Codeforces Round 867 (Div. 3)-D. Super-Permutation

Codeforces 题解 - [Codeforces Round 867 (Div. 3)-D. Super-Permutation]

题目链接

题目描述

A permutation is a sequence \(n\) integers, where each integer from \(1\) to \(n\) appears exactly once. For example, \([1]\), \([3,5,2,1,4]\), \([1,3,2]\) are permutations, while \([2,3,2]\), \([4,3,1]\), \([0]\) are not.

Given a permutation \(a\), we construct(构造) an array \(b\), where \(b_i = (a_1 + a_2 +~\dots~+ a_i) \bmod n\).

A permutation of numbers \([a_1, a_2, \dots, a_n]\) is called a super-permutation if \([b_1 + 1, b_2 + 1, \dots, b_n + 1]\) is also a permutation of length \(n\).

Grisha became interested whether a super-permutation of length \(n\) exists. Help him solve this non-trivial(重要的) problem. Output any super-permutation of length \(n\), if it exists. Otherwise(不然), output \(-1\).

输入格式

The first line contains a single integer \(t\) (\(1 \le t \le 10^4\)) — the number of test cases. The description of the test cases follows.

Each test case consists of a single line containing one integer \(n\) (\(1 \le n \le 2 \cdot 10^5\)) — the length of the desired（希望实现的） permutation.

The sum of \(n\) over all test cases does not exceed \(2 \cdot 10^5\).

输出格式

For each test case, output in a separate line:

• \(n\) integers — a super-permutation of length \(n\), if it exists.
• \(-1\), otherwise.

If there are several suitable permutations, output any of them.

题目大意

构造一个数组A
其中\(b_{i} = (a_{1} + a_2 + \dots + a_i)\bmod n\)
使得数组B和数组A均为1~n的全排列

输入

4
1
2
3
6

输出

1
2 1
-1
6 5 2 3 4 1

解题思路

1

\[b_{i} = (a_{1} + a_2 + \dots + a_i)\bmod n \tag{1} (i \geq 2) \]

\[b_{i - 1} = (a_{1} + a_2 + \dots + a_{i - 1})\bmod n \tag{2}(i \geq 2) \]

由\((1)-(2)\)可得，

\[b_{i} = (b_{i - 1} + a_{i}) \bmod n \]

让数字k代表数字n在a数组排列中的位置，即\(a_k=n\). 当\(k>1\)时，

\[b_k = (b_{k - 1} + a_k) \bmod n = (b_{k - 1} + n) \bmod n = b_{k - 1} \]

即

\[b_k = b_{k - 1} \]

此时一定不是全排列。则k一定为1，即

\[a_1=n \]

则

\[b_1 = 0 \]

2

如果\(n\)为奇数，

\[b_{n} = (a_{1} + a_{2} + \dots + a_{n}) \bmod n \]

\[b_{n} = (1 + 2 + \dots + n) \bmod m = \frac{n(n+1)}{2} \bmod n = 0 = b_1 \]

此时也一定不会是全排列。因此n>1时只有n为偶数有可行方案

3

当n为偶数时，

\[a = [n,~1,~n-2,~3,~n-4,~5,~\dots,~n - 1,~2] \]

对应的

\[b = [0,~1,~n-1,~2,~n-2,~3,~n-3,~\dots,~\frac{n}{2}] \]

代码实现

#include "bits/stdc++.h"
using namespace std;

void Solution()
{
    int n; cin >> n;
    if(n == 1) { cout << 1 << '\n'; return; }
    if (n & 1) { cout << -1 << '\n'; return; }

    for(int i = 1;i <= n;++i)
    {
        if(i & 1)
        {
            cout << n - i + 1 << ' ';
        }
        else
        {
            cout << i - 1 << ' ';
        }
    }
    cout << '\n';
    return;
}//Monotonous

int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);

    int t; cin >> t;
    while (t--)
    {
        Solution();
    }

    return 0;
}