n=k*(k+1)/2+1 in this position consist 1 else 0

I wrote that

Я ничего не понял
Shunmadim
Misunderstanding

I can't understand how you use N in your solution. N is only number of elements and nothing more.
Maybe, I understand you wrong.

if 8k-7 is a squre then a[i]=1 else a[i]=0

Can I ask how could you find that? So beautiful solution

The thing is that the sequence of positions of '1' in 1101001000... can be represented as (N*(N-1)/2+1). You can find this out by yourself or turn to http://oeis.org/classic/

So let's try to check whether an input A is a position of '1'.
N*(N-1)/2 + 1 = A;
N^2 - N + 2 - 2*A = 0;
N is natural, so solution is:
N = 1/2 * (1 + sqrt(8*A - 7));

So, A is a position if sqrt(8*A-7) is natural.

k=sqrt((n-1)*2+0.0);
if (n*(n+1)==(n-1)*2) cout<<1<<" ";
else                  cout<<0<<" ";