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 14:46:22 |
| 言語 | PyPy3 (7.3.15) |
| 結果 |
WA
|
| 実行時間 | - |
| コード長 | 1,252 bytes |
| コンパイル時間 | 398 ms |
| コンパイル使用メモリ | 82,480 KB |
| 実行使用メモリ | 142,128 KB |
| 最終ジャッジ日時 | 2025-06-12 14:48:05 |
| 合計ジャッジ時間 | 4,314 ms |
|
ジャッジサーバーID (参考情報) |
judge5 / judge4 |
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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_alpha = list(map(int, input().split()))
T = input().strip()
# Handle empty T (though problem states T is at least length 1)
if not T:
print(0)
exit()
# Compress T by removing consecutive duplicates
t_compressed = [T[0]]
for c in T[1:]:
if c != t_compressed[-1]:
t_compressed.append(c)
# Check each character in compressed T has at least 1 available
for c in t_compressed:
idx = ord(c) - ord('a')
if s_alpha[idx] < 1:
print(0)
exit()
# Count occurrences of each character in compressed T
from collections import defaultdict
char_counts = defaultdict(int)
for c in t_compressed:
char_counts[c] += 1
# Check if each character has enough total occurrences for its groups
for c in char_counts:
idx = ord(c) - ord('a')
required = char_counts[c]
if s_alpha[idx] < required:
print(0)
exit()
# Calculate the product of contributions for each character
result = 1
for c in char_counts:
cnt = char_counts[c]
idx = ord(c) - ord('a')
s = s_alpha[idx]
q, r = divmod(s, cnt)
part1 = pow(q + 1, r, MOD)
part2 = pow(q, cnt - r, MOD)
contribution = (part1 * part2) % MOD
result = (result * contribution) % MOD
print(result)
gew1fw