班级分组难题：如何算出最大管理方便度之和？

```python # 读取输入 n, k = map(int, input().split()) # 获取每个学生的能力值 abilities = list(map(int, input().split())) # 对能力值进行排序 abilities.sort() # 计算相邻能力值的差值 differences = [] for i in range(n - 1): differences.append(abilities[i + 1] - abilities[i]) # 对差值进行排序 differences.sort() # 所有能力值的最大和最小的差值 total_difference = abilities[-1] - abilities[0] # 计算需要减去的最小差值和 if k > 1: for i in range(k - 1): total_difference -= differences[-i - 1] print(total_difference) ``` ######[AI问答 | 176点数解答 | 2025-12-13 11:17:22]