【SQL牛客在线编程答案】

【01】

select device_id, gender, age, university from user_profile;

[次日平均留存率]

select count(distinct b.device_id,b.date)/count(distinct a.device_id,a.date) as avg_ret 
from question_practice_detail as a 
left join question_practice_detail as b 
on a.device_id = b.device_id and datediff(b.date,a.date)=1;