CF1405A.Permutation Forgery

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题目描述

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$$$.

输入格式

Each test contains multiple test cases. The first line contains the number of test cases tt ( 1t6681 \le t \le 668 ). Description of the test cases follows.

The first line of each test case contains a single integer nn ( 2n1002\le n\le 100 ) — the length of the permutation.

The second line of each test case contains nn integers p1,,pnp_1,\ldots,p_n ( 1pin1\le p_i\le n ). It is guaranteed that pp is a permutation.

输出格式

For each test case, output nn integers p1,,pnp'_1,\ldots, p'_n — a permutation such that ppp'\ne p and F(p)=F(p)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.

输入输出样例

  • 输入#1

    3
    2
    1 2
    6
    2 1 6 5 4 3
    5
    2 4 3 1 5

    输出#1

    2 1
    1 2 5 6 3 4
    3 1 5 2 4

说明/提示

In the first test case, F(p)=sort([1+2])=[3]F(p)=\mathrm{sort}([1+2])=[3] .

And F(p)=sort([2+1])=[3]F(p')=\mathrm{sort}([2+1])=[3] .

In the second test case, F(p)=sort([2+1,1+6,6+5,5+4,4+3])=sort([3,7,11,9,7])=[3,7,7,9,11]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)=sort([1+2,2+5,5+6,6+3,3+4])=sort([3,7,11,9,7])=[3,7,7,9,11]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)=sort([2+4,4+3,3+1,1+5])=sort([6,7,4,6])=[4,6,6,7]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)=sort([3+1,1+5,5+2,2+4])=sort([4,6,7,6])=[4,6,6,7]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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