Codeforces 233A Perfect Permutation（完美序列，水题）

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本文介绍了Codeforces上的一道题目233A，主题是完美序列。完美序列是一种特殊的排列，要求对于每个位置i，序列中的元素pi等于i但位置不同。题目要求给出大小为n的完美序列。当n为偶数时，可以通过交换当前偶数位置与前一个奇数位置的数字来构造完美序列。文章提供了实现思路和AC代码。

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.

   Examples
 
     input
   
1

     output
   
-1

     input
   
2

     output
   
2 1 

     input
   
4

     output
   
2 1 4 3

题意：

构造完美序列。

思路：

只在 n 在偶数的情况下，才可以构造完美序列，把当前的偶数和前一个奇数交换输出即可。

AC CODE：

#include<stdio.h>
#include<cstring>
#include<algorithm>
#define HardBoy main()
#define ForMyLove return 0;
using namespace std;
const int MYDD = 1103;

int HardBoy {
	int n;
	scanf("%d", &n);
	if(n&1) {
		puts("-1");
	} else{
		for(int j = 2; j <= n; j=j+2) {
			printf("%d %d ", j, j-1);
		}
	}
	ForMyLove
}