結果
| 問題 | 
                            No.2146 2 Pows
                             | 
                    
| コンテスト | |
| ユーザー | 
                             gew1fw
                         | 
                    
| 提出日時 | 2025-06-12 18:49:25 | 
| 言語 | PyPy3  (7.3.15)  | 
                    
| 結果 | 
                             
                                TLE
                                 
                             
                            
                         | 
                    
| 実行時間 | - | 
| コード長 | 1,533 bytes | 
| コンパイル時間 | 265 ms | 
| コンパイル使用メモリ | 81,928 KB | 
| 実行使用メモリ | 469,840 KB | 
| 最終ジャッジ日時 | 2025-06-12 18:49:32 | 
| 合計ジャッジ時間 | 5,868 ms | 
| 
                            ジャッジサーバーID (参考情報)  | 
                        judge4 / judge1 | 
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| 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])
    
    max_a = 60  # Precompute up to 2^60 mod N
    
    # Precompute 2^a mod N for a up to max_a
    pow_mod = []
    for a in range(max_a + 1):
        pow_mod.append(pow(2, a, N))
    
    # Initialize distance array: dist[residue][max_a] = minimal cost
    INF = float('inf')
    dist = [ [INF] * (max_a + 1) for _ in range(N) ]
    heap = []
    
    for a in range(max_a + 1):
        r = pow_mod[a]
        cost = A + B + C * a
        if cost < dist[r][a]:
            dist[r][a] = cost
            heapq.heappush(heap, (cost, r, a))
    
    while heap:
        current_cost, current_r, current_m = heapq.heappop(heap)
        if current_cost > dist[current_r][current_m]:
            continue
        for a in range(current_m, max_a + 1):
            r_a = pow_mod[a]
            new_r = (current_r + r_a) % N
            added_B = B if a > current_m else 0
            new_cost = current_cost + A + added_B + C * (a - current_m)
            if new_cost < dist[new_r][a]:
                dist[new_r][a] = new_cost
                heapq.heappush(heap, (new_cost, new_r, a))
    
    # Now compute the answer for each k
    for k in range(N):
        min_cost = INF
        for a in range(max_a + 1):
            if dist[k][a] < min_cost:
                min_cost = dist[k][a]
        print(min_cost)
    
if __name__ == '__main__':
    main()
            
            
            
        
            
gew1fw