結果

問題 No.204 ゴールデン・ウィーク(2)
コンテスト
ユーザー lam6er
提出日時 2025-04-16 01:14:02
言語 PyPy3
(7.3.15)
結果
WA  
実行時間 -
コード長 3,263 bytes
コンパイル時間 360 ms
コンパイル使用メモリ 81,724 KB
実行使用メモリ 54,288 KB
最終ジャッジ日時 2025-04-16 01:15:19
合計ジャッジ時間 3,425 ms
ジャッジサーバーID
（参考情報） 		judge4 / 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 * 24 WA * 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 # 

D = int(input())
s1 = input().strip()
s2 = input().strip()
s = s1 + s2

def create_intervals(s):
    if not s:
        return []
    intervals = []
    current_char = s[0]
    start = 0
    for i in range(1, len(s)):
        if s[i] == current_char:
            continue
        else:
            intervals.append((current_char, start, i-1))
            current_char = s[i]
            start = i
    intervals.append((current_char, start, len(s)-1))
    return intervals

intervals = create_intervals(s)

max_o = 0
for interval in intervals:
    if interval[0] == 'o':
        max_o = max(max_o, interval[2] - interval[1] + 1)

if D == 0:
    print(max_o)
else:
    max_candidate = max_o

    # Process x intervals
    for i in range(len(intervals)):
        interval = intervals[i]
        if interval[0] != 'x':
            continue
        x_len = interval[2] - interval[1] + 1
        available = min(D, x_len)
        prev_o = intervals[i-1] if i > 0 and intervals[i-1][0] == 'o' else None
        next_o = intervals[i+1] if i < len(intervals)-1 and intervals[i+1][0] == 'o' else None
        
        if prev_o and next_o:
            total = (prev_o[2] - prev_o[1] + 1) + available + (next_o[2] - next_o[1] + 1)
            max_candidate = max(max_candidate, total)
        elif prev_o:
            total = (prev_o[2] - prev_o[1] + 1) + available
            max_candidate = max(max_candidate, total)
        elif next_o:
            total = available + (next_o[2] - next_o[1] + 1)
            max_candidate = max(max_candidate, total)
        else:
            max_candidate = max(max_candidate, available)

    # Process o intervals
    for i in range(len(intervals)):
        interval = intervals[i]
        if interval[0] != 'o':
            continue
        o_len = interval[2] - interval[1] + 1
        
        # Previous x part
        if i == 0:
            total = D + o_len
            max_candidate = max(max_candidate, total)
        else:
            prev_interval = intervals[i-1]
            if prev_interval[0] == 'x':
                a = prev_interval[2] - prev_interval[1] + 1
                available = min(D, a)
                total = available + o_len
                max_candidate = max(max_candidate, total)
        
        # Next x part
        if i == len(intervals) - 1:
            total = o_len + D
            max_candidate = max(max_candidate, total)
        else:
            next_interval = intervals[i+1]
            if next_interval[0] == 'x':
                b = next_interval[2] - next_interval[1] + 1
                available = min(D, b)
                total = o_len + available
                max_candidate = max(max_candidate, total)

    # Check front and back infinite x parts
    # Front x part
    if intervals[0][0] == 'o':
        total = D + (intervals[0][2] - intervals[0][1] + 1)
        max_candidate = max(max_candidate, total)
    else:
        max_candidate = max(max_candidate, D)
    
    # Back x part
    if intervals[-1][0] == 'o':
        total = (intervals[-1][2] - intervals[-1][1] + 1) + D
        max_candidate = max(max_candidate, total)
    else:
        max_candidate = max(max_candidate, D)
    
    # Check D alone
    max_candidate = max(max_candidate, D)

    print(max_candidate)
0