求某数为几个2的N次方组成的函数(递归法,循环法)

xiaoer_1982

--递归法
create function f_getnum(@i int)
returns @t table(r int)
as
begin
declare @n int
set @n = 1
while [email=2*@n<=@i]2*@n<=@i[/email]
begin
set @n = @n * 2
end
insert @t select @n
set @n = @i - @n
if @n > 0
insert @t select r from f_getnum(@n)
return;
end
go
select * from f_getnum(10)
--循环法
alter function f_getnum (@i int)
returns @t table(r int)
as
begin
declare @n int
while @i > 0
begin
set @n = 1
while [email=2*@n<=@i]2*@n<=@i[/email]
begin
set @n = @n * 2
end
insert @t select @n
set @i = @i - @n
end
return;
end
go
select * from f_getnum(10)
/*
r
-----------
8
2
(2 行受影响)
*/

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