在这个问题中，我们有一个整数值N。我们的任务是找到前N个自然数的好排列。
排列是对一组对象的全部或部分进行安排，考虑到排列的顺序。
好排列是一个排列，其中$1leqslant{i}leqslant{N}$并满足以下条件：
$P_{pi}:=:i$
$P_{p!}:=:i$
让我们举一个例子来理解这个问题，
Input : N = 1
Output : -1
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Solution Approach
A simple solution to the problem is by finding permutations p such that pi = i.
Then we will reconsider the equation to satisfy pi != i. So, for a value x such that $2x leqslant x$, we have p2x – 1 and p2k. Now, we have an equation that satisfies the permutation equation for n. Here, the solution for the equation.
Example
Program to illustrate the working of our solution
#include
using namespace std;
void printGoodPermutation(int n) {
if (n % 2 != 0)
cout
