Codeforces Round #144 (Div. 2)-A. Perfect Permutation_html/css_WEB-ITnose

Perfect Permutation

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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

Nickolas adores permutations. He likes some permutations more than the others. He calls such permutations perfect. A perfectpermutation is such permutation p that for any i (1?≤?i?≤?n) (n is the permutation size) the following equations hold ppi?=?i and pi?≠?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,?...,?pn ? permutation p, that is perfect. Separate printed numbers by whitespaces.

Sample test(s)

input

output

-1

input

output

2 1 

input

output

2 1 4 3 

解题思路：贪心。直接构造。如果n为奇数，则不能构成满足要求的排列；否则，两个两个的构造，如2,1；4,3。。。以此类推即可。

AC代码：

#include <stdio.h>#include <string.h>#include <iostream>#include <algorithm>#include <vector>#include <queue>#include <set>#include <map>#include <string>#include <math.h>#include <stdlib.h>#include <time.h>using namespace std;#define INF 0x7fffffffint main(){    #ifdef sxk        freopen("in.txt","r",stdin);    #endif    int n;    while(scanf("%d",&n)!=EOF)    {        if(n&1){            printf("-1\n");            continue;        }        else{            for(int i=1; i<n/2; i++)                printf("%d %d ", 2*i, 2*i-1);            printf("%d %d\n", n, n-1);        }    }    return 0;}