OJ Board

ok, let me put my code here, I'll be graceful if you could have a brief look at it :

[pascal]
program P10232;
type Int=LongInt;
var A:array [-1..31,-1..31] of Int;
Bit:array [0..31] of Byte;
function Solve(n:Int):Int;
var i,b,k:Int;
begin
i:=0;
while n>=0 do
begin
n:=n-A[31,i];
Inc(i);
end ...

Thanks for your inputs Larry, but altough it can solve them in the right way, i got WA another time , I'm really getting disappointed of this problem

Shahab Tasharrofi

Hi everybody,

I have solved the problem #10232, but I got WA for it. It solves the sample input and my samples in the right way and I really can't understand the problem behind it.

Can anyone give me some sample input and output plz?

Thanks in advance,
Shahab Tasharrofi

Any idea to solve such a problem, plz?

I have the idea of derivation of the sum of the distances, but it is really hard and unamusing to code.

Thanks in advance

hi, every body

i think that Nak must add this note that the A* is optimally efficient
it means that if u use an unit heuristic in some different algorithms including A*, the A* will use the minimum time to find it

Best regards,
Shahab Tasharrofi

power series r 2 week for such works
at first, u must know that we can not show all the functions with powe series like Maclaurin series, they have their limitation
for ex, consider the following function. there is no power series to show such function

f(x)= 1 if we have two integers p,q such that ...

at first, u need an algorithm to determine whether a point is over a line or under it( u can do it using the line equation)

i mean that if u have the line aX+bY+c=0 and also u have the point (x0,y0), if u replace the point in the equation the result of next expression "sign(a*x0+b*y0+c)" shows ...

As you know the if we have a(k) number of the kth element, and we have n elements, the number of permutations is like this:

C(m,a(1))*C(m-a(1),a(2))*C(m-a(1)-a(2),a(3))*...

where m represents the sum of a(i)s, and C(n,k)=(n!/(k!*(n-k)!))

If we number the permutations, using the above formula, we ...

Hi rfcotta,

I think that your algorithm won't provide the rectangular shape of the answer, it just provides the connectivity of it ( but also it won't be the answer even for the connected numbers , I think ). for example , check your algorithm with the following matrix.

-11 -11 -11 1
-11 -11 1 1 ...

Hi, sir

Are you sure that there is one with O(n^2) or you're just trying find one ? Becuase I think that I must have heard about it if there's one.

And if you find one which its time complexity is lower than O(n^3), I would be happy to hear about that.

Thank you,
Shahab Tasharrofi

hi, every body

I think the "Graph Theory" of the West is the best book. but I think it won't be free online.

Sincerely yours,
Shahab Tasharrofi

I found that article,

you can find it at http://www.research.att.com/~shor/papers/QCjournal.ps ,in this paper there is a random method for factoring numbers, and it is not just for quantum computers.

And Larry, you must read it too, i think. Becuase the Peter Shor ( Writer of this article ) claims ...

Hi, nghiank

I can't understand your definition of the problem. Can you have some example matrices that satisfy this condition and that don't satisfy it?

Sincerely yours,
Shahab Tasharrofi

Hi everybody,

Some monthes ago, I read about factoring integers in quantom computers, and I found a random method of factoring numbers(not only to primes) used in this algorithm, but it wasn't related to quantom computers. The time order of this method was very connected to the number of prime ...

Thank you for your reply,

But I can't understand the proof of your algorithm , and after that have you any idea about a sufficient condition? And if so, i would be happy to hear that.

Sincerely yours,
Shahab Tasharrofi