select u.university , qd.difficult_level,  round(count(u.answer_cnt) / count(distinct u.device_id),4 )  avg_answer_cnt
from 
user_profile as u, question_practice_detail as qpd ,question_detail as qd
where 
qpd.device_id = u.device_id and qpd.question_id = qd.question_id
group by u.university , qd.difficult_level;

-- 方式一 ：隐式内连接