Dado un número N , la tarea es encontrar el N-ésimo término de la serie donde cada término difiere en 6 y 2 alternativamente.
Ejemplos:
Entrada: N = 6
Salida: 24
Explicación:
El término N es 0 + 6 + 2 + 6 + 2 + 6 + 2 = 24
Entrada: N = 3
Salida: 14
Explicación:
El término N es 0 + 6 + 2 + 6 = 14
Enfoque ingenuo: la idea es iterar desde 1 con un incremento de 6 y 2 alternativamente, hasta llegar al N-ésimo término.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to find Nth term
void findNthTerm(int N)
{
int ans = 0;
// Iterate from 1 till Nth term
for (int i = 0; i < N; i++) {
// Check if i is even and
// then add 6
if (i % 2 == 0) {
ans = ans + 6;
}
// Else add 2
else {
ans = ans + 2;
}
}
// Print ans
cout << ans << endl;
}
// Driver Code
int main()
{
int N = 3;
findNthTerm(N);
return 0;
}
Java
// Java program for the above approach
class GFG{
// Function to find Nth term
static void findNthTerm(int N)
{
int ans = 0;
// Iterate from 1 till Nth term
for (int i = 0; i < N; i++) {
// Check if i is even and
// then add 6
if (i % 2 == 0) {
ans = ans + 6;
}
// Else add 2
else {
ans = ans + 2;
}
}
// Print ans
System.out.print(ans +"\n");
}
// Driver Code
public static void main(String[] args)
{
int N = 3;
findNthTerm(N);
}
}
// This code is contributed by PrinciRaj1992
Python3
# Python3 program for the above approach # Function to find Nth term def findNthTerm(N): ans = 0 # Iterate from 1 till Nth term for i in range(N): # Check if i is even and # then add 6 if (i % 2 == 0) : ans = ans + 6 # Else add 2 else : ans = ans + 2 # Print ans print(ans) # Driver Code if __name__=='__main__': N = 3 findNthTerm(N) # This code is contributed by AbhiThakur
C#
// C# program for the above approach
using System;
class GFG{
// Function to find Nth term
static void findNthTerm(int N)
{
int ans = 0;
// Iterate from 1 till Nth term
for (int i = 0; i < N; i++) {
// Check if i is even and
// then add 6
if (i % 2 == 0) {
ans = ans + 6;
}
// Else add 2
else {
ans = ans + 2;
}
}
// Print ans
Console.Write(ans +"\n");
}
// Driver Code
public static void Main()
{
int N = 3;
findNthTerm(N);
}
}
// This code is contributed by AbhiThakur
Javascript
<script>
// JavaScript program for the above approach
// Function to find Nth term
function findNthTerm(N)
{
let ans = 0;
// Iterate from 1 till Nth term
for (let i = 0; i < N; i++) {
// Check if i is even and
// then add 6
if (i % 2 == 0) {
ans = ans + 6;
}
// Else add 2
else {
ans = ans + 2;
}
}
// Print ans
document.write(ans + "<br>");
}
// Driver Code
let N = 3;
findNthTerm(N);
// This code is contributed by Mayank Tyagi
</script>
Producción:
14
Complejidad de tiempo: O(N)
Espacio Auxiliar: O(N)
Enfoque Eficiente: Podemos encontrar el N-ésimo término usando la siguiente fórmula:
- Si N es Impar: El N-ésimo término viene dado por (N/2 + 1)*6 + (N/2)*2.
- Si N es Par: El N-ésimo término viene dado por (N/2)*6 + (N/2)*2.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to find Nth term
void findNthTerm(int N)
{
int ans;
// Check if N is even
if (N % 2 == 0) {
// Formula for n is even
ans = (N / 2) * 6
+ (N / 2) * 2;
}
// Check if N is odd
else {
// Formula for N is odd
ans = (N / 2 + 1) * 6
+ (N / 2) * 2;
}
// Print ans
cout << ans << endl;
}
// Driver Code
int main()
{
int N = 3;
findNthTerm(N);
return 0;
}
Java
// Java program for the above approach
class GFG{
// Function to find Nth term
static void findNthTerm(int N)
{
int ans;
// Check if N is even
if (N % 2 == 0) {
// Formula for n is even
ans = (N / 2) * 6
+ (N / 2) * 2;
}
// Check if N is odd
else {
// Formula for N is odd
ans = (N / 2 + 1) * 6
+ (N / 2) * 2;
}
// Print ans
System.out.print(ans +"\n");
}
// Driver Code
public static void main(String[] args)
{
int N = 3;
findNthTerm(N);
}
}
// This code contributed by PrinciRaj1992
Python3
# Python3 program for the above approach # Function to find Nth term def findNthTerm(N): ans = 0; # Check if N is even if (N % 2 == 0): # Formula for n is even ans = (N // 2) * 6 + (N // 2) * 2; # Check if N is odd else: # Formula for N is odd ans = (N // 2 + 1) * 6 + (N // 2) * 2; # Print ans print(ans); # Driver Code if __name__ == '__main__': N = 3; findNthTerm(N); # This code is contributed by Rajput-Ji
C#
// C# program for the above approach
using System;
class GFG{
// Function to find Nth term
static void findNthTerm(int N)
{
int ans;
// Check if N is even
if (N % 2 == 0) {
// Formula for n is even
ans = (N / 2) * 6
+ (N / 2) * 2;
}
// Check if N is odd
else {
// Formula for N is odd
ans = (N / 2 + 1) * 6
+ (N / 2) * 2;
}
// Print ans
Console.Write(ans +"\n");
}
// Driver Code
public static void Main(String[] args)
{
int N = 3;
findNthTerm(N);
}
}
// This code is contributed by PrinciRaj1992
Javascript
<script>
// JavaScript program for the above approach
// Function to find Nth term
function findNthTerm( N)
{
let ans;
// Check if N is even
if (N % 2 == 0) {
// Formula for n is even
ans = parseInt(N / 2) * 6
+ parseInt(N / 2) * 2;
}
// Check if N is odd
else {
// Formula for N is odd
ans = parseInt(N / 2 + 1) * 6
+ parseInt(N / 2) * 2;
}
// Print ans
document.write(ans);
}
// Driver Function
// get the value of N
let N = 3;
// Calculate and print the Nth term
findNthTerm(N);
// This code is contributed by todaysgaurav
</script>
Producción:
14
Complejidad de tiempo: O(1)
Espacio Auxiliar : O(1) ya que usa variables constantes
Publicación traducida automáticamente
Artículo escrito por rakshitarora y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA