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
