一个很有用的自定义函数(判断自然数是否包含2的指定次幂)
2024-07-21 02:05:39
供稿：网友
/* name : fun_wheincluded function : 判断选定的数字是否在给定的整数中 可以知道任何一个自然数都可以拆分成若干个2的幂的和，如： 1 = 2^0 2 = 2^1 3 = 2^0 + 2^1 4 = 2^2 5 = 2^0 + 2^2 6 = 2^1 + 2^4 7 = 2^0 + 2^1 + 2^2 8 = 2^3 9 = 2^0 + 2^3 10 = 2^1 + 2^3 11 = 2^0 + 2^1 + 2^3 12 = 2^2 + 2^3 13 = 2^0 + 2^2 + 2^3 14 = 2^1 + 2^2 + 2^3 15 = 2^0 + 2^1 + 2^2 + 2^3 16 = 2^4 17 = 2^0 + 2^4 将任意一个数解析为2的幂的和的方法——递归 规律： 如给定 14 ∵ 2^3 < 14 < 2^4 ∴ 14中必有8——2^3 14 - 8 = 6 ∵ 2^2 < 6 < 2^3 ∴ 6中必有4——2^2 6 - 4 = 2 ∵ 2 = 2 ∴ 14 = 2^3 + 2^2 + 2^1
parameters : @totalnum type: int @specifiednum type: int steps : author : waxdoll cheung date : 2005-03-21*/
create function dbo.fun_wheincluded ( @totalnum int, @specifiednum int )returns bit as begin
declare @varret bit
declare @varloop int
set @varloop = 0
while (@totalnum >= cast(power(2, @varloop) as int)) set @varloop = @varloop + 1
set @totalnum = @totalnum - cast(power(2, @varloop - 1) as int)
if (@varloop = @specifiednum + 1) set @varret = 1 else begin if (@totalnum >= 1) return dbo.fun_wheincluded(@totalnum, @specifiednum) else set @varret = 0 end
return @varretend
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