Codeforces Round 668 (Div. 2) |
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Finished |
A permutation of length $$$n$$$ is an array consisting of $$$n$$$ distinct integers from $$$1$$$ to $$$n$$$ in arbitrary order. For example, $$$[2,3,1,5,4]$$$ is a permutation, but $$$[1,2,2]$$$ is not a permutation ($$$2$$$ appears twice in the array) and $$$[1,3,4]$$$ is also not a permutation ($$$n=3$$$ but there is $$$4$$$ in the array).
Let $$$p$$$ be any permutation of length $$$n$$$. We define the fingerprint $$$F(p)$$$ of $$$p$$$ as the sorted array of sums of adjacent elements in $$$p$$$. More formally,
$$$$$$F(p)=\mathrm{sort}([p_1+p_2,p_2+p_3,\ldots,p_{n-1}+p_n]).$$$$$$
For example, if $$$n=4$$$ and $$$p=[1,4,2,3],$$$ then the fingerprint is given by $$$F(p)=\mathrm{sort}([1+4,4+2,2+3])=\mathrm{sort}([5,6,5])=[5,5,6]$$$.
You are given a permutation $$$p$$$ of length $$$n$$$. Your task is to find a different permutation $$$p'$$$ with the same fingerprint. Two permutations $$$p$$$ and $$$p'$$$ are considered different if there is some index $$$i$$$ such that $$$p_i \ne p'_i$$$.
Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 668$$$). Description of the test cases follows.
The first line of each test case contains a single integer $$$n$$$ ($$$2\le n\le 100$$$) — the length of the permutation.
The second line of each test case contains $$$n$$$ integers $$$p_1,\ldots,p_n$$$ ($$$1\le p_i\le n$$$). It is guaranteed that $$$p$$$ is a permutation.
For each test case, output $$$n$$$ integers $$$p'_1,\ldots, p'_n$$$ — a permutation such that $$$p'\ne p$$$ and $$$F(p')=F(p)$$$.
We can prove that for every permutation satisfying the input constraints, a solution exists.
If there are multiple solutions, you may output any.
3 2 1 2 6 2 1 6 5 4 3 5 2 4 3 1 5
2 1 1 2 5 6 3 4 3 1 5 2 4
In the first test case, $$$F(p)=\mathrm{sort}([1+2])=[3]$$$.
And $$$F(p')=\mathrm{sort}([2+1])=[3]$$$.
In the second test case, $$$F(p)=\mathrm{sort}([2+1,1+6,6+5,5+4,4+3])=\mathrm{sort}([3,7,11,9,7])=[3,7,7,9,11]$$$.
And $$$F(p')=\mathrm{sort}([1+2,2+5,5+6,6+3,3+4])=\mathrm{sort}([3,7,11,9,7])=[3,7,7,9,11]$$$.
In the third test case, $$$F(p)=\mathrm{sort}([2+4,4+3,3+1,1+5])=\mathrm{sort}([6,7,4,6])=[4,6,6,7]$$$.
And $$$F(p')=\mathrm{sort}([3+1,1+5,5+2,2+4])=\mathrm{sort}([4,6,7,6])=[4,6,6,7]$$$.
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