Finally AC. Yes brootforce rules here. But it's a little BIT tricky brootforce.

Edited by author 22.03.2011 12:39

"table of covering reducing"- my term(auto trans. is used)
Classical algo for minimal covering had good success in this problem
and under emotion I wrote that helping.
Again: we have NP problem but  can to speed up.
So, in stack data we add matrix of covering for example as
1 0 1 0
0 1 1 1
0 1 1 1
This means that 4 points(columns) are covered by 3 sets(rows). First point is covered by only 1-th row and we must use Set-1 as obligatory.
Thus, the search tree hasn't branching in this-time vertex and we create
only one child vertex with reduced matrix:
1 1
1 1
because 1,3 points are covered by obligated row 1.
Finally: when we   make choice for some set in search tree  we transfer in child
vertex reduced matrix.

P.S. I had expirience in this algo from boolean design(all practice problem are NP).

Input
9800 9999 10
0 0
output
0.98(0098)
or
0.9(8009)

!!!!!!!!!!!!!!!!!!!!!!!!!
What interval at you for T?
And what step of increase?
??????????????????????????

Where is mistake?
read(x,y);
readln(e);
t:=0;
if ((abs (x-sin(sqrt(t)))<=e) and (abs(y-cos(t))<=e))
 then begin  write('0') ; exit; end;

repeat
t:=t+e
until (abs(sin(sqrt(t))-x)<=e) and (abs(cos(t)-y)<=e)  or (t>10e12) ;

if t>10e12 then write ('FAIL') else write(t:0:1);

>  Change
>
> Var
>   n,k                 :longint;
> >>  i,j,s               :longint;
>   m                   :array[0..9] of longint;
>
>
>   And all will be correct!

When N>100,his answer isn't right,either.