Codeforces-233A Perfect Permutation

最新推荐文章于 2020-07-15 13:49:06 发布

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最新推荐文章于 2020-07-15 13:49:06 发布

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原文链接：http://www.cnblogs.com/yyf573462811/archive/2012/11/15/6365153.html

Perfect Permutation

Problem Description

A permutation is a sequence of integersp₁,p₂,...,p_n, consisting ofn distinct positive integers, each of them doesn't exceedn. Let's denote the i-th element of permutation p asp_i. We'll call numbern the size of permutation p₁,p₂,...,p_n.

Nickolas adores permutations. He likes some permutations more than the others. He calls such permutations perfect. Aperfect permutation is such permutation p that for any i (1≤i≤n) (n is the permutation size) the following equations holdp_{p_i}=i andp_i≠i. Nickolas asks you to print any perfect permutation of sizen 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 permutationp, that is perfect. Separate printed numbers by whitespaces.

Sample Input

2

Sample Output

-1

2 1

题目意思：

给你一个n，则会有排列p1,p2,p3,p4...pn，并且pi<=n,；

如果n的某个排列满足：p_{p_i}=i andp_i≠i.（1<=i<=n），则此排列为完美排列。

输出这个排列。。

可以找出规律，当n%2== 1 时，必定找不出符合条件的完美排列。。

所以有：

#include"stdio.h"
int main()
{
	int n; 
    while(scanf("%d", &n) != EOF) 
    { 
        if(n % 2) 
        { 
            printf("-1\n"); 
        } 
        else 
        { 
            int i; 
            for(i = 1; i <= n; i++) 
            { 
                if(i % 2) 
                { 
                    printf("%d", i + 1); 
                } 
                else 
                    printf("%d", i - 1); 
                if(i != n) 
                    printf(" "); 
            } 
            printf("\n"); 
        } 
	}
	return 0;
}

转载于:https://www.cnblogs.com/yyf573462811/archive/2012/11/15/6365153.html