递归
with a (c1, c2, c3, c4, c5, c6)as
(
select 366 as c1, 365 as c2, 65 as c3, 1 as c4, 0.0 as c5, 0 as c6 from dual
union all
select c1, c2, c3, c4 + 1, decode(c5, 0, 1, c5) * (c1 - c4 - 1) / c2, c5 from a
where c4 < 65
)
select c4, decode(c5, 1, 0, 1- c5) from a;
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