解题思路：

1. 用户总数
2. 其中留存成功的人数
3. 判断 留存成功的条件
   • 当用户首次登录后，当天的后一天也再次登录，即判定留存成功

select  round(count(if(to_days(a1.date) = to_days(a1.second_day),true,null))
              / count(distinct user_id),3)
from 
    (select l.user_id, l.date, s1.second_day
     from login l 
     join
        (
             select user_id, date_add(min(date), interval 1 day) as second_day
             from login
             group by user_id
        )s1
     on l.user_id = s1.user_id
    )a1
;