华为od 比赛的冠亚季军 s

### 华为OD比赛冠亚季军详情 在华为OD机试比赛中，冠亚季军的产生遵循特定规则。参赛运动员的ID从0到N-1，其实力由一组整数表示[^2]。比赛过程中，相邻的运动员两两对决，每一轮中实力值较大的选手晋级下一轮；若实力值相等，则ID较小的选手胜出。当选手数量为奇数时，最后一位选手直接轮空进入下一轮。 根据题目描述，最终的比赛结果可以通过模拟实现。例如，在一个具体案例中，冠军被确定为3号选手，亚军为1号选手，而季军则通过2号与0号选手的实力对比得出，由于2号选手的实力值为4大于0号选手的实力值2，因此2号选手获得季军[^1]。 以下是基于上述规则的一个代码示例，用于模拟比赛过程并输出冠亚季军名单： ```python def find_champions(strengths): players = list(range(len(strengths))) # 初始化选手列表 while len(players) > 1: winners = [] n = len(players) for i in range(0, n - 1, 2): # 每两个选手进行比赛 if strengths[players[i]] > strengths[players[i + 1]]: winners.append(players[i]) elif strengths[players[i]] < strengths[players[i + 1]]: winners.append(players[i + 1]) else: # 实力值相等时，ID小的胜出 winners.append(min(players[i], players[i + 1])) if n % 2 == 1: # 如果选手数量为奇数，最后一个选手直接晋级 winners.append(players[-1]) players = winners # 更新选手列表 champion = players[0] # 决定亚军和季军 second_round = [i for i in range(len(strengths)) if i != champion] second_winners = [] n = len(second_round) for i in range(0, n - 1, 2): if strengths[second_round[i]] > strengths[second_round[i + 1]]: second_winners.append(second_round[i]) elif strengths[second_round[i]] < strengths[second_round[i + 1]]: second_winners.append(second_round[i + 1]) else: second_winners.append(min(second_round[i], second_round[i + 1])) if n % 2 == 1: second_winners.append(second_round[-1]) runner_up = second_winners[0] third_place_candidates = [i for i in second_winners if i != runner_up] third_place = third_place_candidates[0] if third_place_candidates else None return champion, runner_up, third_place # 示例输入 strengths = [2, 1, 4, 5] # 各选手的实力值 champion, runner_up, third_place = find_champions(strengths) print(f"冠亚季军分别为：{champion} {runner_up} {third_place}") ```