描述

现有试卷信息表examination_info（exam_id试卷ID, tag试卷类别, difficulty试卷难度, duration考试时长, release_time发布时间）：

id	exam_id	tag	difficulty	duration	release_time
1	9001	SQL	hard	60	2020-01-01 10:00:00
2	9002	C++	hard	80	2020-01-01 10:00:00
3	9003	算法	hard	80	2020-01-01 10:00:00
4	9004	PYTHON	medium	70	2020-01-01 10:00:00

试卷作答记录表exam_record（uid用户ID, exam_id试卷ID, start_time开始作答时间, submit_time交卷时间, score得分）：

id	uid	exam_id	start_time	submit_time	score
6	1003	9001	2020-01-02 12:01:01	2020-01-02 12:31:01	68
9	1001	9001	2020-01-02 10:01:01	2020-01-02 10:31:01	89
1	1001	9001	2020-01-01 09:01:01	2020-01-01 09:21:59	90
12	1002	9002	2021-05-05 18:01:01	(NULL)	(NULL)
3	1004	9002	2020-01-01 12:01:01	2020-01-01 12:11:01	60
2	1003	9002	2020-01-01 19:01:01	2020-01-01 19:30:01	75
7	1001	9002	2020-01-02 12:01:01	2020-01-02 12:43:01	81
10	1002	9002	2020-01-01 12:11:01	2020-01-01 12:31:01	83
4	1003	9002	2020-01-01 12:01:01	2020-01-01 12:41:01	90
5	1002	9002	2020-01-02 19:01:01	2020-01-02 19:32:00	90
11	1002	9004	2021-09-06 12:01:01	(NULL)	(NULL)
8	1001	9005	2020-01-02 12:11:01	(NULL)	(NULL)

在物理学及统计学数据计算时，有个概念叫min-max标准化，也被称为离差标准化，是对原始数据的线性变换，使结果值映射到[0 - 1]之间。转换函数为：

请你将用户作答高难度试卷的得分在每份试卷作答记录内执行min-max归一化后缩放到[0,100]区间，并输出用户ID、试卷ID、归一化后分数平均值；最后按照试卷ID升序、归一化分数降序输出。（注：得分区间默认为[0,100]，如果某个试卷作答记录中只有一个得分，那么无需使用公式，归一化并缩放后分数仍为原分数）。

由示例数据结果输出如下：

uid	exam_id	avg_new_score
1001	9001	98
1003	9001	0
1002	9002	88
1003	9002	75
1001	9002	70
1004	9002	0

解释：高难度试卷有9001、9002、9003；

作答了9001的记录有3条，分数分别为68、89、90，按给定公式归一化后分数为：0、95、100，而后两个得分都是用户1001作答的，因此用户1001对试卷9001的新得分为(95+100)/2≈98（只保留整数部分），用户1003对于试卷9001的新得分为0。最后结果按照试卷ID升序、归一化分数降序输出。

方法一：使用group by聚合函数

#使用group by聚合函数
select
uid,exam_id,ifnull(round(avg((score-min_score)/differ_score)*100,0),min(score)) as avg_new_score
from
(
    select
    exam_id,min(score) as min_score,max(score)-min(score) as differ_score
    from
    exam_record
    where submit_time is not null and
    exam_id in (select exam_id from examination_info where difficulty='hard')
    group by exam_id
)t1
left join
(
    select uid,exam_id,score
    from
    exam_record
    where submit_time is not null and
    exam_id in (select exam_id from examination_info where difficulty='hard')
)t2
using(exam_id)
group by uid,exam_id
order by exam_id,avg_new_score desc

方法二：使用窗口函数

# 使用窗口函数
select
uid,exam_id,
round(avg(if(max_score=min_score,score,(score-min_score)/(max_score-min_score)*100)),0) as avg_new_score
from
(
    select
    uid,exam_id,score,
    max(score) over(partition by exam_id) as max_score,
    min(score) over(partition by exam_id) as min_score
    from
    exam_record
    where submit_time is not null and
    exam_id in (select exam_id from examination_info where difficulty = 'hard')
)t1
group by uid,exam_id
order by exam_id,avg_new_score desc