Perfect Root

The main trap is assuming that a “perfect root” must be rare, like a perfect square. Nope. The statement defines $x$ as a perfect root if there exists an integer $y$ with

$$\sqrt{y}=x.$$

But if $x$ is any positive integer, we can simply choose

$$y=x^2.$$

Then $y$ is definitely an integer, and since $x>0$,

$$\sqrt{x^2}=x.$$

So every positive integer is a perfect root. That completely nukes the problem: we do not need to search, precompute, or do anything fancy.

For a test case with value $n$, output

$$1,2,3,\dots,n.$$

These numbers are distinct, positive, and all at most $20$, because $n \le 20$. Therefore they also satisfy the required bound $1 \le x \le 10^9$.

Proof of correctness:

For each printed number $i$ where $1 \le i \le n$, choose $y=i^2$. Since $i$ is an integer, $y$ is an integer. Also, because $i$ is positive, $\sqrt{i^2}=i$. Therefore every printed number is a perfect root.

The printed numbers $1,2,\dots,n$ are pairwise distinct, so the output contains exactly $n$ distinct perfect roots. Since $n \le 20$, every printed value is within $1 \le x \le 10^9$. Thus the construction always satisfies the statement.

Edge cases:

• If $n=1$, printing 1 works because $\sqrt{1}=1$.
• The maximum $n=20$ is still tiny, and printing 1 through 20 is safely inside the range.

Complexity is $O(n)$ per test case and $O(1)$ extra memory, unless you count the output buffer, which would be some real bookkeeping goblin behavior, so don't.