AtCoder Regular Contest C - Factors of Factorial Difficulty: 925
This theme, prime factorization
It is a problem such as a simplified version of * Solving atCoder ABC057 C prime factorization bit full search with Ruby and Python *.
Although it is called factorial, when you actually factorial, integer overflow is visible in C language system
, so that point may be ** Difficulty ** higher.
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
No actual factorial calculation is required, and the numbers in the factorial sequence are factored into prime factors and inserted into the hash. ** Addendum ** I received a comment, so I corrected it.
perm.rb
puts h.values.map{|i| (i + 1)}.inject{|u, v| u * v % M}
Since there are cases where a certain prime number is not selected (zero is selected), (i + 1)
is used to obtain the number of combinations.
By the way, the factorial of the maximum value 1000
of 1 ≤ N ≤ 1000
is a numerical value of 2568 digits
.
402387260077093773543702433923003985719374864210714632543799910429938512398629020592044208486969404800479988610197196058631666872994808558901323829669944590997424504087073759918823627727188732519779505950995276120874975462497043601418278094646496291056393887437886487337119181045825783647849977012476632889835955735432513185323958463075557409114262417474349347553428646576611667797396668820291207379143853719588249808126867838374559731746136085379534524221586593201928090878297308431392844403281231558611036976801357304216168747609675871348312025478589320767169132448426236131412508780208000261683151027341827977704784635868170164365024153691398281264810213092761244896359928705114964975419909342221566832572080821333186116811553615836546984046708975602900950537616475847728421889679646244945160765353408198901385442487984959953319101723355556602139450399736280750137837615307127761926849034352625200015888535147331611702103968175921510907788019393178114194545257223865541461062892187960223838971476088506276862967146674697562911234082439208160153780889893964518263243671616762179168909779911903754031274622289988005195444414282012187361745992642956581746628302955570299024324153181617210465832036786906117260158783520751516284225540265170483304226143974286933061690897968482590125458327168226458066526769958652682272807075781391858178889652208164348344825993266043367660176999612831860788386150279465955131156552036093988180612138558600301435694527224206344631797460594682573103790084024432438465657245014402821885252470935190620929023136493273497565513958720559654228749774011413346962715422845862377387538230483865688976461927383814900140767310446640259899490222221765904339901886018566526485061799702356193897017860040811889729918311021171229845901641921068884387121855646124960798722908519296819372388642614839657382291123125024186649353143970137428531926649875337218940694281434118520158014123344828015051399694290153483077644569099073152433278288269864602789864321139083506217095002597389863554277196742822248757586765752344220207573630569498825087968928162753848863396909959826280956121450994871701244516461260379029309120889086942028510640182154399457156805941872748998094254742173582401063677404595741785160829230135358081840096996372524230560855903700624271243416909004153690105933983835777939410970027753472000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000
Even if this number is directly Prime.prime_division (1000!)
, Factorization will be performed normally, resulting in ʻAC. Just a little
! `I did.
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)
I added a prime number to defaultdict
with my own factorization function
.
Ruby | Python | |
---|---|---|
Code length(Byte) | 239 | 603 |
Execution time(ms) | 15 | 23 |
memory(KB) | 2300 | 3316 |
Recommended Posts