计算特殊排列的数量

给定两个正整数n和k ，任务是计算特殊排列的数量。特殊排列P被定义为前n 个自然数的排列，其中至少存在(n – k) 个索引，使得P _i = i 。
先决条件：精神错乱
例子：

Input: n = 4, k = 2 
Output: 7 
{1, 2, 3, 4}, {1, 2, 4, 3}, {4, 2, 3, 1}, {2, 1, 3, 4}, {1, 4, 3, 2}, {1, 3, 2, 4} and {3, 2, 1, 4} are the only possible special permutations.
Input: n = 5, k = 3 
Output: 31 
 

方法：让函数f _x表示特殊排列的数量，其中正好存在x 个索引，使得P _i = i 。因此，所需的答案将是：

f_{(n – k)} + f_{(n – k + 1)} + f_{(n – k + 2)} + … + f_{(n – 1)} + f_n 
 

现在， f _x可以通过从n中选择x 个索引来计算，其中P _i = i并计算(n – i)其他索引的乱序数量，因为它们P _i不应该等于i然后将结果乘以ⁿ C _x因为可以有不同的方法从n 中选择x索引。
下面是上述方法的实现：

C++

// C++ program to count the number
// of required permutations
#include 
using namespace std;
#define ll long long int
 
// Function to return the number of ways
// to choose r objects out of n objects
int nCr(int n, int r)
{
    int ans = 1;
    if (r > n - r)
        r = n - r;
    for (int i = 0; i < r; i++) {
        ans *= (n - i);
        ans /= (i + 1);
    }
    return ans;
}
 
// Function to return the number
// of derangements of n
int countDerangements(int n)
{
    int der[n + 1];
 
    der[0] = 1;
    der[1] = 0;
    der[2] = 1;
 
    for (int i = 3; i <= n; i++)
        der[i] = (i - 1) * (der[i - 1]
                         + der[i - 2]);
    return der[n];
}
 
// Function to return the required
// number of permutations
ll countPermutations(int n, int k)
{
    ll ans = 0;
    for (int i = n - k; i <= n; i++) {
 
        // Ways to choose i indices from n indices
        int ways = nCr(n, i);
 
        // Dearangements of (n - i) indices
        ans += ways * countDerangements(n - i);
    }
    return ans;
}
 
// Driver Code to test above functions
int main()
{
    int n = 5, k = 3;
    cout << countPermutations(n, k);
    return 0;
}

Java

// Java program to count the number
// of required permutations
     
public class GFG{
     
    // Function to return the number of ways
    // to choose r objects out of n objects
    static int nCr(int n, int r)
    {
        int ans = 1;
        if (r > n - r)
            r = n - r;
        for (int i = 0; i < r; i++) {
            ans *= (n - i);
            ans /= (i + 1);
        }
        return ans;
    }
     
    // Function to return the number
    // of derangements of n
    static int countDerangements(int n)
    {
        int der[] = new int[ n + 3];
     
        der[0] = 1;
        der[1] = 0;
        der[2] = 1;
     
        for (int i = 3; i <= n; i++)
            der[i] = (i - 1) * (der[i - 1]
                            + der[i - 2]);
        return der[n];
    }
     
    // Function to return the required
    // number of permutations
    static int countPermutations(int n, int k)
    {
        int ans = 0;
        for (int i = n - k; i <= n; i++) {
     
            // Ways to choose i indices from n indices
            int ways = nCr(n, i);
     
            // Dearangements of (n - i) indices
            //System.out.println(ans) ;
            ans += (ways * countDerangements(n- i));
        }
        return ans;
    }
     
    // Driver Code to test above functions
    public static void main(String []args)
    {
        int n = 5, k = 3;
        System.out.println(countPermutations(n, k)) ;
    }
    // This code is contributed by Ryuga
}

Python3

# Python 3 program to count the number
# of required premutation
 
# function to return the number of ways
# to chooser objects out of n objects
def nCr(n, r):
    ans = 1
    if r > n - r:
        r = n - r
    for i in range(r):
        ans *= (n - i)
        ans /= (i + 1)
    return ans
 
# function to return the number of
# degrangemnets of n
def countDerangements(n):
    der = [0 for i in range(n + 3)]
     
    der[0] = 1
    der[1] = 0
    der[2] = 1
     
    for i in range(3, n + 1):
        der[i] = (i - 1) * (der[i - 1] +
                            der[i - 2])
         
    return der[n]
 
# function to return the required
# number of premutations
def countPermutations(n, k):
    ans = 0
    for i in range(n - k, n + 1):
         
        # ways to chose i indices
        # from n indices
        ways = nCr(n, i)
         
        # Dearrangements of (n-i) indices
        ans += ways * countDerangements(n - i)
    return ans
 
# Driver Code
n, k = 5, 3
 
print(countPermutations(n, k))
 
# This code is contributed by
# Mohit kumar 29 (IIIT gwalior)

C#

// C# program to count the number
// of required permutations
using System;   
public class GFG{
      
    // Function to return the number of ways
    // to choose r objects out of n objects
    static int nCr(int n, int r)
    {
        int ans = 1;
        if (r > n - r)
            r = n - r;
        for (int i = 0; i < r; i++) {
            ans *= (n - i);
            ans /= (i + 1);
        }
        return ans;
    }
      
    // Function to return the number
    // of derangements of n
    static int countDerangements(int n)
    {
        int []der = new int[ n + 3];
      
        der[0] = 1;
        der[1] = 0;
        der[2] = 1;
      
        for (int i = 3; i <= n; i++)
            der[i] = (i - 1) * (der[i - 1]
                            + der[i - 2]);
        return der[n];
    }
      
    // Function to return the required
    // number of permutations
    static int countPermutations(int n, int k)
    {
        int ans = 0;
        for (int i = n - k; i <= n; i++) {
      
            // Ways to choose i indices from n indices
            int ways = nCr(n, i);
      
            // Dearangements of (n - i) indices
            //System.out.println(ans) ;
            ans += (ways * countDerangements(n- i));
        }
        return ans;
    }
      
    // Driver Code to test above functions
    public static void Main()
    {
        int n = 5, k = 3;
        Console.WriteLine(countPermutations(n, k)) ;
    }
}
// This code is contributed by 29AjayKumar

PHP

 $n - $r)
        $r = $n - $r;
    for ($i = 0; $i < $r; $i++)
    {
        $ans *= ($n - $i);
        $ans /= ($i + 1);
    }
    return $ans;
}
 
// Function to return the number
// of derangements of n
function countDerangements($n)
{
    $der = array($n + 1);
 
    $der[0] = 1;
    $der[1] = 0;
    $der[2] = 1;
 
    for ($i = 3; $i <= $n; $i++)
        $der[$i] = ($i - 1) *
                   ($der[$i - 1] +
                    $der[$i - 2]);
    return $der[$n];
}
 
// Function to return the required
// number of permutations
function countPermutations($n, $k)
{
    $ans = 0;
    for ($i = $n - $k; $i <= $n; $i++)
    {
 
        // Ways to choose i indices
        // from n indices
        $ways = nCr($n, $i);
 
        // Dearangements of (n - i) indices
        $ans += $ways * countDerangements($n - $i);
    }
    return $ans;
}
 
// Driver Code
$n = 5;
$k = 3;
echo (countPermutations($n, $k));
 
// This code is contributed
// by Shivi_Aggarwal
?>

Javascript

 
      // JavaScript program to count the number
      // of required permutations
      // Function to return the number of ways
      // to choose r objects out of n objects
      function nCr(n, r) {
        var ans = 1;
        if (r > n - r) r = n - r;
        for (var i = 0; i < r; i++) {
          ans *= n - i;
          ans /= i + 1;
        }
        return ans;
      }
 
      // Function to return the number
      // of derangements of n
      function countDerangements(n) {
        var der = [...Array(n + 1)];
 
        der[0] = 1;
        der[1] = 0;
        der[2] = 1;
 
        for (var i = 3; i <= n; i++)
          der[i] = (i - 1) * (der[i - 1] + der[i - 2]);
        return der[n];
      }
 
      // Function to return the required
      // number of permutations
      function countPermutations(n, k) {
        var ans = 0;
        for (var i = n - k; i <= n; i++) {
          // Ways to choose i indices from n indices
          var ways = nCr(n, i);
 
          // Dearangements of (n - i) indices
          ans += ways * countDerangements(n - i);
        }
        return ans;
      }
 
      // Driver Code to test above functions
      var n = 5,
        k = 3;
      document.write(countPermutations(n, k));
       

输出：

31

时间复杂度： O(N^2)
辅助空间： O(N)

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