请问大佬这个题目怎么做

有如下一组数据

求出NAME中每组累加/每组总数的比例大于0.6的ID和NAME

预期的结果应该为

解释：从题目意思可以看出A组的总数为16，从ID为1到5分别累加后的结果分别为1,3,9,13,16，只有13和16除以总数16才大于0.6，所以返回的结果ID为4和5，同样B组为7和8

select t1.id,
       t1.name,
       t1.num
  from (
select t1.*,
       sum(t1.num) over(partition by t1.name order by t1.id) /
       sum(t1.num) over(partition by t1.name) per
  from tab1 t1
) t1
 where t1.per > 0.6

有点粗糙，期待更好的解法，

select t1.id,t1.name,t1.num
-- ,t1.s,t2.s 
from (
select a.id,a.name,a.num,sum(b.num) s from tab1 a 
join tab1 b 
on a.id >= b.id
where a.name = 'A'
group by a.num,a.id,a.name
UNION ALL 
select a.id,a.name,a.num,sum(b.num) s from tab1 a 
join tab1 b 
on a.id >= b.id and b.name ='B'
where a.name = 'B'
group by a.num,a.id,a.name
) t1 
left join 
 (select name,sum(num) s from tab1 group by name  ) t2
on t1.name =t2.name
where (t1.s/t2.s)>0.6
order by t1.id