如何对每个元素进行数组和操作?

Here is what I'm doing:

下面是我所做的:

(1..999).each do |a|
    (1..999).each do |b|
      if Math.sqrt(a**2 + b**2) % 1 == 0 && a + b + Math.sqrt(a**2 + b**2) == 1000 && a >= b
            puts a * b * Math.sqrt(a**2 + b**2) 
        end     
    end
end

What is happening is that a and b are interchangeable in the formulas so there are two matches and thus puts gets outputted twice. To fix this, I added a >= b and now it only gets outputted once. But, if a == bit outputs it twice. I know that a and b will always be different in the example I'm using, but this seems like bad design to me.

现在的情况是a和b在公式中是可以互换的，所以有两个匹配，因此输出两次。为了解决这个问题，我添加了>= b，现在只输出了一次。但是，如果a ==位输出两次。我知道在我使用的例子中a和b总是不同的，但这对我来说似乎是糟糕的设计。

Two questions:

两个问题:

1. Is there a better pattern in Ruby for taking an array and comparing it to it self?

   Ruby中是否有更好的模式来获取数组并将其与它自己进行比较?

2. How can I avoid it outputting twice always. I could set a variable that if changed before the start of the next loop would break out. Is that the proper way to do this?

   我怎样才能避免它两次被输出。我可以设置一个变量，如果在下一个循环开始之前发生了变化。这样做合适吗?

1 个解决方案

#1

#using combination
(1..999).to_a.combination(2).each do |low, high|
  if Math.sqrt(low**2 + high**2) % 1 == 0 && low + hight + Math.sqrt(low**2 + high**2) == 1000
    puts low * high * Math.sqrt(low**2 + high**2)
  end
end

Edited to use a little better practice (each block with first and second for arrays of arrays)

编辑后使用了一个更好的实践(数组数组的每个块都有1和2)