迭代数组(Project Euler#23)

Question

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,每当我尝试对我的d数组进行双重迭代时,我都会出现以下错误:

euler23:21:在块中(2级)来自' 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

由于我不熟悉Ruby,我解决这个问题的各种尝试都是徒劳的.我觉得我需要包含一些库,但我的研究没有提到任何库,我不知所措.此代码用于将所有不能写入的数字相加为两个数字的总和; 这是欧拉计划的第二十三个问题.

Answer 1

当你这样做:

d = Array.new(6965)

您创建一个6965 nil值的数组.

如果在第21行之前添加了此测试代码:

p [c,j,k]

然后你得到结果:

[1, nil, nil]

这表明j和k均nil值.您正在迭代数组中的空项.

如果您将创建更改d为:

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

...然后你的代码运行.(我没有让它运行足够长的时间来查看它是否正确运行,但它至少在相当长的一段时间内没有错误地运行.)

最后,几点随机风格的建议:

这段代码:

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

可以更简洁地重写,更像Ruby-esque:

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

当你问打破了一个循环的,你可以使用break或next适当,或throw与catch首位,而其不用于Ruby-错误处理跳转到一个特定的嵌套循环水平.