Question

如何使用Ruby如下打印矩形：

* = = = *
* = = = *
* * * * *
* = = = *
* = = = *

在这种情况下，行和列的长度相同，并且必须为奇数。

示例：

r =矩形（5）

应打印：

* = = = *
* = = = *
* * * * *
* = = = *
* = = = *

，如果：

r =矩形（7）

应打印：

* = = = = = *
* = = = = = *
* * * * * * *
* = = = = = *
* = = = = = *
* * * * * * *
* = = = = = *

谢谢。

Answer 1

我将从创建以下两行开始：（有很多方法可以实现这一点）

size = 7

a = Array.new(size, '*').fill('=', 1..-2).join(' ') #=> "* = = = = = *"
b = Array.new(size, '*').join(' ')                  #=> "* * * * * * *"

然后我要定义一个重复模式：

pattern = [a, a, b].cycle

最后，我将打印图案 size 次：

puts pattern.take(size)
* = = = = = *
* = = = = = *
* * * * * * *
* = = = = = *
* = = = = = *
* * * * * * *
* = = = = = *

Answer 2

[我自己解决]：D

def square(n)
  begin
    if n.odd?
      1.upto(n) do | row |
        if row % 3 != 0
          puts "#{'*'} #{'= ' * (n - 2)}#{'*'}"
        else
          puts "#{'* ' * n}"
        end
      end
    else
      puts 'Must odd number!'
    end
  rescue
    puts 'Must integer number!'
  end
end

square(5) 输出：

* = = = *
* = = = *
* * * * *
* = = = *
* = = = *

square(7) 输出：

* = = = = = *
* = = = = = *
* * * * * * *
* = = = = = *
* = = = = = *
* * * * * * *
* = = = = = *

square(8) 输出：

Must odd number!

square(8.5) 输出：

Must integer number!

square('blabla') 输出：

Must integer number!

Answer 3

def rectangle(n)
  puts("-----------------For #{n}--------------------")
  if n % 2 == 1
    for i in (1..n)
      for j in (1..n)
        if j == 1 || j == n || 0 == i % 3
          print "* "
        else
          print "= "
        end
      end
      print("\n")
    end
  end
end

rectangle(3)
rectangle(5)
rectangle(7)
rectangle(9)
rectangle(11)

-----------------For 3--------------------
* = * 
* = * 
* * * 

-----------------For 5--------------------
* = = = * 
* = = = * 
* * * * * 
* = = = * 
* = = = * 

-----------------For 7--------------------
* = = = = = * 
* = = = = = * 
* * * * * * * 
* = = = = = * 
* = = = = = * 
* * * * * * * 
* = = = = = * 

-----------------For 9--------------------
* = = = = = = = * 
* = = = = = = = * 
* * * * * * * * * 
* = = = = = = = * 
* = = = = = = = * 
* * * * * * * * * 
* = = = = = = = * 
* = = = = = = = * 
* * * * * * * * * 

-----------------For 11--------------------
* = = = = = = = = = * 
* = = = = = = = = = * 
* * * * * * * * * * * 
* = = = = = = = = = * 
* = = = = = = = = = * 
* * * * * * * * * * * 
* = = = = = = = = = * 
* = = = = = = = = = * 
* * * * * * * * * * * 
* = = = = = = = = = * 
* = = = = = = = = = *

Answer 4

只是为了好玩修改@spn答案

def rectangle(count)
  return 'Must odd number more than 1' unless count.is_a?(Integer) && count.odd? && count > 1
  Array.new(count) { |index| (index + 1) % 3 == 0 ?
                               "#{'* ' * count}".chomp(" ") :
                               "#{'*'} #{'= ' * (count - 2)}#{'*'}" }.join("\n")
end

现在

puts rectangle(2) # will print Must odd number more than 1
puts rectangle("asdf") # will print Must odd number more than 1

puts rectangle(9) # will print:

* = = = = = = = *
* = = = = = = = *
* * * * * * * * *
* = = = = = = = *
* = = = = = = = *
* * * * * * * * *
* = = = = = = = *
* = = = = = = = *
* * * * * * * * *

最好避免使用方法中的puts。因此，您可以在网络，电报机器人等中再次使用它们。同样也不用用自己（但相同）的消息来复制Ruby异常的方法。

Answer 5

我们首先定义一种构造两种类型的线的方法。

def make_line(n, mid_char)
  ['*', *[mid_char]*(n-2), '*'].join(' ')
end

make_line(5, '*') #=> "* * * * *" 
make_line(5, '#') #=> "* # # # *"

现在创建一种方法以所需的图案绘制线条。仅当i等于零时，第(i+1) % 3行（基数为零）由星和空格组成；否则它也是包含英镑符号的行。

def draw(n)
  all_stars = make_line(n, '*')
  two_stars = make_line(n, '#')  
  n.times { |i| puts ((i+1) % 3).zero? ? all_stars : two_stars }
end

draw 5
* # # # *
* # # # *
* * * * *
* # # # *
* # # # *
draw 6
* # # # # *
* # # # # *
* * * * * *
* # # # # *
* # # # # *
* * * * * *
draw 7
* # # # # # *
* # # # # # *
* * * * * * *
* # # # # # *
* # # # # # *
* * * * * * *
* # # # # # *

用Ruby打印矩形

5 个答案: