{$N+}
{ Q = n*(n+1)/2+1 => n^2 + n + 2(1-Q)=0  n must be integer }
var n, i : longint;
  q, nr : extended;
begin
  read(n);
  for i:=1 to n do begin
    read(q);
    nr := (-1 + sqrt(-7+8*q)) / 2;
    if frac(nr) < 1e-8 then write('1 ') else write('0 ');
  end;
end.

I can't understand

that's the obvious solution...

What the f**k is this?

I think you don't need extended,just cardinal.
Here's my program:
var
  n,i,ans:integer;
  k,temp:cardinal;
begin
  readln(n);
  for i:=1 to n do
  begin
    readln(k);
    dec(k);
    if k=0 then ans:=1
    else
    begin
      k:=k*2;
      temp:=trunc(sqrt(k));
      if temp*(temp+1)=k then ans:=1
      else ans:=0;
    end;
    if i=1 then write(ans) else write(' ',ans);
  end;
  writeln;
end.

Its easy.
all 1 digits stand on places of sums of arithmetical progression + 1.
The eq. of sum of the progression is  n*(n+1)/2.
adding 1 you get Q = n*(n+1)/2+1.
Solving this eq. you get (-1 + sqrt(-7+8*q)) / 2.
if nr is integer, you have the position of 1 else the position of 0.

Why nr=(-1 + sqrt(-7+8*q))/2
For (q=3) this nr becomes a floating-point number.
I think, nr should be (sqrt(9+8*q)/2-1.5)  for (q>2).

If 8*q-7 is square there's no need to test is (-1 + sqrt(-7+8*q)) / 2 integer; because discriminant is odd then and this expression is even, so it divides by 2;

I have a similar logic, yet I got WA on test #3.

EDIT: I should've put an explicit conversion from long to double, did it - got AC. Hooray! :)

Edited by author 01.03.2008 01:33

Hooray!
It's work for me too :)