CodeForces 233A

最新推荐文章于 2021-10-27 10:59:54 发布

原创最新推荐文章于 2021-10-27 10:59:54 发布 · 597 阅读

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本文介绍了一个简单的算法问题，即寻找“完美排列”。对于给定的整数n，如果不存在完美排列，则输出-1；若存在，则输出该排列。完美排列是指每个元素与其索引不相等且满足特定条件的排列。文章提供了示例和解决方案。

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CodeForces 233A

Crawling in process... Crawling failed Time Limit:2000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u

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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

为什么我总是做水题！！！

水平有限啊！Orz。。

大致题意，输入一个数，奇数出 -1 偶数颠倒就OK了，输入4，第一个和第二个颠倒，第三个和第四个颠倒。。依次类推

哎，没什么好说的。

#include <iostream>

using namespace std;

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