恨/爱的难题

Question

挑战描述

在这个挑战中，我们只考虑love和hate作为感觉。如果我们想表达顺序N的情感表达，我们可以在这两者之间交替使用(从hate开始)：

order | expression

1       I hate it.
2       I hate that I love it.
3       I hate that I love that I hate it.
4       I hate that I love that I hate that I love it.

对于每个正整数N，模式如下。给定N，输出顺序N的相关情感表达式。

Notes

• 表达式末尾的句号(.)是强制性的，
• 允许尾随和引导空格(包括换行符)，
• 非正或非整数N的输出未定义，
• 这是一个密码-高尔夫挑战，所以让您的代码尽可能短！

Answer 1

Python，54字节

lambda n:("I hate that I love that "*n)[:12*n-5]+"it."

• Python2：把它放进去！
• Python3：把它放进去！

Answer 2

CJam，36字节

ri{"hatlov"3/='IS@"e that "}/2<"it."

在网上试试！

解释

ri            e# Read input and convert to integer N.
{             e# For each i from 0 to N-1...
  "hatlov"3/  e#   Push ["hat" "lov"].
  =           e#   Use i as a cyclic index into this array to alternate between
              e#   "hat" and "lov".
  'IS         e#   Push character 'I' and a space.
  @           e#   Pull "hat" or "lov" on top of it.
  "e that "   e#   Push "e that "
}/
2<            e#   Truncate the last "e that " to just "e ".
"it."         e#   Push "it."

Answer 3

Java 8，91个字节

i->{for(int j=0;j++<i;)System.out.printf("I %se %s",j%2>0?"hat":"lov",j<i?"that ":"it.");};

未高尔夫球测试程序

public static void main(String[] args) {
    Consumer<Integer> c = i -> {
        for (int j = 0; j++ < i;) {
            System.out.printf("I %se %s", j % 2 > 0 ? "hat" : "lov", j < i ? "that " : "it.");
        }
    };

    c.accept(1);
    c.accept(2);
    c.accept(3);
}