配对方式的数量

假设一个聚会有 p 个人。每个人都可以作为一个人加入舞蹈，也可以与任何其他人一起跳舞。任务是找出 p 个人可以加入舞蹈的不同方式的数量。
例子：

Input : p = 3
Output : 4
Let the three people be P1, P2 and P3
Different ways are: {P1, P2, P3}, {{P1, P2}, P3},
{{P1, P3}, P2} and {{P2, P3}, P1}.

Input : p = 2
Output : 2
The groups are: {P1, P2} and {{P1, P2}}.

方法：思路是用动态规划来解决这个问题。有两种情况：一个人单独或成对加入舞蹈。对于第一种情况，问题简化为找到 p-1 人的解决方案。对于第二种情况，有 p-1 个选择来选择一个个体进行配对，在选择一个个体进行配对后，问题简化为找到 p-2 个人的解决方案，因为 p 中的两个人已经配对。
所以dp的公式是：

dp[p] = dp[p-1] + (p-1) * dp[p-2].

下面是上述方法的实现：

C++

// CPP program to find number of ways to
// pair people in party
 
#include 
using namespace std;
 
// Function to find number of ways to
// pair people in party
int findWaysToPair(int p)
{
    // To store count of number of ways.
    int dp[p + 1];
 
    dp[1] = 1;
    dp[2] = 2;
 
    // Using the recurrence defined find
    // count for different values of p.
    for (int i = 3; i <= p; i++) {
        dp[i] = dp[i - 1] + (i - 1) * dp[i - 2];
    }
 
    return dp[p];
}
 
// Driver code
int main()
{
    int p = 3;
    cout << findWaysToPair(p);
    return 0;
}

Java

// Java program to find number of ways to
// pair people in party
 
class GFG
{
     
// Function to find number of ways to
// pair people in party
static int findWaysToPair(int p)
{
    // To store count of number of ways.
    int dp[] = new int[p + 1];
 
    dp[1] = 1;
    dp[2] = 2;
 
    // Using the recurrence defined find
    // count for different values of p.
    for (int i = 3; i <= p; i++)
    {
        dp[i] = dp[i - 1] + (i - 1) * dp[i - 2];
    }
 
    return dp[p];
}
 
// Driver code
public static void main(String args[])
{
    int p = 3;
    System.out.println(findWaysToPair(p));
}
}
 
// This code is contributed by Arnab Kundu

Python3

# Python3 program to find number of
# ways to pair people in party
 
# Function to find number of ways
# to pair people in party
def findWays(p):
 
    # To store count of number of ways.
    dp = [0] * (p + 1)
    dp[1] = 1
    dp[2] = 2
 
    # Using the recurrence defined find
    # count for different values of p.
    for i in range(3, p + 1):
        dp[i] = (dp[i - 1] +
                   (i - 1) * dp[i - 2])
    return dp[p]
 
# Driver code
p = 3
print(findWays(p))
 
# This code is contributed by Shrikant13

C#

// C# program to find number of ways to
// pair people in party
using System;
 
class GFG
{
 
// Function to find number of ways to
// pair people in party
public static int findWaysToPair(int p)
{
    // To store count of number of ways.
    int[] dp = new int[p + 1];
 
    dp[1] = 1;
    dp[2] = 2;
 
    // Using the recurrence defined find
    // count for different values of p.
    for (int i = 3; i <= p; i++)
    {
        dp[i] = dp[i - 1] + (i - 1) * dp[i - 2];
    }
    return dp[p];
}
 
// Driver code
public static void Main(string[] args)
{
    int p = 3;
    Console.WriteLine(findWaysToPair(p));
}
}
 
// This code is contributed by shrikanth13

Javascript

// Javascript program to find number of ways to
// pair people in party
 
// Function to find number of ways to
// pair people in party
function findWaysToPair(p)
{
    // To store count of number of ways.
    var dp = Array(p+1);
 
    dp[1] = 1;
    dp[2] = 2;
 
    // Using the recurrence defined find
    // count for different values of p.
    for (var i = 3; i <= p; i++) {
        dp[i] = dp[i - 1] + (i - 1) * dp[i - 2];
    }
 
    return dp[p];
}
 
// Driver code
var p = 3;
document.write( findWaysToPair(p));
 
// This code is contributed by noob2000.

输出：

4

时间复杂度： O(p)
辅助空间： O(p)

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