查找数字的子阶乘

给定一个整数N，任务是找到表示为!N的数字的子阶乘。一个数的子阶乘是使用以下数N的递归关系定义的：

!N = (N-1) [ !(N-2) + !(N-1) ]

where !1 = 0 and !0 = 1

一些子因子是：

n	0	1	2	3	4	5	6	7	8	9	10	11	12	13
!n	1	0	1	2	9	44	265	1, 854	14, 833	133, 496	1, 334, 961	14, 684, 570	176, 214, 841	2, 290, 792, 932

例子：

Input: N = 4
Output: 9
Explanation: 
!4 = !(4-1)*4 + (-1)⁴ = !3*4 + 1
!3 = !(3 – 1)*3 + (-1)³ = !2*3 – 1
!2 = !(2 – 1)*2 + (-1)² = !1*2 + 1
!1 = !(1 – 1)*1 + (-1)¹ = !0*1 – 1
Since !0 = 1, therefore !1 = 0, !2 = 1, !3 = 2 and !4 = 9. 

Input: N = 0
Output: 1

方法：数N的子阶乘也可以计算为：

${\displaystyle !n={\begin{cases}1&{\text{if }}n=0, \\n\left(!(n-1)\right)+(-1)^{n}&{\text{if }}n>0.\end{cases}}}$

Expanding this gives

${\displaystyle !n=\sum _{i=0}^{n-1}i(!i)+{\frac {1+(-1)^{n}}{2}}}$

=> !N = ( N! )*( 1 – 1/(1!) + (1/2!) – (1/3!)  …….. (1/N!)*(-1)^N )

因此，上述系列可用于查找数字 N 的子阶乘。请按照以下步骤查看如何：

• 初始化变量，比如res = 0 、 fact = 1和count = 0 。
• 使用i遍历从1到N的范围并执行以下操作：
  • 将事实更新为事实*i。
  • 如果计数是偶数，则将 res 更新为res = res – (1 / fact) 。
  • 如果计数是奇数，则将 res 更新为res = res + (1 / fact) 。
  • 将 count 的值增加1。
• 最后，返回fact*(1 + res) 。

下面是上述方法的实现：

C++

/// C++ program for the above approach
#include 
using namespace std;
 
// Function to find the subfactorial
// of the number
double subfactorial(int N)
{
 
    // Initialize variables
    double res = 0, fact = 1;
    int count = 0;
 
    // Iterating over range N
    for (int i = 1; i <= N; i++) {
 
        // Fact variable store
        // factorial of the i
        fact = fact * i;
 
        // If count is even
        if (count % 2 == 0)
            res = res - (1 / fact);
        else
            res = res + (1 / fact);
 
        // Increase the value of
        // count by 1
        count++;
    }
 
    return fact * (1 + res);
}
 
// Driver Code
int main()
{
    int N = 4;
    cout << subfactorial(N);
 
    return 0;
}

Java

/// Java program for the above approach
import java.util.*;
 
class GFG {
 
    // Function to find the subfactorial
    // of the number
    static double subfactorial(int N)
    {
 
        // Initialize variables
        double res = 0, fact = 1;
        int count = 0;
 
        // Iterating over range N
        for (int i = 1; i <= N; i++) {
 
            // Fact variable store
            // factorial of the i
            fact = fact * i;
 
            // If count is even
            if (count % 2 == 0)
                res = res - (1 / fact);
            else
                res = res + (1 / fact);
 
            // Increase the value of
            // count by 1
            count++;
        }
 
        return fact * (1 + res);
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int N = 4;
        System.out.println((int)(subfactorial(N)));
    }
}
 
// This code is contributed by ukasp.

Python3

# python program for the above approach
 
# Function to find the subfactorial
# of the number
def subfactorial(N):
 
        # Initialize variables
    res = 0
    fact = 1
    count = 0
 
    # Iterating over range N
    for i in range(1, N+1):
 
                # Fact variable store
                # factorial of the i
        fact = fact * i
 
        # If count is even
        if (count % 2 == 0):
            res = res - (1 / fact)
 
        else:
            res = res + (1 / fact)
 
            # Increase the value of
            # count by 1
        count += 1
 
    return fact * (1 + res)
 
# Driver Code
if __name__ == "__main__":
 
    N = 4
    print(subfactorial(N))
 
    # This code is contributed by rakeshsahni

C#

/// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG{
 
// Function to find the subfactorial
// of the number
static double subfactorial(int N)
{
 
    // Initialize variables
    double res = 0, fact = 1;
    int count = 0;
 
    // Iterating over range N
    for (int i = 1; i <= N; i++) {
 
        // Fact variable store
        // factorial of the i
        fact = fact * i;
 
        // If count is even
        if (count % 2 == 0)
            res = res - (1 / fact);
        else
            res = res + (1 / fact);
 
        // Increase the value of
        // count by 1
        count++;
    }
 
    return fact * (1 + res);
}
 
// Driver Code
public static void Main()
{
    int N = 4;
    Console.Write(subfactorial(N));
}
}
 
// This code is contributed by ipg2016107.

Javascript

        // JavaScript Program to implement
        // the above approach
 
        // Function to find the subfactorial
        // of the number
        function subfactorial(N) {
 
            // Initialize variables
            let res = 0, fact = 1;
            let count = 0;
 
            // Iterating over range N
            for (let i = 1; i <= N; i++) {
 
                // Fact variable store
                // factorial of the i
                fact = fact * i;
 
                // If count is even
                if (count % 2 == 0)
                    res = res - 1 / fact;
                else
                    res = res + 1 / fact;
 
                // Increase the value of
                // count by 1
                count++;
            }
 
            return fact * (1 + res);
        }
 
        // Driver Code
        let N = 4;
        document.write(subfactorial(N));
 
// This code is contributed by Potta Lokesh
    

输出

9

时间复杂度： O(N)
辅助空间： O(1)