select t3.university ,t4.difficult_level,    count(t3.question_id)/count(distinct t3.device_id) as avg_answer_cntfrom (    select t1.university,t2.question_id,t1.device_id     from user_profile t1 right join question_practice_detail t2    on t1.device_id=t2.device_id    where t1.university='山东大学'   ...