結果
| 問題 |
No.1633 Sorting Integers (Multiple of 2^K)
|
| コンテスト | |
| ユーザー |
gew1fw
|
| 提出日時 | 2025-06-12 15:32:13 |
| 言語 | PyPy3 (7.3.15) |
| 結果 |
TLE
|
| 実行時間 | - |
| コード長 | 1,749 bytes |
| コンパイル時間 | 188 ms |
| コンパイル使用メモリ | 82,608 KB |
| 実行使用メモリ | 91,016 KB |
| 最終ジャッジ日時 | 2025-06-12 15:34:03 |
| 合計ジャッジ時間 | 5,367 ms |
|
ジャッジサーバーID (参考情報) |
judge3 / judge1 |
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| ファイルパターン | 結果 |
|---|---|
| sample | AC * 3 |
| other | AC * 20 TLE * 1 -- * 24 |
ソースコード
import sys
from sys import stdin
from functools import lru_cache
def main():
sys.setrecursionlimit(1 << 25)
N = int(stdin.readline())
c = list(map(int, stdin.readline().split()))
digits = [i+1 for i in range(9) for _ in range(c[i])]
max_k = 46
for k in range(max_k, -1, -1):
s = min(k, N)
required = 1 << k
max_num = (10 ** s) - 1
if required > max_num:
continue
even_digits = [d for d in digits if d % 2 == 0]
if k >= 1 and len(even_digits) < 1:
continue
from collections import defaultdict
counts = defaultdict(int)
for d in digits:
counts[d] += 1
used = [False] * len(digits)
target = required
def backtrack(pos, rem, remaining_counts):
if pos == s:
return rem == 0
for d in remaining_counts:
if remaining_counts[d] == 0:
continue
new_rem = (rem * 10 + d) % target
new_counts = remaining_counts.copy()
new_counts[d] -= 1
if backtrack(pos + 1, new_rem, new_counts):
return True
return False
initial_counts = defaultdict(int)
for d in digits:
initial_counts[d] += 1
success = False
for d in initial_counts:
if initial_counts[d] == 0:
continue
new_counts = initial_counts.copy()
new_counts[d] -= 1
if backtrack(1, d % target, new_counts):
success = True
break
if success:
print(k)
return
print(0)
if __name__ == '__main__':
main()
gew1fw