爱测试码高尔夫

Question

创建一个程序，计算两个名字共有的字母总数，并找到它们长度的乘积，作为一个“爱的测试者”。

条件:你可能得不到1:1的答案(3比3，等等)输出。

输入

两个名字来自STDIN或最近的替代。

输出

将x计算为两个名称之间共有的字母总数，忽略大小写。将y计算为名称长度的乘积。然后输出，到STDOUT或最近的替代，是

Name1 and Name2 have x out of y chances of love.

示例

输入：

Wesley
Polly

输出：

Wesley and Polly have 2 out of 30 chances of love.

卫斯理和波莉有两个共同的字母，y和l，其长度的乘积为6*5= 30。

输入：

Bill
Jill

输出：

Bill and Jill have 3 out of 16 chances of love.

奖金

• 使用简化的分数减去30个字节，即x out of y是完全缩减的形式。

评分

这是密码-高尔夫答案将以字节为单位得分，而更少的字节更好。

Answer 1

Pyth，40字节

jd[z"and"Jw"have"/K-lzl.-rz0rJ0i=J*lzlJK"out of"/JiJK"chances of love.

该代码有70字节长，并有资格获得-30字节加值。

在网上试试。

  [                        Begin an array and fill it with the following:
   z                         The first line of input.
   "and"                     That string.
   Jw                        The second line of input, saved in J.
   "have"                    That string.
           rz0rJ0              Convert both lines to lowercase.
         .-                    Remove the characters form the second string
                               from the first, counting multiplicities.
        l                      Get the length.
     -lz                       Subtract it from the length of the first line.
    K                          Save in K.
                  =J*lzlJ      Save the lines' lengths' product in J.
                 i       K     Compute the GCD of J and K.
   /                         The quotient of K and the GCD.
   "out of"                  That string.
   /JiJK                     The quotient of J and the GCD of J and K.
   "chances of love.         That string.
jd                         Join the elements, separated by spaces.

Answer 2

CJam，55字节

ll]_~@_{'~,\elfe=}/.e<:+\:,:*]_2>~{_@\%}h;f/"and
have
out of
chances of love."N/.{}S*

代码长85字节，并有资格获得-30字节加值。

在CJam解释器网上试一试。

是如何工作的

ll]      e# Read two lines and wrap them in an array.
_~       e# Copy the array and dump the lines on the stack.
@_       e# Rotate the array on top and push a copy.
{        e# For each line of the copy.
  '~,    e#   Push the array of all ASCII charcters up to '}'.
  \el    e#   Convert the line to lowercase.
  fe=    e#   Count the occurrences of each character in the line.
}/       e#
.e<      e# Vectorized minimum of the occurrences.
:+       e# Add to find the number of shared charaters.
\:,      e# Compute the length of each line.
:*       e# Push the product.
]_       e# Wrap the stack in an array and push a copy.
2>~      e# Discard the lines of the copy and dump the calculated integers.
{_@\%h}; e# Compute their GCD, using the Euclidean algorithm.
f/       e# Divide each element of the array by the GCD.
         e# This also splits the names, which won't affect printing.

"and
have
out of
chances of love."

N/       e# Split the above string at linefeeds.
.{}      e# Vectorized no-op. Interleaves the arrays.
S*       e# Join the results elements, separated by spaces.

Answer 3

斯塔克斯，42-30= 12 字节数

±à>mª≈◄ΣÇyⁿ`à>²CΔ7ù«▀╦X╠;7╣ⁿ¡♣ë╩■B§F»é▲Y∩«

运行并调试它

分数是很棒的。