結果

問題 No.1292 パタパタ三角形
コンテスト
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
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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 #
raw source code 

# coding:UTF-8
import sys

MOD = 10 ** 9 + 7
INF = float('inf')

S = input()     # 文字列

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)))
0