結果
問題 |
No.1963 Subset Mex
|
ユーザー |
![]() |
提出日時 | 2025-06-12 15:43:10 |
言語 | PyPy3 (7.3.15) |
結果 |
TLE
|
実行時間 | - |
コード長 | 2,632 bytes |
コンパイル時間 | 267 ms |
コンパイル使用メモリ | 82,176 KB |
実行使用メモリ | 85,504 KB |
最終ジャッジ日時 | 2025-06-12 15:43:19 |
合計ジャッジ時間 | 8,434 ms |
ジャッジサーバーID (参考情報) |
judge1 / judge4 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 4 |
other | AC * 2 TLE * 1 -- * 23 |
ソースコード
import sys from collections import deque MOD = 998244353 def main(): N = int(sys.stdin.readline()) S = list(map(int, sys.stdin.readline().split())) from collections import defaultdict initial = defaultdict(int) for x in S: initial[x] += 1 # Each state is a dictionary: key is the number, value is the count visited = set() queue = deque() # Convert the initial state to a tuple of sorted items for hashing initial_tuple = tuple(sorted(initial.items())) visited.add(initial_tuple) queue.append(initial_tuple) count = 0 while queue: current = queue.popleft() count += 1 # Convert back to a dictionary for manipulation current_dict = dict(current) # Generate all possible subsets (non-empty) # To do this, we consider each element and how many times it can be taken # This part is tricky because the multiset allows multiple instances # But for mex calculation, the presence or absence matters, not the count # So, for each element, we can choose to include it or not in the subset, regardless of count # But the subset must be non-empty # We can generate all possible subsets by considering each element's presence # For each number in current_dict, we can choose to include it or not # But we need to avoid the empty subset # Generate all possible subsets (non-empty) elements = list(current_dict.keys()) n = len(elements) for mask in range(1, 1 << n): subset = [] for i in range(n): if (mask >> i) & 1: subset.append(elements[i]) # Compute mex of subset mex = 0 present = set(subset) while mex in present: mex += 1 # Create the next state next_dict = current_dict.copy() for num in subset: if next_dict[num] > 0: next_dict[num] -= 1 if next_dict[num] == 0: del next_dict[num] if mex in next_dict: next_dict[mex] += 1 else: next_dict[mex] = 1 # Convert to a tuple for hashing next_tuple = tuple(sorted(next_dict.items())) if next_tuple not in visited: visited.add(next_tuple) queue.append(next_tuple) print(count % MOD) if __name__ == '__main__': main()