中等困难题 | #牛客每个人最近的登录日期(五)#

select t0.date date,
round(ifnull(count(distinct t2.user_id) / count(t1.user_id), 0), 3)
from
(select date from login group by date) t0
left join (select user_id, min(date) date from login group by user_id) t1
on t0.date = t1.date
left join login t2
on t1.user_id = t2.user_id and datediff(t2.date, t1.date) = 1
group by t0.date

求每日活跃用户的次日留存率已是准困难题，求每日新用户的次日留存率可以算作中等困难题。