結果

問題 No.1292 パタパタ三角形
ユーザー manini
提出日時 2021-01-25 17:58:38
言語 PyPy3
(7.3.15)
結果
AC  
実行時間 173 ms / 2,000 ms
コード長 1,393 bytes
コンパイル時間 363 ms
コンパイル使用メモリ 82,432 KB
実行使用メモリ 107,588 KB
最終ジャッジ日時 2024-06-22 10:53:54
合計ジャッジ時間 3,000 ms
ジャッジサーバーID
(参考情報)
judge2 / judge5
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 AC * 3
other AC * 14
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 #
プレゼンテーションモードにする

# coding:UTF-8
import sys
MOD = 10 ** 9 + 7
INF = float('inf')
S = input() #
mp = {(0, 0)}
pos = [0, 0]
mode = [0, 0]
for s in S:
if (mode == [0, 0] and s == "c") or (mode == [0, 1] and s == "a") or (mode == [0, 2] and s == "b"):
pos[1] += 1
mode[0] = 1
mp.add((pos[0], pos[1]))
elif (mode == [0, 0] and s == "b") or (mode == [0, 1] and s == "c") or (mode == [0, 2] and s == "a"):
pos[0] += 1
mode[0] = 1
mode[1] = (mode[1] + 1) % 3
mp.add((pos[0], pos[1]))
elif (mode == [0, 0] and s == "a") or (mode == [0, 1] and s == "b") or (mode == [0, 2] and s == "c"):
pos[0] -= 1
mode[0] = 1
mode[1] = (mode[1] - 1) % 3
mp.add((pos[0], pos[1]))
elif (mode == [1, 0] and s == "c") or (mode == [1, 1] and s == "a") or (mode == [1, 2] and s == "b"):
pos[1] -= 1
mode[0] = 0
mp.add((pos[0], pos[1]))
elif (mode == [1, 0] and s == "b") or (mode == [1, 1] and s == "c") or (mode == [1, 2] and s == "a"):
pos[0] += 1
mode[0] = 0
mode[1] = (mode[1] + 1) % 3
mp.add((pos[0], pos[1]))
elif (mode == [1, 0] and s == "a") or (mode == [1, 1] and s == "b") or (mode == [1, 2] and s == "c"):
pos[0] -= 1
mode[0] = 0
mode[1] = (mode[1] - 1) % 3
mp.add((pos[0], pos[1]))
print("{}".format(len(mp)))
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