找到前N个自然数的好排列 C++

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$1\leqslant{i}\leqslant{N}$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

Solution Approach

A simple solution to the problem is by finding permutations p such that p_i = i.

Then we will reconsider the equation to satisfy p_i != i. So, for a value x such that $2x \leqslant x$, we have p^{2x - 1} and p^2k. 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 <iostream>
using namespace std;
void printGoodPermutation(int n) {
   if (n % 2 != 0)
      cout<<-1;
   else
      for (int i = 1; i <= n / 2; i++)
         cout<<(2*i)<<"\t"<<((2*i) - 1)<<"\t";
}
int main() {
   int n = 4;
   cout<<"Good Permutation of first N natural Numbers : \n"; printGoodPermutation(n);
   return 0;
}

输出

Good Permutation of first N natural Numbers :
2 1 4 3

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