select a.university, c.difficult_level, count(b.question_id) / count(distinct(a.device_id)) avg_answer_cnt
from question_practice_detail b
join user_profile a
on a.device_id = b.device_id and a.university='山东大学'
join question_detail c
on b.question_id = c.question_id
group by a.university,c.difficult_level

第一次连表直接筛选出山东大学数据