Python：dict_perform函数实现学生成绩字典多模式操作

```python def dict_perform(a, b, c, d=1): dict1 = {} if d == 1: for i in range(1, a + 1): key = f'第{i}位学生' value = 5 * i - a dict1[key] = value print(dict1) elif d == 2: total_score = 0 for i in range(1, a + 1): key = f'第{i}位学生' value = 5 * i - a total_score += value dict1[key] = value dict1['总分'] = total_score return dict1['总分'] elif d == 3: total_score = 0 for i in range(1, a + 1): key = f'第{i}位学生' value = 5 * i - a if value < b: value = b elif value > c: value = c total_score += value dict1[key] = value dict1['总分'] = total_score average_score = total_score // a dict1['平均分'] = average_score return dict1['平均分'] ``` ######[AI写代码神器 | 293点数解答 | 2024-11-09 12:28:38]