結果
| 問題 | No.2738 CPC To F | 
| コンテスト | |
| ユーザー |  lam6er | 
| 提出日時 | 2025-03-31 17:35:41 | 
| 言語 | PyPy3 (7.3.15) | 
| 結果 | 
                                WA
                                 
                             | 
| 実行時間 | - | 
| コード長 | 1,056 bytes | 
| コンパイル時間 | 193 ms | 
| コンパイル使用メモリ | 82,128 KB | 
| 実行使用メモリ | 93,216 KB | 
| 最終ジャッジ日時 | 2025-03-31 17:36:27 | 
| 合計ジャッジ時間 | 2,413 ms | 
| ジャッジサーバーID (参考情報) | judge2 / judge3 | 
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| ファイルパターン | 結果 | 
|---|---|
| sample | AC * 2 | 
| other | AC * 5 WA * 17 | 
ソースコード
n = int(input())
s = input().strip()
# Step 1: Count original CPCTF substrings
original_count = 0
for i in range(n - 4):
    if s[i] == 'C' and s[i+1] == 'P' and s[i+2] == 'C' and s[i+3] == 'T' and s[i+4] == 'F':
        original_count += 1
# Step 2: Collect all valid CPC positions that can generate CPCTF when converted
valid_cpc = []
for i in range(n - 2):
    if s[i] == 'C' and s[i+1] == 'P' and s[i+2] == 'C':
        # Check if the previous characters form T followed by CPC
        if i >= 4:
            if (s[i-1] == 'T' and
                s[i-4] == 'C' and
                s[i-3] == 'P' and
                s[i-2] == 'C'):
                valid_cpc.append( (i, i+2) )
# Step 3: Find the maximum non-overlapping intervals using greedy algorithm
conversion_count = 0
if valid_cpc:
    # Sort intervals by their end points
    valid_cpc.sort(key=lambda x: x[1])
    last_end = -1
    for start, end in valid_cpc:
        if start > last_end:
            conversion_count += 1
            last_end = end
print(original_count + conversion_count)
            
            
            
        