CF1405A.Permutation Forgery

传统题 时间 2000 ms 内存 256 MiB 3 尝试 1 已通过 1 标签

Permutation Forgery

A permutation of length nn is an array consisting of nn distinct integers from 11 to nn in arbitrary order. For example, [2,3,1,5,4][2,3,1,5,4] is a permutation, but [1,2,2][1,2,2] is not a permutation (22 appears twice in the array) and [1,3,4][1,3,4] is also not a permutation (n=3n=3 but there is 44 in the array).

Let pp be any permutation of length nn. We define the fingerprint F(p)F(p) of pp as the sorted array of sums of adjacent elements in pp. More formally,

$$F(p)=\mathrm{sort}([p_1+p_2,p_2+p_3,\ldots,p_{n-1}+p_n]).$$</p><p>For example, if$n=4$and$p=[1,4,2,3],$then the fingerprint is given by$F(p)=\mathrm{sort}([1+4,4+2,2+3])=\mathrm{sort}([5,6,5])=[5,5,6]$.</p><p>You are given a permutation$p$of length$n$. Your task is to find a <span class="tex-font-style-bf">different</span> permutation$p'$with the same fingerprint. Two permutations$p$and$p'$are considered different if there is some index$i$such that$p\_i \ne p'\_i$. ## Input 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. ## Output 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. ## Note 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 $F(p')=\mathrm{sort}([3+1,1+5,5+2,2+4])=\mathrm{sort}([4,6,7,6])=[4,6,6,7]$. ## Samples ```input1 3 2 1 2 6 2 1 6 5 4 3 5 2 4 3 1 5 ``` ```output1 2 1 1 2 5 6 3 4 3 1 5 2 4 ```$$

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