Dado un número entero N , la tarea es encontrar la suma de los primeros N números naturales con signos alternos, es decir, 1 – 2 + 3 – 4 + 5 – 6 + ….
Ejemplos:
Entrada: N = 6
Salida: -3
Explicación:
1 – 2 + 3 – 4 + 5 – 6 = -3
Por lo tanto, la salida requerida es -3.Entrada: N = 5
Salida: 3
Explicación:
1 – 2 + 3 – 4 + 5 = 3
Por lo tanto, la salida requerida = 3
Enfoque ingenuo: siga los pasos a continuación para resolver el problema:
- Inicialice una variable, diga alternateSum para almacenar la suma del signo alternativo de los primeros N números naturales.
- Itere sobre el rango [1, N] usando la variable i y verifique si i es par o no . Si se determina que es cierto, actualice la suma alternativa += -i .
- De lo contrario, actualice alternateSum += i .
- Finalmente, imprima el valor de alternateSum .
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to find the sum of first
// N natural numbers with alternate signs
int alternatingSumOfFirst_N(int N)
{
// Stores sum of alternate sign
// of first N natural numbers
int alternateSum = 0;
for (int i = 1; i <= N; i++) {
// If is an even number
if (i % 2 == 0) {
// Update alternateSum
alternateSum += -i;
}
// If i is an odd number
else {
// Update alternateSum
alternateSum += i;
}
}
return alternateSum;
}
// Driver Code
int main()
{
int N = 6;
cout<<alternatingSumOfFirst_N(N);
return 0;
}
Java
// Java program to implement
// the above approach
import java.util.*;
class GFG{
// Function to find the sum of first
// N natural numbers with alternate signs
static int alternatingSumOfFirst_N(int N)
{
// Stores sum of alternate sign
// of first N natural numbers
int alternateSum = 0;
for(int i = 1; i <= N; i++)
{
// If is an even number
if (i % 2 == 0)
{
// Update alternateSum
alternateSum += -i;
}
// If i is an odd number
else
{
// Update alternateSum
alternateSum += i;
}
}
return alternateSum;
}
// Driver Code
public static void main(String[] args)
{
int N = 6;
System.out.print(alternatingSumOfFirst_N(N));
}
}
// This code is contributed by Amit Katiyar
Python3
# Python3 program to implement # the above approach # Function to find the sum of # First N natural numbers with # alternate signs def alternatingSumOfFirst_N(N): # Stores sum of alternate sign # of First N natural numbers alternateSum = 0 for i in range(1, N + 1): # If is an even number if (i % 2 == 0): # Update alternateSum alternateSum += -i # If i is an odd number else: alternateSum += i return alternateSum # Driver Code if __name__ == "__main__" : N = 6 print(alternatingSumOfFirst_N(N)) # This code is contributed by Virusbuddah_
C#
// C# program to implement
// the above approach
using System;
class GFG{
// Function to find the sum of first
// N natural numbers with alternate signs
static int alternatingSumOfFirst_N(int N)
{
// Stores sum of alternate sign
// of first N natural numbers
int alternateSum = 0;
for(int i = 1; i <= N; i++)
{
// If is an even number
if (i % 2 == 0)
{
// Update alternateSum
alternateSum += -i;
}
// If i is an odd number
else
{
// Update alternateSum
alternateSum += i;
}
}
return alternateSum;
}
// Driver Code
public static void Main(String[] args)
{
int N = 6;
Console.Write(alternatingSumOfFirst_N(N));
}
}
// This code is contributed by 29AjayKumar
Javascript
<script>
// Javascript program to implement
// the above approach
// Function to find the sum of first
// N natural numbers with alternate signs
function alternatingSumOfFirst_N(N)
{
// Stores sum of alternate sign
// of first N natural numbers
var alternateSum = 0;
for (i = 1; i <= N; i++) {
// If is an even number
if (i % 2 == 0) {
// Update alternateSum
alternateSum += -i;
}
// If i is an odd number
else {
// Update alternateSum
alternateSum += i;
}
}
return alternateSum;
}
// Driver Code
var N = 6;
document.write(alternatingSumOfFirst_N(N));
// This code is contributed by Rajput-Ji
</script>
Producción:
-3
Complejidad temporal: O(N)
Espacio auxiliar: O(1)
Enfoque eficiente: Para optimizar el enfoque anterior, la idea se basa en las siguientes observaciones:
Si N es un número par, la suma de los signos alternos de los primeros N números naturales es = (-N) / 2.
Si N es un número impar, la suma de los signos alternos de los primeros N números naturales es = (N + 1) / 2.
Siga los pasos a continuación para resolver el problema:
- Inicialice una variable, diga alternateSum para almacenar la suma del signo alternativo de los primeros N números naturales.
- Comprueba si N es un número par o no . Si se determina que es cierto, actualice la suma alternativa = (-N) / 2 .
- De lo contrario, actualice alternateSum = (N + 1) / 2 .
- Finalmente, imprima el valor de alternateSum .
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to find the sum of first
// N natural numbers with alternate signs
int alternatingSumOfFirst_N(int N)
{
// Stores sum of alternate sign
// of first N natural numbers
int alternateSum = 0;
// If N is an even number
if (N % 2 == 0) {
// Update alternateSum
alternateSum = (-N) / 2;
}
// If N is an odd number
else {
// Update alternateSum
alternateSum = (N + 1) / 2;
}
return alternateSum;
}
// Driver Code
int main()
{
int N = 6;
cout<<alternatingSumOfFirst_N(N);
return 0;
}
Java
// Java program to implement
// the above approach
import java.util.*;
class GFG
{
// Function to find the sum of first
// N natural numbers with alternate signs
static int alternatingSumOfFirst_N(int N)
{
// Stores sum of alternate sign
// of first N natural numbers
int alternateSum = 0;
// If N is an even number
if (N % 2 == 0)
{
// Update alternateSum
alternateSum = (-N) / 2;
}
// If N is an odd number
else
{
// Update alternateSum
alternateSum = (N + 1) / 2;
}
return alternateSum;
}
// Driver Code
public static void main(String[] args)
{
int N = 6;
System.out.print(alternatingSumOfFirst_N(N));
}
}
// This code is contributed by 29AjayKumar
Python3
# Python program to implement # the above approach # Function to find the sum of first # N natural numbers with alternate signs def alternatingSumOfFirst_N(N): # Stores sum of alternate sign # of first N natural numbers alternateSum = 0; # If N is an even number if (N % 2 == 0): # Update alternateSum alternateSum = (-N) // 2; # If N is an odd number else: # Update alternateSum alternateSum = (N + 1) // 2; return alternateSum; # Driver Code if __name__ == '__main__': N = 6; print(alternatingSumOfFirst_N(N)); # This code contributed by shikhasingrajput
C#
// C# program to implement
// the above approach
using System;
class GFG
{
// Function to find the sum of first
// N natural numbers with alternate signs
static int alternatingSumOfFirst_N(int N)
{
// Stores sum of alternate sign
// of first N natural numbers
int alternateSum = 0;
// If N is an even number
if (N % 2 == 0)
{
// Update alternateSum
alternateSum = (-N) / 2;
}
// If N is an odd number
else
{
// Update alternateSum
alternateSum = (N + 1) / 2;
}
return alternateSum;
}
// Driver Code
public static void Main(String[] args)
{
int N = 6;
Console.Write(alternatingSumOfFirst_N(N));
}
}
// This code is contributed by 29AjayKumar
Javascript
<script>
// Javascript program to implement
// the above approach
// Function to find the sum of first
// N natural numbers with alternate signs
function alternatingSumOfFirst_N(N)
{
// Stores sum of alternate sign
// of first N natural numbers
var alternateSum = 0;
// If N is an even number
if (N % 2 == 0) {
// Update alternateSum
alternateSum = (-N) / 2;
}
// If N is an odd number
else {
// Update alternateSum
alternateSum = (N + 1) / 2;
}
return alternateSum;
}
// Driver Code
var N = 6;
document.write(alternatingSumOfFirst_N(N));
// This code contributed by Rajput-Ji
</script>
Producción:
-3
Tiempo Complejidad: O(1)
Espacio Auxiliar: O(1)
Publicación traducida automáticamente
Artículo escrito por karangargabc y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA