题解 | #考试分数(四)#
考试分数(四)
https://www.nowcoder.com/practice/502fb6e2b1ad4e56aa2e0dd90c6edf3c
我是觉得逻辑很清晰的，直接case when每种情况
select job,
case when count(*) <= 2 then 1
when count(*) > 2 and count(*) % 2 = 1 then ceiling(count(*)/2)
else floor(count(*)/2)
end as start,
case when count(*) <= 2 then count(*)
when count(*) > 2 and count(*) % 2 = 1 then ceiling(count(*)/2)
else cast((count(*)/2 + 1) as unsigned)
end as 'end'
from grade
group by job
order by job
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