找到一个数字的子工厂

原文:https://www . geesforgeks . org/find-subfactory-of-a-number/

给定一个整数 N，任务是找到表示为的数字的子工厂！编号使用编号 N 的递归关系定义编号的子因子:

！N = (N-1) [！(N-2) +！(N-1) ]

在哪里！1 = 0 和！0 = 1

一些子工厂是:

| *n* | Zero | one | Two | three | four | five | six | seven | eight | nine | Ten | Eleven | Twelve | Thirteen | | ！ *n* | one | Zero | one | Two | nine | forty-four | Two hundred and sixty-five | 1, 854 | 14, 833 | 133, 496 | 1, 334, 961 | 14, 684, 570 | 176, 214, 841 | 2, 290, 792, 932 |

示例:

输入: N = 4 输出: 9 解释: ！4 = !(4-1)4 + (-1) 4 =！34 + 1 ！3 = !(3–1) 3+(-1)3=！2 * 3–1 ！2 = !(2–1) 2+(-1)2=！12 + 1 ！1 = !(1–1) 1+(-1)1=！0 * 1–1 自！因此，0 = 1！1 = 0, !2 = 1, !3 = 2 和！4 = 9.

输入:N = 0 T3】输出: 1

方法:数量 N 的子工厂也可以计算为:

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

扩展这一点

{\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，上面的系列可以用来寻找数字 n 的子工厂。按照下面的步骤，看看如何:

  • 初始化变量，比如 res = 0 、 fact = 1 和 count = 0 。
  • 使用 i 迭代从 1 到NT5 的范围，并执行以下操作:
    • 将事实更新为事实*i.
    • 如果计数为偶数，则将 res 更新为RES = RES–(1/事实)。
    • 如果计数为奇数，则将 res 更新为 res = res + (1 /事实)。
    • 将计数值增加 1。
  • 最后，返回事实*(1 + res) 。

下面是上述方法的实现:

C++

/// C++ program for the above approach
#include <iostream>
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 语言(一种计算机语言，尤用于创建网站)

/// 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.

Python 3

# 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.

java 描述语言

<script>
        // 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
    </script>

Output

9

时间复杂度:O(N) T5辅助空间:** O(1)

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