用字典保存输入的学生科目及该科成绩，求总成绩和平均成绩。 用input()语句输入科目及该科的成绩（逗号隔开,输入“#”结束），再使用split()方法分割，分别赋值给两个变量（科目、成绩），然后再添加到事先定义好的字典中。 再用for循环、keys()方法求总成绩，然后求平均成绩，并将结果输出。（参考例8-3） 2.把下面列表中大学的类型及其数量的结果统计出来。 ["综合","理工","综合","综合","军事","综合","师范","综合","理工","综合","综合","综合","军事","师范","综合","师范","理工","民族","综合","理工","综合","民族","综合","师范","综合","农林","综合","综合","理工","理工","军事","理工","理工","师范","综合","农林","理工","综合","理工","理工","理工","综合","理工","民族","综合","综合","理工","农林","民族","师范","军事"] 提示：使用集合去重，得到各大学的类型；再使用列表的count()方法统计集合中每个元素出现的次数，即该

以下是解决这两个问题的 Python 代码： ```python ''' 专业：计算机科学与技术 学号：123456 姓名：张三 ''' # 问题 1 student_scores = {} while True: input_str = input("请输入科目及成绩（逗号隔开，输入'#'结束）：") if input_str == '#': break subject, score = input_str.split(',') student_scores[subject] = int(score) total_score = 0 for score in student_scores.values(): total_score += score average_score = total_score / len(student_scores) print("总成绩为：", total_score) print("平均成绩为：", average_score) # 问题 2 university_types = ["综合","理工","综合","综合","军事","综合","师范","综合","理工","综合","综合","综合","军事","师范","综合","师范","理工","民族","综合","理工","综合","民族","综合","师范","综合","农林","综合","综合","理工","理工","军事","理工","理工","师范","综合","农林","理工","综合","理工","理工","理工","综合","理工","民族","综合","综合","理工","农林","民族","师范","军事"] unique_types = set(university_types) counts = {} for type_ in unique_types: counts[type_] = university_types.count(type_) for type_, count in counts.items(): print(f"{type_}: {count}") ``` 额外需要注意的逻辑点： 1. 输入的成绩要确保能够转换为整数。 2. 处理输入时要考虑用户输入不符合格式的情况。 3. 计算平均成绩时要注意除数不能为 0 。 [2025-05-09 10:52:51 | AI写代码神器 | 388点数解答]