Home >Web Front-end >HTML Tutorial >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;}