結果

問題 No.3042 拡大コピー
コンテスト
ユーザー qwewe
提出日時 2025-05-14 13:13:00
言語 PyPy3
(7.3.15)
結果
WA  
実行時間 -
コード長 2,193 bytes
コンパイル時間 231 ms
コンパイル使用メモリ 82,304 KB
実行使用メモリ 316,844 KB
最終ジャッジ日時 2025-05-14 13:14:01
合計ジャッジ時間 5,711 ms
ジャッジサーバーID
（参考情報） 		judge1 / judge2
このコードへのチャレンジ
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ファイルパターン 結果
sample -- * 1
other WA * 1 TLE * 1 -- * 22
権限があれば一括ダウンロードができます

ソースコード

diff # 

import sys

# Define a function to read input and process
def solve():
    # Read the number of books (N)
    # We don't strictly need N if we read until EOF or use readlines,
    # but reading it first is standard practice.
    # However, for golfing, reading all lines and slicing is shorter.
    
    # Read all lines from standard input starting from the second line
    # (skipping the first line which contains N).
    # map(str.split, ...) applies the split() method to each line read.
    # sys.stdin.readlines() reads all lines, including the trailing newline '\n'.
    # str.split() handles splitting by whitespace and ignores leading/trailing whitespace,
    # effectively removing the newline as well.
    # *L, = ... unpacks the resulting map object into a list L.
    # Each element of L will be a list like ['title', 'id'].
    *L,=map(str.split,sys.stdin.readlines()[1:])

    # Sort the list L in-place.
    # The key for sorting is a lambda function that takes an element x from L
    # (which is a list like ['title', 'id']).
    # The key function returns a tuple: (x[0], x[1][0]).
    # x[0] is the title (string).
    # x[1] is the id (string like "iti", "ni", "san", "yon").
    # x[1][0] is the first character of the id ('i', 'n', 's', 'y').
    # Python sorts tuples lexicographically. It will first compare the titles (x[0]).
    # If the titles are the same, it will compare the second element of the key tuple,
    # which is the first character of the id.
    # The required order for ids is "iti", "ni", "san", "yon".
    # Their first characters are 'i', 'n', 's', 'y'.
    # Since 'i' < 'n' < 's' < 'y' lexicographically, sorting by the first character
    # correctly implements the secondary sorting criterion for the ids.
    L.sort(key=lambda x:(x[0],x[1][0]))

    # Iterate through the sorted list L.
    # Each element `x` is a list containing the title and id (e.g., ['a', 'iti']).
    # `print(*x)` uses the splat operator (*) to unpack the list `x`.
    # `print('a', 'iti')` prints the elements separated by a space, followed by a newline.
    # This matches the required output format.
    for x in L:print(*x)

# Execute the solve function
solve()
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