quote:

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Open Contest 2007 consists of 10 problems. Tourney will last for several days, from 15 April 2007 to 27 April 2007. Solutions for any problem can be sent for many times but not more than 256. Each solution will be executed with set of tests. Results received with your program will be processed with special Judge Program that will estimate the correctness/effectiveness of your solution. In some tasks you will receive some score for your solution. The score will be proportional to efficiency of the algorithm offered by you. Score will define your place in rating table for given problem. Only the best solution of each participant for given problem will be considered at rating tables. Current results and the information about all sent solutions will be automatically generated by Sphere Online Judge system and will be accessible for all participants of competition.

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quote:

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Originally posted by Peso

I want prime numbers?
https://www.spoj.pl/problems/PRIME1.pdf

I would use http://www.sqlteam.com/forums/topic.asp?TOPIC_ID=69646


Peter Larsson
Helsingborg, Sweden

---

/*
To calculate digits of sqrt(2)
As many as possible in 10 seconds
Results appx 1500 digits in 10 seconds, appx 5000 digits in 2 minutes. 2.4GHz PC
Problem from: 
Algorithm references:
http://www.nist.gov/dads/HTML/squareRoot.html
http://antwrp.gsfc.nasa.gov/htmltest/gifcity/sqrt2.1mil
*/

set nocount on
go

-- helper functions (2)
create function mult(@n varchar(8000),@p int) returns varchar(8000) as
begin
	if @p = 0 return '0'; if @p = 1 return @n
	declare	@i int ,@res varchar(8000) ,@rem int ,@m int
	select	@i = len(@n) ,@res = '' ,@rem = 0
	while @i > 0 select @m = @rem + @p*substring(@n,@i,1) ,@res = ltrim(@m%10) + @res ,@rem = @m/10, @i = @i - 1
	return case when @rem = 0 then '' else ltrim(@rem) end + @res
end
go

create function sub(@n1 varchar(8000),@n2 varchar(8000)) returns varchar(8000) as
begin
	declare	@i int, @j int, @res varchar(8000), @rem int, @m int
	select	@i = len(@n1), @j = len(@n2), @res = '', @rem = 0
	--if @i < @j or (@n1 < @n2 and @i = @j) return '-' + dbo.sub(@n2,@n1)
	while @i > 0 select @m = 10+cast(substring(@n1,@i,1) as int)-cast(substring(@n2,@j,1) as int)+@rem ,@res = ltrim(@m%10) + @res ,@rem = @m/10-1, @i = @i - 1, @j = @j - 1
	return substring(@res,patindex('%[^0]%',@res),8000)
end
go

declare @s int; set @s = 10 -- seconds to run

-- start calc of sqrt(2)
declare @i datetime; set @i = dateadd(second,@s,getdate())

declare	@res varchar(8000)
	,@rem varchar(8000)
	,@c varchar(8000)
	,@try int
	,@tst varchar(8000)
	,@lenres int
select	@res = '1414'
	,@rem = '60400'
	,@lenres = len(@res)
	

while getdate() <= @i
begin
	set @try = case len(@rem)-@lenres
			when 0 then 0
			when 1 then cast(left(@rem,5) as int) / 28280
			else cast(left(@rem,5) as int) / 2828 end
	tst:
	set @tst = dbo.mult(dbo.mult(@res,2)+ltrim(@try),@try)
	if len(@tst)>len(@rem) or (len(@tst)=len(@rem) and @tst>@rem)
	begin
		set @try = @try-1
		goto tst
	end
		select @rem = dbo.sub(@rem,@tst)+'00', @res = @res + ltrim(@try), @lenres = @lenres + 1
end

select @res as sq2 into #sq2

select len(sq2) as digits from #sq2
select sq2 from #sq2
go


drop table #sq2
go

drop function mult,sub
go

quote:

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Just thirty-nine decimal places would be enough to compute the circumference of a circle surrounding the known universe to within the ra dius of a hydrogen atom.

---

-- Set up initialization
DECLARE	@Number INT
SET @Number = 14

-- Show the expected output
IF @Number < 1 OR @Number > 2000000000 OR @Number IS NULL
	SELECT NULL
ELSE
	IF @Number % 10 = 0
		SELECT	2
	ELSE
		SELECT	1 UNION ALL
		SELECT	@Number % 10

quote:

---

Originally posted by Peso

Is this challenge for real?
http://www.spoj.pl/ZFUN07/problems/NGM/

-- Set up initialization
DECLARE	@Number INT
SET @Number = 14

-- Show the expected output
IF @Number < 1 OR @Number > 2000000000 OR @Number IS NULL
	SELECT NULL
ELSE
	IF @Number % 10 = 0
		SELECT	2
	ELSE
		SELECT	1 UNION ALL
		SELECT	@Number % 10

Peter Larsson
Helsingborg, Sweden

---

/*
Problem : http://www.spoj.pl/ZFUN07/problems/GEOM/
*/

declare	@dx int
	,@dy int
	,@side numeric(9,2)
	,@px int
	,@py int

select	@dx = 10
	,@dy = 10
	,@side = 20
	,@px = 5
	,@py = 12


create table #v(v varchar(8) primary key, x numeric(9,2), y numeric(9,2))

-- vectors from each corner to given point P in the square
insert #v(v,x,y)
select 'A->P', @px-(@dx-@side/2), @py-(@dy-@side/2)
union all
select 'B->P', @px-(@dx-@side/2), @py-(@dy+@side/2)
union all
select 'C->P', @px-(@dx+@side/2), @py-(@dy+@side/2)
union all
select 'D->P', @px-(@dx+@side/2), @py-(@dy-@side/2)

-- p'' is the hypothetical point where the lines will meet
-- the line drawn through vertex A : AP'' is perpendicular to line BP
-- using the fact that two vectors are perpendicular, if their dotproduct is 0, (EDIT: not cross!)
-- we create a set of linear equations
select	linearEquationToSolve = 
	ltrim(#v.x)+'*(@x-'+ltrim(u.x)+') + ' + ltrim(#v.y)+'*(@y-'+ltrim(u.y)+') = 0'
from	#v
	cross join
	(
		select 'A->P''' v, (@dx-@side/2) x, (@dy-@side/2) y
		union all
		select 'B->P''', (@dx-@side/2), (@dy+@side/2)
		union all
		select 'C->P''', (@dx+@side/2), (@dy+@side/2)
		union all
		select 'D->P''', (@dx+@side/2), (@dy-@side/2)
	) u
where	(u.v = 'A->P''' and #v.v = 'B->P')
	or
	(u.v = 'B->P''' and #v.v = 'C->P')
	or
	(u.v = 'C->P''' and #v.v = 'D->P')
	or
	(u.v = 'D->P''' and #v.v = 'A->P')

drop table #v


======================================================================================

linearEquationToSolve                 
-----------------------------------------------
5.00*(@x-20.000000) + 12.00*(@y-0.000000) = 0
5.00*(@x-0.000000) + -8.00*(@y-0.000000) = 0
-15.00*(@x-0.000000) + -8.00*(@y-20.000000) = 0
-15.00*(@x-20.000000) + 12.00*(@y-20.000000) = 0


/* test
declare @x int, @y int; select @x=8,@y=5
select
5.00*(@x-20.000000) + 12.00*(@y-0.000000)
,5.00*(@x-0.000000) + -8.00*(@y-0.000000)
,-15.00*(@x-0.000000) + -8.00*(@y-20.000000) 
,-15.00*(@x-20.000000) + 12.00*(@y-20.000000)
*/

quote:

---

Originally posted by Peso

quote:

---

Just thirty-nine decimal places would be enough to compute the circumference of a circle surrounding the known universe to within the radius of a hydrogen atom.

---

http://www.math.rutgers.edu/~cherlin/History/Papers2000/wilson.html


Peter Larsson
Helsingborg, Sweden

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quote:

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Originally posted by Michael Valentine Jones
I hate when these topics deteriorate into discussions of theoretical physics.

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