結果
| 問題 | No.603 hel__world (2) | 
| コンテスト | |
| ユーザー |  gew1fw | 
| 提出日時 | 2025-06-12 19:40:20 | 
| 言語 | PyPy3 (7.3.15) | 
| 結果 | 
                                WA
                                 
                             | 
| 実行時間 | - | 
| コード長 | 1,009 bytes | 
| コンパイル時間 | 168 ms | 
| コンパイル使用メモリ | 82,608 KB | 
| 実行使用メモリ | 141,016 KB | 
| 最終ジャッジ日時 | 2025-06-12 19:40:29 | 
| 合計ジャッジ時間 | 4,154 ms | 
| ジャッジサーバーID (参考情報) | judge5 / judge1 | 
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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()))
T = input().strip()
# Create T_comp by removing consecutive duplicates
prev_char = None
T_comp = []
for c in T:
    if c != prev_char:
        T_comp.append(c)
        prev_char = c
# Count occurrences of each character in T_comp
from collections import defaultdict
char_counts = defaultdict(int)
for c in T_comp:
    char_counts[c] += 1
# Check if all characters in T_comp have sufficient counts in S
result = 1
for char in char_counts:
    required = char_counts[char]
    available = S_counts[ord(char) - ord('a')]
    if available < required:
        print(0)
        exit()
    k = available - required
    base, remainder = divmod(k, required)
    # Calculate contribution (base+1)^(required - remainder) * (base+2)^remainder mod MOD
    part1 = pow(base + 1, required - remainder, MOD)
    part2 = pow(base + 2, remainder, MOD)
    contribution = (part1 * part2) % MOD
    result = (result * contribution) % MOD
print(result)
            
            
            
        