求数列 1 的和！– 2!+ 3!– 4!+ 5!。。。直到第 n 个学期
原文:https://www . geesforgeks . org/find-sum-the-series-1-2-3-4-5-till-n-term/
给定一个正整数 N ，任务是求数列 1 的和！– 2!+ 3!– 4!+ 5!…直到第 n 个学期。
示例:
输入: N = 6 输出: -619 说明:到第 5 项的级数之和可以计算为 1！– 2!+ 3!– 4!+ 5!-6!= 1 -2 +6 -24 +120 -720 = -619
输入:N = 5 T3】输出: 101
天真法:解决这个问题最简单的方法就是在【1，N】范围内找到所有数字的阶乘，用它们各自的正负符号计算它们的和，即奇数位置为(+)ve，偶数位置为负数。
下面是上述方法的实现:
C++
// C++ program of the above approach
#include <iostream>
using namespace std;
// Function to find factorial
// of a given number
int factorial(int N)
{
if (N == 1) {
return 1;
}
// Recursive Call
return N * factorial(N - 1);
}
// Function to find the sum of
// the series 1! - 2! + 3! - 4!
// + 5!... till the Nth term
int SeriesSum(int N)
{
// Stores Required Sum
int Sum = 0;
// Loop to calculate factorial
// and sum them with their
// respective sign
for (int i = 1; i <= N; i++) {
// Factorial of cur integer
int fact = factorial(i);
// Stores the sign
int sign = fact;
// If position is even sign
// must be negative
if (i % 2 == 0) {
sign = sign * -1;
}
// Adding value in sum
Sum += sign;
}
// Return Answer
return Sum;
}
// Driver Code
int main()
{
int N = 6;
cout << SeriesSum(N);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java implementation for the above approach
import java.util.*;
class GFG{
// Function to find factorial
// of a given number
static int factorial(int N)
{
if (N == 1)
{
return 1;
}
// Recursive Call
return N * factorial(N - 1);
}
// Function to find the sum of
// the series 1! - 2! + 3! - 4!
// + 5!... till the Nth term
static int SeriesSum(int N)
{
// Stores Required Sum
int Sum = 0;
// Loop to calculate factorial
// and sum them with their
// respective sign
for(int i = 1; i <= N; i++)
{
// Factorial of cur integer
int fact = factorial(i);
// Stores the sign
int sign = fact;
// If position is even sign
// must be negative
if (i % 2 == 0)
{
sign = sign * -1;
}
// Adding value in sum
Sum += sign;
}
// Return Answer
return Sum;
}
// Driver Code
public static void main(String args[])
{
int N = 6;
System.out.print(SeriesSum(N));
}
}
// This code is contributed by sanjoy_62
Output
-619
时间复杂度:O(N2) 辅助空间: O(1)
有效方法:上述解决方案可以通过保持前一个数的阶乘的值，并使用该值计算当前数的阶乘，以及用它们各自的正负符号计算它们的和来优化。
下面是上述方法的实现:
C++
// C++ program of the above approach
#include <iostream>
using namespace std;
// Function to find the sum of
// the series 1! - 2! + 3! - 4!
// + 5!... till the Nth term
int SeriesSum(int N)
{
// Stores the required sum
// and factorial value
int Sum = 0, fact = 1;
// Loop to calculate factorial
// and sum them with their
// respective sign
for (int i = 1; i <= N; i++) {
// Calculate factorial
fact = fact * i;
// Stores the sign
int sign = fact;
// If position is even sign
// must be negative
if (i % 2 == 0) {
sign = sign * -1;
}
// Adding value in sum
Sum += sign;
}
// Return Answer
return Sum;
}
// Driver Code
int main()
{
int N = 6;
cout << SeriesSum(N);
return 0;
}
Output
-619
时间复杂度:O(N) T5辅助空间:** O(1)
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