Perfect Permutation

最新推荐文章于 2019-12-02 19:39:19 发布

踢树叶

最新推荐文章于 2019-12-02 19:39:19 发布

阅读量504

点赞数 1

CC 4.0 BY-SA版权

分类专栏： AC路漫漫

本文链接：https://blog.youkuaiyun.com/geekdoe/article/details/50457074

AC路漫漫专栏收录该内容

180 篇文章

订阅专栏

该问题要求构造一种特殊的排列——完美排列，即一个长度为n的序列，其中每个元素等于其索引且不与其所在位置相同。当n为偶数时存在完美排列，此时可以将1到n的数字两两配对，交换每对数字的位置。输入一个整数n，输出一个完美排列或-1表示不存在。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Description

A permutation is a sequence of integers p₁, p₂, ..., p_n, consisting of n distinct positive integers, each of them doesn't exceed n. Let's denote the i-th element of permutation p as p_i. We'll call number n the size of permutation p₁, p₂, ..., p_n.

Nickolas adores permutations. He likes some permutations more than the others. He calls such permutations perfect. A perfect permutation is such permutation p that for any i(1 ≤ i ≤ n) (n is the permutation size) the following equations hold p_{p_i} = i and p_i ≠ i. Nickolas asks you to print any perfect permutation of size n for the given n.

Input

A single line contains a single integer n (1 ≤ n ≤ 100) — the permutation size.

Output

If a perfect permutation of size n doesn't exist, print a single integer -1. Otherwise print n distinct integers from 1 to n, p₁, p₂, ..., p_n — permutation p, that is perfect. Separate printed numbers by whitespaces.

Sample Input

Input

Output

-1

Input

Output

2 1

Input

Output

2 1 4 3

只有偶数有答案，两两翻转。

#include<cstdio>
#include<iostream>
using namespace std;

int main(){
    int n,i;
    while(cin>>n){
        if(n%2){
            cout<<"-1"<<endl;
        }
        else{
            for(i=1;i<=n;i+=2){
                if(i==n-1){
                    cout<<i+1<<" "<<i<<endl;
                }
                else{
                    cout<<i+1<<" "<<i<<" ";
                }
            }
        }
    }
    return 0;
}