How do you use only 2x500 numbers???

In order to know how many d digits numbers which their sum is s, you
just have to know how many d-1 digits numbers which their sums are
s , s-1, s-2...s-9 , right?

Yes, this is how I solve the problem-> using DP, i count these cases.
But, I have to use long numbers. And there are two variants :
1) To have an array NxS of long numbers to store the possibilities
2) To use just recursion and not to remember anything.
But, bot cases are impossible. Because, the first will get Memory
Limit, the second- Time Limit. So, I tried to allocate memory
dinamically. However, timus (as well as my PC) does not work all
right with realloc() (in C). So, I cannot solve it that way. My
quesiton is, how do you manage to solve the problem with a 2x500
array???

10x in advance

You don't have to store all NxS.
All you have to store is S bignums for current d and another S for d-1.
And current d will be d-1 to next d (i.e. d+1)

should i show you pseudo code ?

"Ha ha , how could a stupid person solved many problems that I can't :)"
i don't want to say like this , it's like chatting
because this board mainly for questions and answers
... anyway , :)