結果
問題 |
No.1481 Rotation ABC
|
ユーザー |
![]() |
提出日時 | 2025-06-12 15:48:57 |
言語 | PyPy3 (7.3.15) |
結果 |
TLE
|
実行時間 | - |
コード長 | 2,185 bytes |
コンパイル時間 | 351 ms |
コンパイル使用メモリ | 83,036 KB |
実行使用メモリ | 77,248 KB |
最終ジャッジ日時 | 2025-06-12 15:49:03 |
合計ジャッジ時間 | 5,413 ms |
ジャッジサーバーID (参考情報) |
judge4 / judge3 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 3 |
other | AC * 10 TLE * 1 -- * 27 |
ソースコード
import sys from collections import deque MOD = 998244353 def main(): N = int(sys.stdin.readline()) S = list(sys.stdin.readline().strip()) # We can model each state as a tuple for hashing initial = tuple(S) visited = set() queue = deque() visited.add(initial) queue.append(initial) count = 0 while queue: current = queue.popleft() count += 1 # For each possible triplet, check if it's in A,B,C state # Consider all possible orderings # We need to look for triplets (i,j,k) in any of the three orderings where S_i=A, S_j=B, S_k=C # For each such triplet, apply the operation and enqueue the new state if not visited for i in range(N): for j in range(N): for k in range(N): if i == j or j == k or k == i: continue # Check the three orderings valid = False # Check order 1: i < j < k if i < j < k: if current[i] == 'A' and current[j] == 'B' and current[k] == 'C': valid = True # Check order 2: j < k < i elif j < k < i: if current[i] == 'A' and current[j] == 'B' and current[k] == 'C': valid = True # Check order 3: k < i < j elif k < i < j: if current[i] == 'A' and current[j] == 'B' and current[k] == 'C': valid = True if valid: # Create a new state new_state = list(current) new_state[i] = 'B' new_state[j] = 'C' new_state[k] = 'A' new_state_tuple = tuple(new_state) if new_state_tuple not in visited: visited.add(new_state_tuple) queue.append(new_state_tuple) print(count % MOD) if __name__ == "__main__": main()