結果
問題 |
No.2146 2 Pows
|
ユーザー |
![]() |
提出日時 | 2025-03-26 15:50:34 |
言語 | PyPy3 (7.3.15) |
結果 |
TLE
|
実行時間 | - |
コード長 | 1,803 bytes |
コンパイル時間 | 242 ms |
コンパイル使用メモリ | 82,540 KB |
実行使用メモリ | 74,420 KB |
最終ジャッジ日時 | 2025-03-26 15:51:45 |
合計ジャッジ時間 | 5,590 ms |
ジャッジサーバーID (参考情報) |
judge1 / judge2 |
(要ログイン)
ファイルパターン | 結果 |
---|---|
sample | AC * 2 |
other | AC * 3 TLE * 1 -- * 28 |
ソースコード
import heapq def main(): import sys input = sys.stdin.read data = input().split() N = int(data[0]) A = int(data[1]) B = int(data[2]) C = int(data[3]) m_max = 60 # Sufficiently large to cover possible exponents s = [pow(2, m, N) for m in range(m_max)] INF = float('inf') dist = [[INF] * m_max for _ in range(N)] heap = [] # Initialize for each possible exponent m for m in range(m_max): sm = s[m] cost = A + B + C * m if dist[sm][m] > cost: dist[sm][m] = cost heapq.heappush(heap, (cost, sm, m)) while heap: current_cost, r, m = heapq.heappop(heap) if current_cost > dist[r][m]: continue # Transition a: add another element of current exponent m new_r = (r + s[m]) % N new_cost = current_cost + A if new_cost < dist[new_r][m]: dist[new_r][m] = new_cost heapq.heappush(heap, (new_cost, new_r, m)) # Transition b: add a new exponent m' > m for m_prime in range(m + 1, m_max): sm_p = s[m_prime] new_r_b = (r + sm_p) % N added_cost = A + B + C * (m_prime - m) new_cost_b = current_cost + added_cost if new_cost_b < dist[new_r_b][m_prime]: dist[new_r_b][m_prime] = new_cost_b heapq.heappush(heap, (new_cost_b, new_r_b, m_prime)) # Compute the minimal cost for each residue ans = [INF] * N for r in range(N): min_val = min(dist[r][m] for m in range(m_max)) ans[r] = min_val if min_val != INF else -1 # Ensure no -1 as problem guarantees a solution for k in range(N): print(ans[k]) if __name__ == '__main__': main()