Dado un número N, la tarea es imprimir el siguiente patrón hasta N términos:
2, 1, 4, 3, 6, 5, …. N términos
Ejemplos:
Input: N = 4 Output: 2, 1, 4, 3 Explanation: Nth Term = (N % 2 == 0) ? (N - 1) : (N + 1) 1st term = (1 % 2 == 0) ? (1 - 1) : (1 + 1) = (1 + 1) = 2 2nd term = (2 % 2 == 0) ? (2 - 1) : (2 + 1) = (2 - 1) = 1 3rd term = (3 % 2 == 0) ? (3 - 1) : (3 + 1) = (3 + 1) = 4 4th term = (4 % 2 == 0) ? (4 - 1) : (4 + 1) = (4 - 1) = 3 Therefore, Series = 2, 1, 4, 3 Input: N = 7 Output: 2, 1, 4, 3, 6, 5, 8
Fórmula:
Nth Term = (N % 2 == 0) ? (N - 1) : (N + 1)
A continuación se muestra la solución al problema anterior:
C++
// C++ program to print the series // 2, 1, 4, 3, 6, 5, …. up to N terms #include <iostream> using namespace std; // Function to print the series void printPattern(int N) { for (int i = 1; i <= N; i++) { // Find and print the ith term cout <<" "<<((i % 2 == 0) ? (i - 1) : (i + 1)); } } // Driver code int main() { // Get the value of N int N = 10; // Print the Series printPattern(N); return 0; }
Java
// Java Program to print the series // 2, 1, 4, 3, 6, 5, …. up to N terms import java.util.*; import java.lang.*; import java.io.*; class GFG { // Function to print the series static void printPattern(int N) { for (int i = 1; i <= N; i++) { // Find and print the ith term System.out.print(" "+((i % 2 == 0) ? (i - 1) : (i + 1))); } } // Driver code public static void main(String args[]) { // Get the value of N int N = 10; // Print the Series printPattern(N); } }
Python3
# Python program to print the series # 2, 1, 4, 3, 6, 5, …. up to N terms # Function to print the series def printPattern(N): for i in range(1, N + 1): # Find and print the ith term print(i - 1 if i % 2 == 0 else i + 1, end = " ") # Driver code N = 10 printPattern(N) # This code is contributed by Shrikant13
C#
// C# program to print the series // 2, 1, 4, 3, 6, 5, …. up to N terms using System; class GFG { // Function to print the series public void printPattern(int N) { for (int i = 1; i <= N; i++) { // Find and print the ith term int a = ((i % 2 == 0) ? (i - 1) : (i + 1)); Console.Write(" {0}", a); } } // Driver code public static void Main() { GFG g = new GFG(); // Get the value of N int N = 10; // Print the Series g.printPattern(N); } } // This code is contributed // by Soumik
PHP
<?php // PHP program to print the series // 2, 1, 4, 3, 6, 5, …. up to N terms // Function to print the series function printPattern($N) { for ($i = 1; $i <= $N; $i++) { // Find and print the ith term echo " " . (($i % 2 == 0) ? ($i - 1) : ($i + 1)); } } // Driver code // Get the value of N $N = 10; // Print the Series printPattern($N); // This code is contributed // by ChitraNayal ?>
Javascript
<script> // javascript program to print the series // 2, 1, 4, 3, 6, 5, …. up to N terms // Function to print the series function printPattern( N) { for (let i = 1; i <= N; i++) { // Find and print the ith term document.write(" " + ((i % 2 == 0) ? (i - 1) : (i + 1))); } } // Driver code // Get the value of N let N = 10; // Print the Series printPattern(N); // This code is contributed by Rajput-Ji </script>
Producción:
2 1 4 3 6 5 8 7 10 9