1

我有以下代码

#!/usr/bin/ruby -w
c = 1
d = Array.new(6965)  #6965 is the amount of abundant numbers below 28123 of which all numbers greater than that can be written as the sum of two abundant numbers
f = 0
while c < 28124      # no need to go beyond 28123 for this problem
  a = 0
  b = 1
  i = true           # this will be set to false if a number can be written as the sum of two abundant numbers
  while b <= c/2 + 1 # checks will go until they reach just over half of a number
    if c % b == 0    # checks for integer divisors
      a += b         # sums integer divisors
    end
    b += 1           # iterates to check for new divisor
  end
  if a > c           # checks to see if sum of divisors is greater than the original number
    d << c           # if true it is read into an array
  end
  d.each{|j|         # iterates through array
    d.each{|k|       # iterates through iterations to check all possible sums for number
                     # false is declared if a match is found. does ruby have and exit statement i could use here?
      i = false if c - j - k == 0
    }
  }
  c+=1               # number that we are checking is increased by one
                     # if a number cannot be found as the sum of two abundant number it is summed into f
  f += c if i == true
end
puts f

对于以下代码，每当我尝试对d数组进行双重迭代时，都会出现以下错误：

euler23:21:in block (2 levels) in ' from euler23:20:in block in ' from euler23:19:in ' -': nil can't be coerced into Fixnum (TypeError)
from euler23:21:in
each'
from euler23:20:in
each'
from euler23:19:in

由于我对 Ruby 不熟悉，因此我为解决此问题所做的各种尝试都是徒劳的。我感觉有些库我需要包含，但我的研究没有提到任何库，我很茫然。这段代码旨在将所有不能写成两个丰富数字之和的数字相加；这是欧拉计划的第二十三道题。

4

2 回答 2

6

当你这样做时：

d = Array.new(6965)

nil您创建一个包含 6965 个值的数组。

如果在第 21 行之前添加此测试代码：

p [c,j,k]

然后你得到结果：

[1, nil, nil]

这表明j和k都是nil值。您正在遍历数组中的空项目。

如果您将创建更改d为：

d = [] # an empty array, which in Ruby can change size whenever you want

...然后您的代码运行。（我没有让它运行足够长的时间来查看它是否正确运行，但它至少运行了相当长一段时间没有错误。）

最后，一些随机风格的建议：

这段代码：

while b <= c/2 + 1
  if c % b == 0
    a += b
  end
  b += 1
end

可以更简洁和更 Ruby-esque 重写为：

b.upto(c/2+1){ a+=b if c%b==0 }

同样，这个循环：

c=1
while c < 28124
  # ...
  c += 1
end

可以改写为：

1.upto(28123) do |c|
  # ...
end

当您询问是否要跳出循环时，您可以根据需要使用breakornext或throwandcatch（在 Ruby 中不用于错误处理）来跳转到特定的嵌套循环级别。

于 2013-04-02T16:33:29.223 回答
3

下面的代码有问题：

d.each{|j|                     
d.each{ |k|             
p c,j,k  #1,nil,nil
i = false if c - j - k == 0 }}

因为：

1 - nil - nil
#TypeError: nil can't be coerced into Fixnum
#      from (irb):2:in `-'
#      from (irb):2
#     from C:/Ruby193/bin/irb:12:in `<main>'
于 2013-04-02T16:33:42.977 回答