select
up.university,
qd.difficult_level,
count(qpd.question_id)/count(distinct up.device_id) #注意,问题数量可有重复,但设备(即用户)不能重复
from
user_profile as up
inner join
question_practice_detail as qpd on up.device_id = qpd.device_id
inner join
question_detail as qd on qpd.question_id = qd.question_id
group by up.university , qd.difficult_level
order by up.university asc