Dado un entero positivo N , ¡la tarea es encontrar la suma de la serie 1! – 2! + 3! – 4! + 5!… hasta el N-ésimo término.
Ejemplos:
Entrada: N = 6
Salida: -619
Explicación: ¡ La suma de la serie hasta el quinto término se puede calcular como
1! – 2! + 3! – 4! + 5! -6! = 1 -2 +6 -24 +120 -720 = -619Entrada: N = 5
Salida: 101
Enfoque Nativo: La forma más sencilla de resolver este problema es encontrar el factorial de todos los números en el rango [1, N] y calcular su suma con sus respectivos signos positivo y negativo, es decir, la posición impar será (+) ve y par la posición será negativa.
A continuación se muestra la implementación del enfoque anterior:
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
// 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
Python
# Python program of the above approach # Function to find factorial # of a given number def factorial(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 def SeriesSum(N): # Stores Required Sum Sum = 0 # Loop to calculate factorial # and sum them with their # respective sign for i in range(1, N + 1): # Factorial of cur integer fact = factorial(i); # Stores the sign 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 N = 6 print(SeriesSum(N)) # This code is contributed by Samim Hossain Mondal.
C#
// C# implementation for the above approach
using System;
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()
{
int N = 6;
Console.Write(SeriesSum(N));
}
}
// This code is contributed by Samim Hossain Mondal.
Javascript
<script>
// JavaScript code for the above approach
// Function to find factorial
// of a given number
function factorial(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
function SeriesSum(N) {
// Stores Required Sum
let Sum = 0;
// Loop to calculate factorial
// and sum them with their
// respective sign
for (let i = 1; i <= N; i++) {
// Factorial of cur integer
let fact = factorial(i);
// Stores the sign
let 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
let N = 6;
document.write(SeriesSum(N));
// This code is contributed by Potta Lokesh
</script>
Producción
-619
Complejidad de Tiempo: O(N 2 )
Espacio Auxiliar: O(1)
Enfoque eficiente: la solución anterior se puede optimizar manteniendo el valor del factorial del número anterior y calculando el factorial del número actual usando ese valor y calculando su suma con su respectivo signo positivo y negativo.
A continuación se muestra la implementación del enfoque anterior:
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;
}
Java
// Java program of the above approach
import java.util.*;
class GFG{
// Function to find the sum of
// the series 1! - 2! + 3! - 4!
// + 5!... till the Nth term
static 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
public static void main(String[] args)
{
int N = 6;
System.out.print(SeriesSum(N));
}
}
// This code is contributed by 29AjayKumar
Python3
# Python 3 program of the above approach # Function to find the sum of # the series 1! - 2! + 3! - 4! # + 5!... till the Nth term def SeriesSum(N): # Stores the required sum # and factorial value Sum = 0 fact = 1 # Loop to calculate factorial # and sum them with their # respective sign for i in range(1, N + 1): # Calculate factorial fact = fact * i # Stores the sign 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 if __name__ == "__main__": N = 6 print(SeriesSum(N)) # This code is contributed by ukasp.
C#
// C# program of the above approach
using System;
public class GFG{
// Function to find the sum of
// the series 1! - 2! + 3! - 4!
// + 5!... till the Nth term
static 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
public static void Main(String[] args)
{
int N = 6;
Console.Write(SeriesSum(N));
}
}
// This code is contributed by 29AjayKumar
Javascript
<script>
// javascript program of the above approach
// Function to find the sum of
// the series 1! - 2! + 3! - 4!
// + 5!... till the Nth term
function SeriesSum(N)
{
// Stores the required sum
// and factorial value
var Sum = 0, fact = 1;
// Loop to calculate factorial
// and sum them with their
// respective sign
for (var i = 1; i <= N; i++) {
// Calculate factorial
fact = fact * i;
// Stores the sign
var 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
var N = 6;
document.write(SeriesSum(N));
// This code is contributed by 29AjayKumar
</script>
Producción
-619
Complejidad temporal: O(N)
Espacio auxiliar: O(1)
Publicación traducida automáticamente
Artículo escrito por athakur42u y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA