# 仅查看山东大学的用户在不同难度下的每个用户的平均答题题目数情况 select university, difficult_level, count(q.question_id) / count(distinct q.device_id) from question_practice_detail q join user_profile u on u.device_id = q.device_id join question_detail qd on q.question_id = qd.question_id where university = '山东大学' group by university, difficult_level ;