如何找到阶乘之和

Question

使用最小测试，我正在尝试通过此测试，要求获得5!的总和：

def test_sum_of_factorials

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


  assert_equal 153, sum_of_factorials
end

我写了一个通过的解决方案，但是它很长且重复：

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

看来您已获得：

Answer 2

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