是否有更好的方法来查找阶乘之和？

Question

使用最小测试，我正在尝试使测试波纹管通过，要求总和为5！

def test_sum_of_factorials

  sum_of_factorials = 0
  numbers = [1, 2, 3, 4, 5]


  assert_equal 153, sum_of_factorials
end

我已经写了一个合格的解决方案，但是它又长又重复​​：

    fractorial_5 = numbers.inject(1) {|aggregate, num| aggregate * num}
    fractorial_4 = numbers[0..3].inject(1) {|aggregate, num| aggregate * num}
    fractorial_3 = numbers[0..2].inject(1) {|aggregate, num| aggregate * num}
    fractorial_2 = numbers[0..1].inject(1) {|aggregate, num| aggregate * num}
    fractorial_1 = 1

    fractorials_array = [fractorial_1, fractorial_2, fractorial_3, fractorial_4, fractorial_5]

    fractorials_array.each {|fractorial| sum_of_factorials += fractorial}

有人愿意解释的更干净，更简单的解决方案吗？

Answer 1

我会这样写：

(1..5).sum{ |n| (1..n).inject(1, :*) }
#=> 153

Answer 2

def sum_of_factorials(n)
  (1..n).reduce([1, 0]) { |(f, sum), e| [(f *= e), (sum + f)] }.last
end

产生

pry(main)> sum_of_factorials(5)
=> 153

说明：

def sum_of_factorials(n)
  (1..n).             # n times
    reduce([1, 0]) do # iterate and use [factorial, sum] as an accumulator
    |(f, sum), e|     # splat the accumulator into f and sum variables
      [               # each time recalculate the accumulator as
        f *= e,       # product of the element and previous factorial,
        sum + f       # sum of current factorial and previous sum
      ]
    end.last          # and return only last part(sum) of the accumulator
end

Answer 3

由于您可以写5！+4！+3！+2！+1 !，通常为1！+2！+3！+ ... + n！作为1 + 2（1 + 3（1 + 4（1 + 5（1 + ...（1 + n）））））），您可以在O（n）]中执行此操作

sum_of_factorials = 1
numbers.reverse_each { |i| sum_of_factorials = 1 + i * sum_of_factorials }