AOJ0052 Factorial II

問題

例によってRubyで書いた。

末尾に0がつく===10が掛かっている===2と5がかかっている、だから
素因数の2と5の少ない方が0の個数になる、という方針で解いた。

以下ソースと、テスト。

# encoding: utf-8
class ZeroCounter
  attr_accessor:num
  attr_accessor:num_zero
  def initialize(i)
    @num = i
    @num_zero = [count_num_of_primes(2), count_num_of_primes(5)].min
    p @num_zero
 end

  def count_num_of_primes(prime)
    num_pf = 0
    i = prime
    while @num / i > 0 do
      num_pf += @num / i
      i = i * prime
    end
    return num_pf
  end
end

while (line = gets) do
  if line.to_i != 0 then
    ZeroCounter.new(line.to_i)
  end
end

#encoding: utf-8
require 'rspec'
require './a'

describe ZeroCounter do
  it "should return the num of prime factor 2" do
    zc = ZeroCounter.new(19)
    zc.count_num_of_primes(2).should == 16
  end
  it "should return the num of zeros of factorial" do
    zc = ZeroCounter.new(19)
    zc.num_zero.should == 3
  end
  it "should return the num of zeros of factorial" do
    zc = ZeroCounter.new(45)
    zc.num_zero.should == 10
  end
  it "should return the num of zeros of factorial" do
    zc = ZeroCounter.new(50)
    zc.num_zero.should == 12
  end
end