just put all the numbers in a array of boolean.

var
  n,i,caint:longint;
  b:array[-100000..100000]of boolean;
begin
  readln(n);
  fillchar(b,sizeof(b),false);
  for i := 1 to n do
    begin
      readln(caint);
      b[caint]:=true;
    end;
  readln(n);
  for i := 1 to n do
    begin
      readln(caint);
      if b[10000-caint] then
        begin
          writeln('YES');
          halt;
        end;
    end;
  writeln('NO');
end.

Edited by author 10.09.2009 08:34

I think it is O(n2)

It is O(n), not O(1)

And there's another O(n) solution for this problem, which requires only O(n) memory instead of O(v) in your case, where v is max value of a_i.

Yes , SkidanovAlex right. O(n).

thanks