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题解 | 考试分数(四)

with count_temp as ( select job, count(*) as count from grade group by job ) select job, floor((count + 1) / 2) as start, flo...

2025-12-21

0 9

题解 | 考试分数(四)

with count_temp as ( select job,count(*) as count from grade group by job ) select job,round(count/2) as start, case when count%2 != 0 then round(co...

2025-12-21

0 8

题解 | 考试分数(三)

select t.id,name,score from ( select id,language_id,score, dense_rank() over(partition by language_id order by score desc) as rn from grade ) t join l...

2025-12-21

0 10

题解 | 牛客每个人最近的登录日期(五)

with fd as ( select first_day, round(count(distinct case when datediff(date,first_day)=1 then user_id end)*1.0/ count(distinct user_id),3) as p from ...

2025-12-20

0 9

题解 | 牛客每个人最近的登录日期(四)

select date, count(distinct case when first_day=date then user_id end) as new from ( select user_id, date, min(date) over(partition by user_id ) as fi...

2025-12-20

0 9

题解 | 牛客每个人最近的登录日期(四)

select date, count(distinct case when datediff(date,first_day)=0 then user_id end) as new from ( select user_id, date, min(date) over(partition by use...

2025-12-20

0 12

题解 | 牛客每个人最近的登录日期(三)

with rank_date as ( select user_id,date ,row_number() over(partition by user_id order by date) as rn from login ), first_date as ( select user_id,...

2025-12-20

0 10

题解 | 牛客每个人最近的登录日期(三)

with rank_date as ( select user_id,date ,row_number() over(partition by user_id order by date) as rn from login ), first_date as ( select user_id,...

2025-12-20

0 11

题解 | 异常的邮件概率

with no_black as ( select id from ( select id from user where is_blacklist=0 ) t ), send as ( select send_id,receive_id,type,date from em...

2025-12-20

0 9

题解 | 给出employees表中排名为奇数行的first_name

select first_name as first from ( select first_name ,emp_no, row_number() over(order by first_name ) as rn from employees ) t where rn%2 != 0 order by...

2025-12-19

0 10