DECLARE @c int = 1000;
DECLARE @numbers TABLE (n int NOT NULL PRIMARY KEY);
DECLARE @products TABLE (p int NOT NULL PRIMARY KEY);
DECLARE @primes TABLE (p int NOT NULL PRIMARY KEY);
-- The 'composite exclusion' approach
-- 1. list all n = 2, 3, 4, ... c
WITH numbers AS
(
SELECT 2 AS n
UNION ALL
SELECT n + 1 FROM numbers
WHERE n <= @c - 1
)
INSERT INTO @numbers SELECT n FROM numbers OPTION(MAXRECURSION 0);
-- 2. find all products n x n <= c
WITH products AS
(
SELECT DISTINCT m.n * n.n AS p
FROM @numbers m LEFT OUTER JOIN
@numbers n ON 1 = 1
WHERE m.n * n.n <= @c
)
INSERT INTO @products SELECT p FROM products;
-- 3. numbers with no matching products are not composite, i.e, they're prime numbers.
INSERT INTO @primes
SELECT n.n FROM @numbers n LEFT JOIN @products p ON n.n = p.p WHERE p.p IS NULL;
这有点像埃拉托舍内斯方法的一次过筛。
我见过循环、存储过程等，以及伪代码和其他语言实现，但在我看来，这种源于素数定义的简单、基于集合的方法应该足够了。
请记住，我现在不关心性能、内存消耗或优化，我也没有用更大的数字来测试它。我只想发布算法，让人们确认（或质疑）从列表中排除复合数字就足够了。
最佳答案
递归cte（rcte）很少是性能最好的解决方案。下面是一个使用计数表的方法，它是hugo kornelis在这里发布的方法的一个稍微修改过的版本：https://sqlserverfast.com/blog/hugo/2006/09/the-prime-number-challenge-great-waste-of-time/
让我们将计数表解与rcte解进行比较：
SET STATISTICS TIME ON;
PRINT 'tally table approach'+char(13)+char(10)+replicate('-',50);
DECLARE @primes TABLE (p int NOT NULL PRIMARY KEY);
DECLARE @limit bigint = 10000;
WITH E(x) AS (SELECT * FROM (VALUES (1),(1),(1),(1),(1),(1),(1),(1),(1),(1)) t(x)),
iTally(N) AS (SELECT TOP(@limit) ROW_NUMBER() OVER (ORDER BY (SELECT 1)) FROM E a, E b, E c, E d, E f)
INSERT @primes
SELECT n1.N
FROM itally AS n1
WHERE n1.N > 1
AND n1.N < @Limit
AND NOT EXISTS
(SELECT *
FROM itally AS n2
WHERE n2.N < @limit
AND n2.N BETWEEN 2 AND n1.N-1
AND n1.n % n2.N = 0)
--ORDER BY N
GO
PRINT 'rCTE approach'+char(13)+char(10)+replicate('-',50);
DECLARE @c int = 10000;
DECLARE @numbers TABLE (n int NOT NULL PRIMARY KEY);
DECLARE @products TABLE (p int NOT NULL PRIMARY KEY);
DECLARE @primes TABLE (p int NOT NULL PRIMARY KEY);
WITH numbers AS
(
SELECT 2 AS n
UNION ALL
SELECT n + 1 FROM numbers
WHERE n <= @c - 1
)
INSERT INTO @numbers SELECT n FROM numbers OPTION(MAXRECURSION 0);
-- 2. find all products n x n <= c
WITH products AS
(
SELECT DISTINCT m.n * n.n AS p
FROM @numbers m LEFT OUTER JOIN
@numbers n ON 1 = 1
WHERE m.n * n.n <= @c
)
INSERT INTO @products SELECT p FROM products;
-- 3. numbers with no matching products are not composite, i.e, they're prime numbers.
INSERT INTO @primes
SELECT n.n FROM @numbers n LEFT JOIN @products p ON n.n = p.p WHERE p.p IS NULL;
SET STATISTICS TIME OFF;
结果是：
tally table approach
--------------------------------------------------
SQL Server Execution Times:
CPU time = 3042 ms, elapsed time = 3241 ms.
SQL Server parse and compile time:
CPU time = 0 ms, elapsed time = 10 ms.
rCTE approach
--------------------------------------------------
SQL Server Execution Times:
CPU time = 14976 ms, elapsed time = 15757 ms.
如您所见，针对10000的计数表方法速度快了5倍，而且也不会产生任何读操作（rCTE产生一吨！）
如果你真的在使用素数，最快的方法就是把它们存储在一个表中，这样你就不必每次需要素数时都计算它们了。
关于sql - 这是列出质数的好算法吗？，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/40642383/