如何找到阶乘的总和

Question

使用 minitest，我试图让这个测试通过，它要求 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}

有没有人愿意解释更清晰、更简单的解决方案？

Answer 1

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

生产

sum_of_factorials(5) #=> 153

解释：

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 2

既然可以写5!+4!+3!+2!+1!，一般是1!+2!+3!+...+n!作为 1 + 2(1 + 3(1 + 4(1 + 5(1+...(1+n)))))，你可以在 O(n)

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

Answer 3

您似乎得到了：

def test_sum_of_factorials
  sum_of_factorials = 0
  numbers = [1, 2, 3, 4, 5]
  <...missing bits...>
  assert_equal 153, sum_of_factorials
end

并被要求填写缺失的位。我认为正在要求类似以下内容。

def test_sum_of_factorials
  sum_of_factorials = 0
  numbers = [1, 2, 3, 4, 5]
  fac = 1
  numbers.each do |n|
    fac *= n
    sum_of_factorials += fac
  end
  assert_equal 153, sum_of_factorials
end

我们可以这样写：

def test_sum_of_factorials
  numbers = [1, 2, 3, 4, 5]      
  assert_equal 153, sum_of_factorials(numbers)
end

def sum_of_factorials(numbers)
  fac_sum = 0
  fac = 1
  numbers.each do |n|
    fac *= n
    fac_sum += fac
  end
  fac_sum
end

哪里

sum_of_factorials([1,2,3,4,5])
  #=> 153

要多Ruby-like，不过，用Array#sum写sum_of_factorials如下：

def sum_of_factorials(numbers)
  fac = 1
  numbers.sum { |n| fac *= n }
end

连同您的问题标题，这就是为什么其他答案的作者假设您在询问如何改进方法 sum_of_factorials 的原因。首先，它可以传递参数 numbers.max，而不是数组 numbers.

另一种写 sum_of_factorials 的方法是使用方法 Enumerator::produce，它在 v2.7 中首次亮相。

def sum_of_factorials(n)
  enum = Enumerator.produce([1,1]) { |n0, n1| [n0+1, (n0+1)*n1] }
  n.times.sum { enum.next.last }
end

(1..8).each { |n| puts "%d: %6d" % [n, sum_of_factorials(n)] }
1:      1
2:      3
3:      9
4:     33
5:    153
6:    873
7:   5913
8:  46233

注意如果：

enum = Enumerator.produce([1,1]) { |n0, n1| [n0+1, (n0+1)*n1] }
  #=> #<Enumerator: #<Enumerator::Producer:0x000059d490c742a0>:each>

然后

enum.next #=> [1, 1] 
enum.next #=> [2, 2] 
enum.next #=> [3, 6] 
enum.next #=> [4, 24]

所以 1 到 4 的阶乘由下式给出（在重新定义或倒带 enum 之后）：

enum.next.last #=>  1 
enum.next.last #=>  2 
enum.next.last #=>  6 
enum.next.last #=> 24

如果 n 可以等于零，请在方法开头添加行 return 1 if n.zero?。

Answer 4

另一种选择是定义阶乘函数（此处被盗：），可能作为 Integer class:

的补丁

module MyMathFunctions
  def factorial
    (1..self).reduce(1, :*)
  end
end

Integer.include MyMathFunctions

所以你可以这样使用它：10.factorial #=> 3628800

然后，调用：

[1, 2, 3, 4, 5].sum &:factorial
#=> 153

Answer 5

我想简单的如下，

[1, 2, 3, 4, 5].map { |x| (1..x).inject(1, :*) }.sum
# => 153

如何找到阶乘的总和

How to find the sum of factorials

ruby

minitest