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;
}