結果
| 問題 |
No.2019 Digits Filling for All Substrings
|
| コンテスト | |
| ユーザー |
gew1fw
|
| 提出日時 | 2025-06-12 14:07:24 |
| 言語 | PyPy3 (7.3.15) |
| 結果 |
AC
|
| 実行時間 | 89 ms / 2,000 ms |
| コード長 | 1,391 bytes |
| コンパイル時間 | 201 ms |
| コンパイル使用メモリ | 82,216 KB |
| 実行使用メモリ | 77,364 KB |
| 最終ジャッジ日時 | 2025-06-12 14:07:57 |
| 合計ジャッジ時間 | 3,313 ms |
|
ジャッジサーバーID (参考情報) |
judge3 / judge2 |
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| ファイルパターン | 結果 |
|---|---|
| sample | AC * 4 |
| other | AC * 30 |
ソースコード
MOD = 998244353
def main():
import sys
input = sys.stdin.read().split()
N = int(input[0])
S = input[1]
cnt = [4, 3, 3] # cnt[0] = 4 digits (0,3,6,9), cnt[1] = 3 (1,4,7), cnt[2] = 3 (2,5,8)
prev_dp = [0, 0, 0]
total = 0
for c in S:
current_dp = [0, 0, 0]
if c == '?':
# For each previous remainder, distribute to new remainders based on t
for r_prev in range(3):
for t in range(3):
new_r = (r_prev + t) % 3
current_dp[new_r] = (current_dp[new_r] + prev_dp[r_prev] * cnt[t]) % MOD
# Add the single character case: all possible digits
for r in range(3):
current_dp[r] = (current_dp[r] + cnt[r]) % MOD
else:
d = int(c)
for r_prev in range(3):
new_r = (r_prev + d) % 3
current_dp[new_r] = (current_dp[new_r] + prev_dp[r_prev]) % MOD
# Add single character case: d mod 3
new_r = d % 3
current_dp[new_r] = (current_dp[new_r] + 1) % MOD
# Add the number of valid substrings ending at this position to the total
total = (total + current_dp[0]) % MOD
# Update previous DP
prev_dp = current_dp.copy()
print(total % MOD)
if __name__ == '__main__':
main()
gew1fw