找到一个排列，使得gcd(p [i]，i)> 1的索引数恰好是K

给定两个整数N和K ，任务是从范围[1，N]中查找整数的排列，以使gcd(p [i]，i)> 1的索引数(从1开始的索引)正好ķ。如果无法进行此排列，则打印-1 。

例子：

Input: N = 4, K = 3
Output: 1 2 3 4
gcd(1, 1) = 1
gcd(2, 2) = 2
gcd(3, 3) = 3
gcd(4, 4) = 4
Therefore, there are exactly 3 indices where gcd(p[i], i) > 1

Input: N = 1, K = 1
Output: -1

方法：可以在此处进行一些观察：

1. gcd(i，i + 1)= 1
2. gcd(1，i)= 1
3. gcd(i，i)= i

由于1的gcd与其他任何数字始终将为1，因此K最多可以为N – 1 。考虑其中p [i] = i的置换，其中gcd(p [i]，i)> 1的索引数将为N – 1 。请注意，交换2个连续元素(不包括1个)将使此类索引的数目减少2。而与1交换将使计数的数目减少1。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
  
// Function to print the required permutation
void printPermutation(int n, int k)
{
  
    // Not possible
    if (k >= n || (n % 2 == 0 && k == 0)) {
        cout << -1;
        return;
    }
    int per[n + 1], i;
  
    // Initial permutation
    for (i = 1; i <= n; i++)
        per[i] = i;
  
    // Indices for which gcd(p[i], i) > 1
    // in the initial permutation
    int cnt = n - 1;
    i = 2;
    while (i < n) {
  
        // Reduce the count by 2
        if (cnt - 1 > k) {
            swap(per[i], per[i + 1]);
            cnt -= 2;
        }
  
        // Reduce the count by 1
        else if (cnt - 1 == k) {
            swap(per[1], per[i]);
            cnt--;
        }
        else
            break;
        i += 2;
    }
  
    // Print the permutation
    for (i = 1; i <= n; i++)
        cout << per[i] << " ";
}
  
// Driver code
int main()
{
    int n = 4, k = 3;
    printPermutation(n, k);
  
    return 0;
}

Java

// Java implementation of the approach 
class GFG
{
  
// Function to print the required permutation 
    static void printPermutation(int n, int k)
    {
  
        // Not possible 
        if (k >= n || (n % 2 == 0 && k == 0)) 
        {
            System.out.print(-1);
            return;
        }
        int per[] = new int[n + 1], i;
  
        // Initial permutation 
        for (i = 1; i <= n; i++)
        {
            per[i] = i;
        }
  
        // Indices for which gcd(p[i], i) > 1 
        // in the initial permutation 
        int cnt = n - 1;
        i = 2;
        while (i < n) 
        {
  
            // Reduce the count by 2 
            if (cnt - 1 > k)
            {
                swap(per, i, i + 1);
                cnt -= 2;
            } 
              
            // Reduce the count by 1 
            else if (cnt - 1 == k) 
            {
                swap(per, 1, i);
                cnt--;
            } 
            else
            {
                break;
            }
            i += 2;
        }
  
        // Print the permutation 
        for (i = 1; i <= n; i++)
        {
            System.out.print(per[i] + " ");
        }
    }
  
    static int[] swap(int[] arr, int i, int j)
    {
        int temp = arr[i];
        arr[i] = arr[j];
        arr[j] = temp;
        return arr;
    }
      
    // Driver code 
    public static void main(String[] args)
    {
        int n = 4, k = 3;
        printPermutation(n, k);
    }
}
  
/* This code contributed by PrinciRaj1992 */

Python3

# Python3 implementation of the approach
  
# Function to print the required permutation
def printPermutation(n, k):
  
    # Not possible
    if (k >= n or (n % 2 == 0 and
                   k == 0)):
        print(-1);
        return;
    per = [0] * (n + 1);
  
    # Initial permutation
    for i in range(1, n + 1):
        per[i] = i;
  
    # Indices for which gcd(p[i], i) > 1
    # in the initial permutation
    cnt = n - 1;
    i = 2;
    while (i < n): 
  
        # Reduce the count by 2
        if (cnt - 1 > k):
              
            t = per[i];
            per[i] = per[i + 1];
            per[i + 1] = t;
              
            # swap(per[i], per[i + 1]);
            cnt -= 2;
  
        # Reduce the count by 1
        elif (cnt - 1 == k):
              
            # swap(per[1], per[i]);
            t = per[1];
            per[1] = per[i];
            per[i] = t;
            cnt -= 1;
        else:
            break;
        i += 2;
  
    # Print the permutation
    for i in range(1, n + 1):
        print(per[i], end = " ");
  
# Driver code
n = 4;
k = 3;
printPermutation(n, k);
  
# This code is contributed by mits

C#

// C# implementation of the approach 
using System;
  
class GFG 
{ 
  
    // Function to print the required permutation 
    static void printPermutation(int n, int k) 
    { 
  
        // Not possible 
        if (k >= n || (n % 2 == 0 && k == 0)) 
        { 
            Console.Write(-1); 
            return; 
        } 
        int []per = new int[n + 1] ;
        int i ; 
  
        // Initial permutation 
        for (i = 1; i <= n; i++) 
        { 
            per[i] = i; 
        } 
  
        // Indices for which gcd(p[i], i) > 1 
        // in the initial permutation 
        int cnt = n - 1; 
        i = 2; 
        while (i < n) 
        { 
  
            // Reduce the count by 2 
            if (cnt - 1 > k) 
            { 
                swap(per, i, i + 1); 
                cnt -= 2; 
            } 
              
            // Reduce the count by 1 
            else if (cnt - 1 == k) 
            { 
                swap(per, 1, i); 
                cnt--; 
            } 
            else
            { 
                break; 
            } 
            i += 2; 
        } 
  
        // Print the permutation 
        for (i = 1; i <= n; i++) 
        { 
            Console.Write(per[i] + " "); 
        } 
    } 
  
    static int[] swap(int[] arr, int i, int j) 
    { 
        int temp = arr[i]; 
        arr[i] = arr[j]; 
        arr[j] = temp; 
        return arr; 
    } 
      
    // Driver code 
    public static void Main() 
    { 
        int n = 4, k = 3; 
        printPermutation(n, k); 
    } 
} 
  
/* This code contributed by Ryuga */

PHP

= $n || ($n % 2 == 0 && 
                     $k == 0)) 
    {
        echo -1;
        return;
    }
    $per[$n + 1] = array();
    $i;
  
    // Initial permutation
    for ($i = 1; $i <= $n; $i++)
        $per[$i] = $i;
  
    // Indices for which gcd(p[i], i) > 1
    // in the initial permutation
    $cnt = $n - 1;
    $i = 2;
    while ($i < $n) 
    {
  
        // Reduce the count by 2
        if ($cnt - 1 > $k)
        {
              
            list($per[$i], 
                 $per[$i + 1]) = array($per[$i + 1],
                                       $per[$i]);
            //swap(per[i], per[i + 1]);
            $cnt -= 2;
        }
  
        // Reduce the count by 1
        else if ($cnt - 1 == $k) 
        {
              
            // swap(per[1], per[i]);
            list($per[1], 
                 $per[$i]) = array($per[$i],
                                   $per[1]);
            $cnt--;
        }
        else
            break;
        $i += 2;
    }
  
    // Print the permutation
    for ($i = 1; $i <= $n; $i++)
        echo $per[$i], " ";
}
  
// Driver code
$n = 4;
$k = 3;
printPermutation($n, $k);
  
// This code is contributed by ajit.
?>

输出：

1 2 3 4

时间复杂度： O(N)