結果
問題 |
No.603 hel__world (2)
|
ユーザー |
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提出日時 | 2025-03-31 17:33:10 |
言語 | PyPy3 (7.3.15) |
結果 |
WA
|
実行時間 | - |
コード長 | 1,021 bytes |
コンパイル時間 | 188 ms |
コンパイル使用メモリ | 82,348 KB |
実行使用メモリ | 141,172 KB |
最終ジャッジ日時 | 2025-03-31 17:33:44 |
合計ジャッジ時間 | 4,202 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 * 3 |
other | AC * 10 WA * 20 |
ソースコード
MOD = 10**6 + 3 # Read input s_counts = list(map(int, input().split())) s = {chr(ord('a') + i): s_counts[i] for i in range(26)} T = input().strip() # Process T to get runs and count occurrences per character Run_T = [] prev_char = None for c in T: if c != prev_char: Run_T.append(c) prev_char = c from collections import defaultdict count_runs = defaultdict(int) for c in Run_T: count_runs[c] += 1 # Check if any character in T's runs has insufficient count in S for c, m in count_runs.items(): if s[c] < m: print(0) exit() result = 1 # Calculate contribution for each character in the runs for c, m in count_runs.items(): sc = s[c] k = sc - m # Remaining characters after accounting for minimum needed per run base = k // m rem = k % m a = (base + 2) % MOD b = (base + 1) % MOD # Compute (a^rem * b^(m - rem)) mod MOD term = (pow(a, rem, MOD) * pow(b, m - rem, MOD)) % MOD result = (result * term) % MOD print(result)