阶乘结果换算进制后得到数字的尾部有几个0

原题地址：
https://www.codewars.com/kata/trailing-zeros-in-factorials-in-any-given-integer-base/train/ruby
最开始没有仔细思考，先计算阶乘，再统计0的个数，但测试用例中包括大数情况，例如：
trailing_zeros(93326215443944152681699238856266700490715968264381621468592963895217599993229915608941463976156518286253697920827223758251185210916864000000000000000000000000, 100)
，很明显这种算法无法实现。
后来经过查询，在https://www.cnblogs.com/hutonm/p/5624996.html 发现了新思路；
最新算法：

def trailing_zeros(num, base)  start = Time.now  arr = list(base)  # puts "time_list:#{(Time.now - start) * 1000}"  brr = Hash.new(0)  arr.each do |item|    brr[item] = figure(num, item)  end  # puts "time_figure:#{Time.now - start}"  uniq_arr = arr.uniq  counter = Hash.new(0)  arr.each {|val| counter[val] += 1}  # puts "time_counter:#{Time.now - start}"  crr = []  brr.each do |key, value|    crr.push(value / counter[key])  end  # puts "time_crr:#{Time.now - start}"  crr.min  # puts "time_total:#{Time.now - start}"enddef figure(num, item)  num >= item ? num / item + figure(num / item, item) : 0enddef list(base)  i = 2  j = 0  arr = []  while (base > 1) do    if base % i == 0      max = i      while base % i == 0        base /= i        arr[j] = max        j += 1      end    end    i += 1  end  return arrend

经过测试：

require 'benchmark'   Benchmark.bmbm(10) do |t|   t.report{puts trailing_zeros(93326215443944152681699238856266700490715968264381621468592963895217599993229915608941463976156518286253697920827223758251185210916864000000000000000000000000, 100)}   t.report{puts trailing_zeros(93326215443944152681699238856266700490715968264381621468592963895217599993229915608941463976156518286253697920827223758251185210916864000000000000000000000000, 100)}   t.report{puts trailing_zeros(93326215443944152681699238856266700490715968264381621468592963895217599993229915608941463976156518286253697920827223758251185210916864000000000000000000000000, 100)} end

结果如下：

Rehearsal ----------------------------------------------           11665776930493019085212404857033337561339496033047702683574120486902199999153739451117682997019564785781712240103402969781398151364607999999999999999999999950  0.000000   0.000000   0.000000 (  0.003088)           11665776930493019085212404857033337561339496033047702683574120486902199999153739451117682997019564785781712240103402969781398151364607999999999999999999999950  0.010000   0.000000   0.010000 (  0.002103)           11665776930493019085212404857033337561339496033047702683574120486902199999153739451117682997019564785781712240103402969781398151364607999999999999999999999950  0.000000   0.000000   0.000000 (  0.002209)------------------------------------- total: 0.010000sec                 user     system      total        real           11665776930493019085212404857033337561339496033047702683574120486902199999153739451117682997019564785781712240103402969781398151364607999999999999999999999950  0.000000   0.000000   0.000000 (  0.001200)           11665776930493019085212404857033337561339496033047702683574120486902199999153739451117682997019564785781712240103402969781398151364607999999999999999999999950  0.000000   0.000000   0.000000 (  0.001220)           11665776930493019085212404857033337561339496033047702683574120486902199999153739451117682997019564785781712240103402969781398151364607999999999999999999999950  0.000000   0.000000   0.000000 (  0.001380)

每次运行时间在1.5ms以内。
但是还是不符合题目要求，提交反馈结果如下：

Process was terminated. It took longer than 12000ms to complete

---

尝试优化list()方法，优化效果不明显：

# return the number of trailing zeros in num! (base)def trailing_zeros(num, base)  arr = list(base)  brr = Hash.new(0)  arr.each_key do |key|    brr[key] = figure(num, key)  end  crr = []  brr.each do |key, value|    crr.push(value / arr[key])  end  crr.minenddef figure(num, item)  num >= item ? num / item + figure(num / item, item) : 0enddef list(base)  i = 2  hash = Hash.new(0)  while (base > 1) do    count = 0    if base % i == 0      while base % i == 0        base /= i        count += 1      end      hash[i] = count    end    i += 1  end  return hashend

待更新。。。。