994808558901323829669944590997424504087073759918823627727188732519779505950995276120874975462497043601418278094646496291056393887437886487337119181045825783647849977012476632889835955735432513185323958463075557409114262417474349347553428646576611667797396668820291207379143853719588249808126867838374559731746136085379534524221586593201928090878297308431392844403281231558611036976801357304216168747609675871348312025478589320767169132448426236131412508780208000261683151027341827977704784635868170164365024153691398281264810213092761244896359928705114964975419909342221566832572080821333186116811553615836546984046708975602900950537616475847728421889679646244945160765353408198901385442487984959953319101723355556602139450399736280750137837615307127761926849034352625200015888535147331611702103968175921510907788019393178114194545257223865541461062892187960223838971476088506276862967146674697562911234082439208160153780889893964518263243671616762179168909779911903754031274622289988005195444414282012187361745992642956581746628302955570299024324153181617210465832036786906117260158783520751516284225540265170483304226143974286933061690897968482590125458327168226458066526769958652682272807075781391858178889652208164348344825993266043367660176999612831860788386150279465955131156552036093988180612138558600301435694527224206344631797460594682573103790084024432438465657245014402821885252470935190620929023136493273497565513958720559654228749774011413346962715422845862377387538230483865688976461927383814900140767310446640259899490222221765904339901886018566526485061799702356193897017860040811889729918311021171229845901641921068884387121855646124960798722908519296819372388642614839657382291123125024186649353143970137428531926649875337218940694281434118520158014123344828015051399694290153483077644569099073152433278288269864602789864321139083506217095002597389863554277196742822248757586765752344220207573630569498825087968928162753848863396909959826280956121450994871701244516461260379029309120889086942028510640182154399457156805941872748998094254742173582401063677404595741785160829230135358081840096996372524230560855903700624271243416909004153690105933983835777939410970027753472000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000

この数値を直接Prime.prime_division(1000!)しても正常に因数分解を行い、結果としてACとなります。
ちょっとだけしました。

Python

python.py
from math import sqrt
from collections import defaultdict

n = int(input())
h = defaultdict(int)
def factorization(arg):
    while arg % 2 == 0:
        h[2] += 1
        arg //= 2
    for i in range(3, int(sqrt(arg)) + 1, 2):
        while arg % i == 0:
            h[i] += 1
            arg //= i
    if arg > 1:
        h[arg] += 1
if n == 1:
    print(1)
else:
    m = 10 ** 9 + 7
    for i in range(1, n + 1):
        factorization(i)
    cnt = 1
    for key, v in h.items():
        cnt *= (v + 1)
        cnt %= m
    print(cnt)

自作のfactorization関数defaultdictに素数を追加しています。

Ruby Python
コード長 (Byte) 239 603
実行時間 (ms) 15 23
メモリ (KB) 2300 3316

まとめ

  • ARC 067 C を解いた
  • Ruby に詳しくなった
  • Python に詳しくなった
2
0
6

Register as a new user and use Qiita more conveniently

  1. You get articles that match your needs
  2. You can efficiently read back useful information
  3. You can use dark theme
What you can do with signing up
2
0

Delete article

Deleted articles cannot be recovered.

Draft of this article would be also deleted.

Are you sure you want to delete this article?