select a.university, c.difficult_level, -- 平均刷题数 round(count(b.question_id)/count(distinct (b.device_id)),4) as avg_answer_cnt from question_practice_detail b left join user_profile a on a.device_id=b.device_id left join question_detail c on b.question_id=c.question_id group by a.university, c.diffi...