- Category: Misc
- Score: 496/500
- Solves: 9/286
JeetQode - The World's Leading Online JQ Programming Learning Platform
You don't have to read the source code to solve this challenge. The source code is provided in case that the description isn't clear.
這題是使用 jq 解三個簡單問題就能拿到 flag 的 PPC 題目。輸入一律以 json 表示，然後需要寫個轉換的 jq program 去把輸入轉換成輸出，類似 Online Judge 的一樣有多筆測資去評斷答案是否正確。輸入有限制 512 字元，但是以這邊出的三個題目來說 512 都是相當充裕的限制，基本上可以忽略不計。
輸入是一個字串，輸出是 boolean 代表它是否是迴文而已。
Input: "aba"
Output: true
Input: "peko"
Output: false
解法就是直接比對 reverse 之後是否相同即可:
(.|split("")|reverse|join(""))==.輸入是個 json 的 binary tree，需要將 left 和 right 交換，左右的子樹也要分別 invert 過才行，
Input: {"left": 1, "right": 3}
Output: {"left": 3, "right": 1}
Input: {"left": 1, "right": {"left": 1, "right": 3}}
Output: {"left": {"left": 3, "right": 1}, "right": 1}
做法就是寫個 invert 函數讓它遞迴下去處理即可，和 tree traversal 一樣:
def invert: if type=="number" then . else {left:.right|invert,right:.left|invert} end; .|invert
type和.|type是一樣的意思，單純判斷參數的類型而已
輸入是個精簡過的 Python AST，只包含 +-*/ 的四則運算而已，輸出就是四則運算出來的結果。
Input: {"body": {"left": {"value": 1, "kind": null, "lineno": 1, "col_offset": 0, "end_lineno": 1, "end_col_offset": 1}, "op": "<_ast.Add object at 0x7f0387ccde20>", "right": {"value": 2, "kind": null, "lineno": 1, "col_offset": 2, "end_lineno": 1, "end_col_offset": 3}, "lineno": 1, "col_offset": 0, "end_lineno": 1, "end_col_offset": 3}}
Output: 3
Input: {"body": {"left": {"left": {"value": 8, "kind": null, "lineno": 1, "col_offset": 1, "end_lineno": 1, "end_col_offset": 2}, "op": "<_ast.Mult object at 0x7f20eb76aee0>", "right": {"value": 7, "kind": null, "lineno": 1, "col_offset": 3, "end_lineno": 1, "end_col_offset": 4}, "lineno": 1, "col_offset": 1, "end_lineno": 1, "end_col_offset": 4}, "op": "<_ast.Sub object at 0x7f20eb76ae80>", "right": {"left": {"value": 6, "kind": null, "lineno": 1, "col_offset": 7, "end_lineno": 1, "end_col_offset": 8}, "op": "<_ast.Mult object at 0x7f20eb76aee0>", "right": {"value": 3, "kind": null, "lineno": 1, "col_offset": 9, "end_lineno": 1, "end_col_offset": 10}, "lineno": 1, "col_offset": 7, "end_lineno": 1, "end_col_offset": 10}, "lineno": 1, "col_offset": 0, "end_lineno": 1, "end_col_offset": 11}}
Output: 38
解法其實和前題幾乎一樣，只是要透過 value 判斷是否是數字，不是的話再判斷 op 去檢測 operator 然後做相對應的計算。一樣是用遞迴的方法寫 tree traversal 就很簡單:
def eval: if .|has("value") then .value elif .op|contains("Add") then (.left|eval)+(.right|eval) elif .op|contains("Sub") then (.left|eval)-(.right|eval) elif .op|contains("Mult") then (.left|eval)*(.right|eval) elif .op|contains("Div") then (.left|eval)/(.right|eval) else null end;.body|eval