Codeforces Round #144 (Div. 2) A. Perfect Permutation

题目:http://codeforces.com/contest/233/problem/A

输出 p_pi = i且p_i ≠ i的数列

思路：p_i=x p_x=i

#include <iostream>

using namespace std;

int main()
{
    int n;
    cin>>n;

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

转载于:https://www.cnblogs.com/danielqiu/archive/2013/01/23/2872879.html