GolfScript 54 53 52
Chỉnh sửa 1:
Tôi chỉ phát hiện ra một lỗi trong mã. Nó không phát hiện các thẻ trùng lặp nếu các bản sao là hai đầu tiên trong đầu vào (vì tôi đang sử dụng *toán tử gấp chứ không phải /mỗi toán tử cho vòng lặp đầu tiên).
Bây giờ tôi đã sửa mã và cũng quản lý để loại bỏ 1 char trong quá trình này. Đây là phiên bản mới:
' '/{1$1$?){]?}{\+}if}/2%{"UOK0D"\?).0>+.4>5*+}%{+}*
Đầu vào phải nằm trên ngăn xếp dưới dạng chuỗi, theo định dạng đã chỉ định (ví dụ '7A UA DA':).
Trong trường hợp đầu vào hợp lệ, chương trình sẽ in tổng giá trị của thẻ.
Trong trường hợp có ít nhất một thẻ trùng lặp, chương trình sẽ ném ngoại lệ sau:
(eval):1:in `block in initialize': undefined method `class_id' for nil:NilClass (NoMethodError)
Chỉnh sửa 2:
Sau khi thấy bài đăng này trên trang web meta , tôi quyết định đăng một mô tả về mã. Điều này cũng giúp tôi tìm và sửa lỗi. Vì vậy, ở đây đi:
# Initially, we epect the input string to be on the stack
# Example: "7A UA DA"
' '/ # split the input string by spaces
# now we have on the stack an array of strings
# (in our example: ["7A" "UA" "DA"])
# {1$1$?)!{\+}{]?}if}/ -> this piece of code checks for duplicate cards
#
# The trailing symbol (/) is the 'each' operator, meaning that the
# preceding code block (enclosed in curly brackets) will be executed
# for every cards in the previous array.
#
# Before each execution of the code block, the current card value
# is pushed on the stack.
#
# Basically what this code does is concatenate cards into a string
# and checks whether the current card is contained in the accumulated
# value.
#
# So, for each card, this is what we execute:
1$ # copies the concatenated string on top of the stack
# (initially this is an empty string)
1$ # copies the current card on top of the stack
? # returns (places on the stack) the 0-based index where
# the current card is found in the concatenated string
# or -1 if not found
) # increments the topmost stack value
# Now we have 0 if the card is not a duplicate
# or a value greater than 0 otherwise
{]?}{\+}if # if the current stack value is non-0 (duplicate)
# then execute the first code {]?} (generates an error)
# Otherwise, if the card is valid, execute the {\+} block.
# What this code does is essentially concatenate the current
# card value:
# \ -> swaps the two topmost stack values; now we have
# the concatenated string and the current card value
# + -> this is the concatenation operator
# After the previous code block finishes execution (in case the input is)
# valid, we end up having the concatenated card values on the stack
# In our example, this value is "DAUAUB7A".
# The next code fragment is the one that computes the card values
# This is the code: 2%{"UOK0D"\?).0>+.4>5*+}%{+}*
# And this is how it can be broken down:
2% # takes only the even indexed chars from the existing string
# in our case, "DAUA7A" -> "DU7"
# Only these characters are important for determining the
# card values.
# The following piece of code is:
# {"UOK0D"\?).0>+.4>5*+}%
# This code performs a map; it takes the individual chars,
# computes the corresponding numeric value for each of them and outputs an
# array containing those values
# This is achieved using the map operator (%) which evaluates the preceding
# code block, delimited by curly braces, so, essentially this is the code that
# computes the value for a card:
# "UOK0D"\?).0>+.4>5*+
# It can be broken down like this:
"UOK0D" # pushes the "UOK0D" string on the stack
\ # swaps the two topmost stack values
# Now, these values are: "UOK0D" and "l"
# (where "l" represents the current letter
# that gives the card its value: U,O,K,0,D,7,8...)
? # Find the index of the card's letter in the
# "UOK0D" string.
# Remember, this is 0-based index, or -1 if not found.
) # increment the index value
# Now we have the following value:
# 1 if the card is U
# 2 if the card is O
# 3 if the card is K
# 4 if the card is 0
# 5 if the card is D
# 0 if it is anything else
.0>+ # if the current value is greater than 0,
# add 1 to it.
.4>5*+ # if the current value is greater than 4,
# add 5 to it.
# Passing through these steps, we now have the following value:
# 2 if the card is U
# 3 if the card is O
# 4 if the card is K
# 10 if the card is 0
# 11 if the card is D
# 0 if it is anything else
# This is the exact value we were looking for.
# Now we have an array containing the value of each card.
# in our example, [0, 2, 11]
# The next piece of code is easy:
{+}* # uses the * (fold) operator to add up all the
# values in the array.
# This leaves the total value of the cards on the stack,
# which is exactly what we were looking for (0+2+11=13).
# Golfscript is awesome! :-)