Dado un número N. La tarea es escribir un programa para encontrar el N-ésimo término en la siguiente serie:
1, 3, 12, 60, 360…
Ejemplos:
Input: 2 Output: 3 Input: 4 Output: 60
Enfoque: ¡ La idea es encontrar primero el factorial del número (N+1), es decir (N+1)!
Ahora, el N-ésimo término de la serie anterior será:
N-th term = (N+1)!/2
A continuación se muestra la implementación del enfoque anterior:
C++
// CPP program to find N-th term of the series: // 1, 3, 12, 60, 360… #include <iostream> using namespace std; // Function to find factorial of N int factorial(int N) { int fact = 1; for (int i = 1; i <= N; i++) fact = fact * i; // return factorial of N+1 return fact; } // calculate Nth term of series int nthTerm(int N) { return (factorial(N + 1) / 2); } // Driver Function int main() { // Taking n as 6 int N = 6; // Printing the nth term cout << nthTerm(N); return 0; }
C
// C program to find N-th term of the series: // 1, 3, 12, 60, 360… #include <stdio.h> // Function to find factorial of N int factorial(int N) { int fact = 1; for (int i = 1; i <= N; i++) fact = fact * i; // return factorial of N+1 return fact; } // calculate Nth term of series int nthTerm(int N) { return (factorial(N + 1) / 2); } // Driver Function int main() { // Taking n as 6 int N = 6; // Printing the nth term printf("%d",nthTerm(N)); return 0; } // This code is contributed by kothavvsaakash.
Java
// Java program to find N-th // term of the series: // 1, 3, 12, 60, 360 import java.util.*; import java.lang.*; import java.io.*; class GFG { // Function to find factorial of N static int factorial(int N) { int fact = 1; for (int i = 1; i <= N; i++) fact = fact * i; // return factorial of N return fact; } // calculate Nth term of series static int nthTerm(int N) { return (factorial(N + 1) / 2); } // Driver Code public static void main(String args[]) { // Taking n as 6 int N = 6; // Printing the nth term System.out.println(nthTerm(N)); } }
Python3
# Python 3 program to find # N-th term of the series: # 1, 3, 12, 60, 360… # Function for finding # factorial of N def factorial(N) : fact = 1 for i in range(1, N + 1) : fact = fact * i # return factorial of N return fact # Function for calculating # Nth term of series def nthTerm(N) : # return nth term return (factorial(N + 1) // 2) # Driver code if __name__ == "__main__" : N = 6 # Function Calling print(nthTerm(N))
C#
// C# program to find N-th // term of the series: // 1, 3, 12, 60, 360 using System; class GFG { // Function to find factorial of N static int factorial(int N) { int fact = 1; for (int i = 1; i <= N; i++) fact = fact * i; // return factorial of N return fact; } // calculate Nth term of series static int nthTerm(int N) { return (factorial(N + 1) / 2); } // Driver Code static void Main() { int N = 6 ; // Printing the nth term Console.WriteLine(nthTerm(N)); } } // This code is contributed // by ANKITRAI1
PHP
<?php // PHP program to find N-th term // of the series: 1, 3, 12, 60, 360… // Function to find factorial of N function factorial($N) { $fact = 1; for ($i = 1; $i <= $N; $i++) $fact = $fact * $i; // return factorial of N+1 return $fact; } // calculate Nth term of series function nthTerm($N) { return (factorial($N + 1) / 2); } // Driver Code // Taking n as 6 $N = 6; // Printing the nth term echo nthTerm($N); // This code is contributed // by chandan_jnu.. ?>
Javascript
<script> // JavaScript program to find N-th term of the series: // 1, 3, 12, 60, 360… // Function to find factorial of N function factorial(N) { let fact = 1; for (let i = 1; i <= N; i++) fact = fact * i; // return factorial of N+1 return fact; } // calculate Nth term of series function nthTerm(N) { return (Math.floor(factorial(N + 1) / 2)); } // Driver Function // Taking n as 6 let N = 6; // Printing the nth term document.write(nthTerm(N)); // This code is contributed by Surbhi Tyagi </script>
Producción:
2520
Complejidad de tiempo: O(n), donde n representa la entrada dada.
Espacio auxiliar: O(1), no se requiere espacio adicional, por lo que es una constante.
Publicación traducida automáticamente
Artículo escrito por pawan_asipu y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA