結果
| 問題 | 
                            No.2131 Concon Substrings (COuNt Version)
                             | 
                    
| コンテスト | |
| ユーザー | 
                             lam6er
                         | 
                    
| 提出日時 | 2025-03-20 20:52:14 | 
| 言語 | PyPy3  (7.3.15)  | 
                    
| 結果 | 
                             
                                AC
                                 
                             
                            
                         | 
                    
| 実行時間 | 634 ms / 2,000 ms | 
| コード長 | 1,763 bytes | 
| コンパイル時間 | 177 ms | 
| コンパイル使用メモリ | 81,992 KB | 
| 実行使用メモリ | 83,556 KB | 
| 最終ジャッジ日時 | 2025-03-20 20:52:31 | 
| 合計ジャッジ時間 | 5,163 ms | 
| 
                            ジャッジサーバーID (参考情報)  | 
                        judge2 / judge5 | 
(要ログイン)
| ファイルパターン | 結果 | 
|---|---|
| sample | AC * 4 | 
| other | AC * 16 | 
ソースコード
import sys
from collections import defaultdict
MOD = 998244353
def main():
    N = int(sys.stdin.readline())
    prev_dp = [defaultdict(int) for _ in range(3)]
    prev_dp[0][0] = 1  # Initial state: 0 chars processed, state 0, k=0
    for _ in range(N):
        next_dp = [defaultdict(int) for _ in range(3)]
        for s in range(3):
            for k in prev_dp[s]:
                count = prev_dp[s][k]
                if count == 0:
                    continue
                if s == 0:
                    # Transition for choosing 'c'
                    next_s = 1
                    next_k = k
                    next_dp[next_s][next_k] = (next_dp[next_s][next_k] + count) % MOD
                    # Transition for others (25 possibilities)
                    next_dp[0][k] = (next_dp[0][k] + count * 25) % MOD
                elif s == 1:
                    # Transition for choosing 'o'
                    next_s = 2
                    next_k = k
                    next_dp[next_s][next_k] = (next_dp[next_s][next_k] + count) % MOD
                    # Transition for others (25 possibilities)
                    next_dp[1][k] = (next_dp[1][k] + count * 25) % MOD
                elif s == 2:
                    # Transition for choosing 'n'
                    next_s = 0
                    next_k = k + 1
                    next_dp[next_s][next_k] = (next_dp[next_s][next_k] + count) % MOD
                    # Transition for others (25 possibilities)
                    next_dp[2][k] = (next_dp[2][k] + count * 25) % MOD
        prev_dp = next_dp
    result = 0
    for s in range(3):
        for k in prev_dp[s]:
            result = (result + k * prev_dp[s][k]) % MOD
    print(result)
if __name__ == "__main__":
    main()
            
            
            
        
            
lam6er