计算1/1系列和的有效程序！ + 1/2！ + 1/3！ + 1/4！ + .. + 1 / n！

给定一个正整数n，编写一个函数来计算1/1的和！ + 1/2！ + .. + 1 / n！
一个简单的解决方案是将sum初始化为0，然后运行一个循环并在循环内调用阶乘函数。
以下是简单解决方案的实现。

C++

// A simple C++ program to compute sum of series 1/1! + 1/2! + .. + 1/n!
#include 
using namespace std;
 
//  Utility function to find
int factorial(int n)
{
    int res = 1;
    for (int i=2; i<=n; i++)
       res *= i;
    return res;
}
 
// A Simple Function to return value of 1/1! + 1/2! + .. + 1/n!
double sum(int n)
{
    double sum = 0;
    for (int i = 1; i <= n; i++)
        sum += 1.0/factorial(i);
    return sum;
}
 
// Driver program to test above functions
int main()
{
    int n = 5;
    cout << sum(n);
    return 0;
}

Java

// A simple Java program to compute
// sum of series 1/1! + 1/2! + .. + 1/n!
import java.io.*;
 
class GFG {
     
    // Utility function to find
    static int factorial(int n)
    {
        int res = 1;
        for (int i = 2; i <= n; i++)
        res *= i;
        return res;
    }
     
    // A Simple Function to return value
    // of 1/1! + 1/2! + .. + 1/n!
    static double sum(int n)
    {
        double sum = 0;
        for (int i = 1; i <= n; i++)
            sum += 1.0/factorial(i);
        return sum;
    }
 
    // Driver program
    public static void main (String[] args)
    {
        int n = 5;
        System.out.println(sum(n));
    }
}
 
// This code is contributed by Ajit.

Python3

# Python3 program to compute sum of series
# 1/1! + 1/2! + .. + 1/n!
 
# Function to find factorail of a number
def factorial(n):
    res = 1
    for i in range(2, n + 1):
            res *= i
    return res
         
# A Simple Function to return value
# of 1/1! + 1/2! + .. + 1/n!
def sum(n):
    s = 0.0
     
    for i in range(1, n + 1):
        s += 1.0 / factorial(i)
    print(s)
 
# Driver program to test above functions
n = 5
sum(n)
 
# This code is contributed by Danish Raza

C#

// A simple C# program to compute sum
// of series 1/1! + 1/2! + .. + 1/n!
using System;
 
class GFG {
     
    // Utility function to find
    static int factorial(int n)
    {
        int res = 1;
        for (int i = 2; i <= n; i++)
            res *= i;
             
        return res;
    }
     
    // A Simple Function to return value
    // of 1/1! + 1/2! + .. + 1/n!
    static double sum(int n)
    {
        double sum = 0;
        for (int i = 1; i <= n; i++)
            sum += 1.0/factorial(i);
             
        return sum;
    }
 
    // Driver program
    public static void Main ()
    {
        int n = 5;
         
        Console.WriteLine(sum(n));
    }
}
 
// This code is contributed by Sam007.

Javascript

 
//Javascript program to compute
// sum of series 1/1! + 1/2! + .. + 1/n!
 
// Utility function to find
function factorial(n)
{
    let res = 1;
    for (let i = 2;  i <= n; i++)
        res *= i;
    return res;
}
 
// A Simple Function to return
// value of 1/1! + 1/2! + .. + 1/n!
function sum(n)
{
    let sum = 0;
    for (let i = 1; i <= n; i++)
        sum += 1.0 / factorial(i);
    return sum;
}
 
// Driver Code
 
let n = 5;
document.write(sum(n).toFixed(5));
 
// This code is contributed by sravan kumar
 

输出：

1.71667

上述解决方案的时间复杂度是O(n * n！)，这是巨大的。
一个有效的解决方案可以找到O(n)时间的总和。这个想法是在与求和相同的循环中计算阶乘。以下是该想法的实现。

C++

// A simple C++ program to compute sum of series 1/1! + 1/2! + .. + 1/n!
#include 
using namespace std;
 
// An Efficient Function to return value of 1/1! + 1/2! + .. + 1/n!
double sum(int n)
{
    double sum = 0;
    int fact = 1;
    for (int i = 1; i <= n; i++)
    {
       fact *= i;         // Update factorial
       sum += 1.0/fact;   // Update series sum
    }
    return sum;
}
 
// Driver program to test above functions
int main()
{
    int n = 5;
    cout << sum(n);
    return 0;
}

Java

// A simple Java program to compute
// sum of series 1/1! + 1/2! + .. + 1/n!
import java.io.*;
 
class GFG {
     
    // An Efficient Function to return
    // value of 1/1! + 1/2! + .. + 1/n!
    static double sum(int n)
    {
        double sum = 0;
        int fact = 1;
        for (int i = 1; i <= n; i++)
        {
            // Update factorial
            fact *= i;
             
            // Update series sum
            sum += 1.0/fact;
        }
        return sum;
    }
 
    // Driver program
    public static void main (String[] args)
    {
        int n = 5;
        System.out.println(sum(n));
    }
}
 
// This code is contributed by Ajit.

Python3

# Python3 program to compute sum of series
# 1/1! + 1/2! + .. + 1/n!
 
# Function to return value of
# 1/1! + 1/2! + .. + 1/n!
def sum(n):
    sum = 0
    fact = 1
 
    for i in range(1, n + 1):
 
        # Update factorial
        fact *= i
 
        # Update series sum
        sum += 1.0/fact
 
    print(sum)
 
# Driver program to test above functions
n = 5
sum(n)
 
# This code is contributed by Danish Raza

C＃

// A simple C# program to compute sum
// of series 1/1! + 1/2! + .. + 1/n!
using System;
 
class GFG {
     
    // An Efficient Function to return
    // value of 1/1! + 1/2! + .. + 1/n!
    static double sum(int n)
    {
        double sum = 0;
        int fact = 1;
         
        for (int i = 1; i <= n; i++)
        {
             
            // Update factorial
            fact *= i;
             
            // Update series sum
            sum += 1.0 / fact;
        }
        return sum;
    }
 
    // Driver program
    public static void Main ()
    {
        int n = 5;
         
        Console.WriteLine(sum(n));
    }
}
 
// This code is contributed by Sam007.

输出：

1.71667