统计每个学校各难度的用户平均刷题数

count(distinct up.device_id)

select university,difficult_level,count(qpd.question_id)/count(distinct up.device_id) avg_answer_cnt
from user_profile up
inner join question_practice_detail qpd
on up.device_id=qpd.device_id
inner join question_detail qd
on qpd.question_id=qd.question_id
group by university,difficult_level