結果
| 問題 | No.1694 ZerOne |
| コンテスト | |
| ユーザー |
lam6er
|
| 提出日時 | 2025-04-15 22:52:40 |
| 言語 | PyPy3 (7.3.15) |
| 結果 |
TLE
|
| 実行時間 | - |
| コード長 | 1,945 bytes |
| 記録 | |
| コンパイル時間 | 358 ms |
| コンパイル使用メモリ | 82,004 KB |
| 実行使用メモリ | 106,320 KB |
| 最終ジャッジ日時 | 2025-04-15 22:54:51 |
| 合計ジャッジ時間 | 3,964 ms |
|
ジャッジサーバーID (参考情報) |
judge4 / judge2 |
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| ファイルパターン | 結果 |
|---|---|
| sample | AC * 3 |
| other | TLE * 1 -- * 30 |
ソースコード
from collections import deque
def count_possible_strings(S):
visited = set()
queue = deque([S])
visited.add(S)
n = len(S)
while queue:
current = queue.popleft()
# Generate all possible pairs of non-overlapping intervals t and u
intervals = []
# Precompute all possible intervals and their counts
for i in range(n):
for j in range(i, n):
cnt0 = current[i:j+1].count('0')
cnt1 = (j+1 - i) - cnt0
intervals.append((i, j, cnt0, cnt1))
# Check all pairs of intervals
for idx1 in range(len(intervals)):
i1, j1, c0_1, c1_1 = intervals[idx1]
for idx2 in range(idx1 + 1, len(intervals)):
i2, j2, c0_2, c1_2 = intervals[idx2]
# Check if intervals are non-overlapping and have the same counts
if (j1 < i2 or j2 < i1) and (c0_1 == c0_2 and c1_1 == c1_2):
# Ensure t is before u
if j1 < i2:
t_start, t_end = i1, j1
u_start, u_end = i2, j2
elif j2 < i1:
t_start, t_end = i2, j2
u_start, u_end = i1, j1
else:
continue # overlapping
# Create new string by swapping t and u
t_part = current[t_start:t_end+1]
u_part = current[u_start:u_end+1]
before_t = current[0:t_start]
between = current[t_end+1:u_start]
after_u = current[u_end+1:]
new_str = before_t + u_part + between + t_part + after_u
if new_str not in visited:
visited.add(new_str)
queue.append(new_str)
return len(visited)
S = input().strip()
print(count_possible_strings(S))
lam6er