求阶乘之和直到 N 阶乘(1！+ 2!+ 3!+ … + N！)
原文:https://www . geesforgeks . org/find-阶乘和-till-n-阶乘-1-2-3-n/
给定正整数 N 。任务是从 1 开始计算阶乘的和！去 N！， 1！+ 2!+ 3!+ … + N！。
示例:
输入 : N = 5 输出 : 153 解释 : 1！+ 2!+ 3!+ 4!+ 5!= 1 + 2 + 6 + 24 + 120 = 153.
输入 : N = 1 输出 : 1
天真的做法:解决这个问题的基本方法是找到所有数字直到 1 到 N 的阶乘，并计算它们的和。 时间复杂度 : O(N^2) 辅助空间 : O(1)
方法:一个有效的方法是在同一个循环中计算阶乘和和，使时间为 O(N)。遍历从 1 到 N 的数字，对于每个数字 i:
- 将 I 乘以先前的阶乘(最初为 1)。
- 将这个新阶乘加到一个集合和上
最后，打印这个集体总和。
下面是上述方法的实现。
C++
// C++ program to compute sum of series
// 1! + 2! + 3! + ... + N!
#include <iostream>
using namespace std;
// Function to return sum
// of 1!, 2! upto N!
int findFactSum(int N)
{
// Initializing the variables
int f = 1, Sum = 0;
// Calculate the factorial and sum
// in the same loop
for (int i = 1; i <= N; i++) {
f = f * i;
Sum += f;
}
// Return Sum as the final result.
return Sum;
}
// Driver Code
int main()
{
int N = 5;
// Function call
cout << findFactSum(N);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java code to implement above approach
class GFG {
// Function to return sum
// of 1!, 2! upto N!
static int findFactSum(int N)
{
// Initializing the variables
int f = 1, Sum = 0;
// Calculate the factorial and sum
// in the same loop
for (int i = 1; i <= N; i++) {
f = f * i;
Sum += f;
}
// Return Sum as the final result.
return Sum;
}
// Driver code
public static void main(String[] args)
{
int N = 5;
System.out.print(findFactSum(N));
}
}
// This code is contributed ukasp.
Python 3
# python program to compute sum of series
# 1! + 2! + 3! + ... + N!
# Function to return sum
# of 1!, 2! upto N!
def findFactSum(N):
# Initializing the variables
f = 1
Sum = 0
# Calculate the factorial and sum
# in the same loop
for i in range(1, N + 1):
f = f * i
Sum += f
# Return Sum as the final result.
return Sum
# Driver Code
if __name__ == "__main__":
N = 5
# Function call
print(findFactSum(N))
# This code is contributed by rakeshsahni
C
// C# code to implement above approach
using System;
class GFG
{
// Function to return sum
// of 1!, 2! upto N!
static int findFactSum(int N)
{
// Initializing the variables
int f = 1, Sum = 0;
// Calculate the factorial and sum
// in the same loop
for (int i = 1; i <= N; i++) {
f = f * i;
Sum += f;
}
// Return Sum as the final result.
return Sum;
}
// Driver code
public static void Main()
{
int N = 5;
Console.Write(findFactSum(N));
}
}
// This code is contributed by Samim Hossain Mondal.
java 描述语言
<script>
// JavaScript code for the above approach
// Function to return sum
// of 1!, 2! upto N!
function findFactSum(N)
{
// Initializing the variables
let f = 1, Sum = 0;
// Calculate the factorial and sum
// in the same loop
for (let i = 1; i <= N; i++) {
f = f * i;
Sum += f;
}
// Return Sum as the final result.
return Sum;
}
// Driver Code
let N = 5;
// Function call
document.write(findFactSum(N));
// This code is contributed by Potta Lokesh
</script>
Output:
153
时间复杂度 : O(N) 辅助空间 : O(1)
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