Hints: A Simple Sequence

You need consecutive remainders $a_i \bmod a_{i+1}$ to form a non-increasing sequence. Rather than searching over all permutations, think about what values $a \bmod b$ can take for specific relationships between $a$ and $b$.

What is $k \bmod (k-1)$ for any integer $k \ge 2$?

Answer to Hint 1: $k \bmod (k-1) = 1$ for every $k \ge 2$. So if you place consecutive integers in descending order — say $n, n-1, n-2, \ldots, 2$ — every adjacent remainder is exactly $1$, which is trivially non-increasing.

But you still need to place the element $1$ somewhere. What remainder does $k \bmod 1$ produce, and where should $1$ go?

Answer to Hint 2: $k \bmod 1 = 0$ for every $k$. Since $0 \le 1$, the element $1$ works perfectly at the end of the permutation: all the remainders of $1$ are at most the remainders before it. A descending sequence $n, n-1, \ldots, 2, 1$ gives remainders $1, 1, \ldots, 1, 0$ — non-increasing.

That already solves the problem for a pure descending order. But can you do something slightly different that also works and matches the sample output? Look at where the element $2$ sits in the sample.

Answer to Hint 3: The sample uses $2$ as the first element: $[2, n, n-1, \ldots, 3, 1]$. Since $2 < n$, the first remainder is $2 \bmod n = 2$. After that, the descending block $n, n-1, \ldots, 3$ gives remainders all equal to $1$, and $3 \bmod 1 = 0$ finishes it off. The full remainder sequence is $2, 1, 1, \ldots, 1, 0$ — non-increasing.

Either construction (pure descending, or $2$ first then descending rest) runs in $O(n)$ per test case, well within limits.