思路感觉没错，相邻的 1 会合并为一个操作，相邻的 2 如果偶数个不操作，奇数个变一下，但还是超时， 45% N, M = [int(_) for _ in sys.stdin.readline().strip().split()] t_ls = [int(_) for _ in sys.stdin.readline().strip().split()] P = [str(_) for _ in range(1, N+1)] i = 0 while i < M: if t_ls[i] == 1: count = 0 while i < M and t_ls[i] == 1: count += 1 i += 1 count = count % N if count != 0: P = P[count:] + P[:count] else: count = 0 while i < M and t_ls[i] == 2: count += 1 i += 1 if count % 2 == 1: for j in range(0, N, 2): P[j], P[j+1] = P[j+1], P[j] print(" ".join(P))