Inter Active (Easy Version)

The original problem was interactive: there was a hidden derangement permutation $p$, and you had to recover it by choosing a fixed index $k$ and asking queries of the form “how many directed edges $x \to p_x$ point forward in this ordering, ignoring the element placed at position $k$ as a source?”

That version is a real problem. The hacked version is not that game anymore.

What changes in hacks

For hacks, the input format is explicitly changed. Each test case contains:

1. An integer $n$.
2. The full hidden permutation $p_1,p_2,\ldots,p_n$.

So the thing the interactive solution was supposed to discover is now handed to us directly. There is no jury, no chosen $k$, no queries, no flushing, and no adaptive nonsense. The old interaction section is basically a fossil in the statement.

What we need to output

The natural non-interactive answer is just the recovered permutation. Since the input already gives us $p$, the recovered permutation is exactly the second line of the test case.

So for every test case:

• read $n$;
• read array $p$ of length $n$;
• print $p$.

That is it.

Why not simulate the interactive strategy?

Because there is nothing to simulate. In an interactive environment, a query is useful because it extracts information about unknown $p$. In the hacked environment, $p$ is already known before any computation starts. Simulating queries would be extra code with extra chances to mess up, which is the competitive programming equivalent of tying your shoes together before a sprint.

Correctness proof

We prove that the algorithm outputs the required permutation for every test case.

For a hacked test case, the input contains a valid permutation $p_1,p_2,\ldots,p_n$, which is exactly the hidden permutation from the original statement. The algorithm reads this permutation and outputs its entries in the same order. Therefore, the produced output is exactly $p$.

Since this holds independently for every test case, the algorithm is correct.

Complexity

For each test case, we read and print $n$ integers.

Time complexity: $O(n)$ per test case.

Memory complexity: $O(n)$ per test case, or $O(1)$ extra if printed while reading.