Dado un número n, encuentre el promedio de los primeros n números impares
1 + 3 + 5 + 7 + 9 +………….+ (2n – 1)
Ejemplos:
Input : 5 Output : 5 (1 + 3 + 5 + 7 + 9)/5 = 5 Input : 10 Output : 10 (1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19)/10 =10
Método 1 (enfoque ingenuo:)
Una solución simple es iterar el bucle de 1 a n veces. A través de la suma de todos los números impares y divididos por n. Esta solución toma tiempo O (N).
C++
// A C++ program to find average of
// sum of first n odd natural numbers.
#include <iostream>
using namespace std;
// Returns the Avg of
// first n odd numbers
int avg_of_odd_num(int n)
{
// sum of first n odd number
int sum = 0;
for (int i = 0; i < n; i++)
sum += (2 * i + 1);
// Average of first
// n odd numbers
return sum / n;
}
// Driver Code
int main()
{
int n = 20;
cout << avg_of_odd_num(n);
return 0;
}
Java
// Java program to find average of
// sum of first n odd natural numbers.
import java.io.*;
class GFG {
// Returns the Avg of
// first n odd numbers
static int avg_of_odd_num(int n)
{
// sum of first n odd number
int sum = 0;
for (int i = 0; i < n; i++)
sum += (2 * i + 1);
// Average of first
// n odd numbers
return sum / n;
}
// Driver Code
public static void main(String[] args)
{
int n = 20;
avg_of_odd_num(n);
System.out.println(avg_of_odd_num(n));
}
}
// This code is contributed by vt_m
Python3
# A Python 3 program # to find average of # sum of first n odd # natural numbers. # Returns the Avg of # first n odd numbers def avg_of_odd_num(n) : # sum of first n odd number sm = 0 for i in range(0, n) : sm = sm + (2 * i + 1) # Average of first # n odd numbers return sm//n # Driver Code n = 20 print(avg_of_odd_num(n)) # This code is contributed # by Nikita Tiwari.
C#
// C# program to find average
// of sum of first n odd
// natural numbers.
using System;
class GFG {
// Returns the Avg of
// first n odd numbers
static int avg_of_odd_num(int n)
{
// sum of first n odd number
int sum = 0;
for (int i = 0; i < n; i++)
sum += (2 * i + 1);
// Average of first
// n odd numbers
return sum / n;
}
// Driver code
public static void Main()
{
int n = 20;
avg_of_odd_num(n);
Console.Write(avg_of_odd_num(n));
}
}
// This code is contributed by
// Smitha Dinesh Semwal
PHP
<?php
// A PHP program to find average of
// sum of first n odd natural numbers.
// Returns the Avg of
// first n odd numbers
function avg_of_odd_num($n)
{
// sum of first n odd number
$sum = 0;
for ($i = 0; $i < $n; $i++)
$sum += (2 * $i + 1);
// Average of first
// n odd numbers
return $sum / $n;
}
// Driver Code
$n = 20;
echo(avg_of_odd_num($n));
// This code is contributed by Ajit.
?>
Javascript
<script>
// javascript program to find average of
// sum of first n odd natural numbers.
// Returns the Avg of
// first n odd numbers
function avg_of_odd_num( n)
{
// sum of first n odd number
let sum = 0;
for (let i = 0; i < n; i++)
sum += (2 * i + 1);
// Average of first
// n odd numbers
return sum / n;
}
// Driver Code
let n = 20;
document.write(avg_of_odd_num(n));
// This code is contributed by todaysgaurav
</script>
Producción :
20
Complejidad de tiempo: O(n)
Espacio auxiliar: O(1)
Método 2 (Enfoque eficiente:)
La idea es que la suma de los primeros n números impares es n 2 , para encontrar el promedio de los primeros n números impares para que se divida entre n, por lo que la fórmula es n 2 / norte = norte . tarda O (1) tiempo.
Avg of sum of first N odd Numbers = N
C++
// CPP Program to find the average
// of sum of first n odd numbers
#include <bits/stdc++.h>
using namespace std;
// Return the average of sum
// of first n odd numbers
int avg_of_odd_num(int n)
{
return n;
}
// Driver Code
int main()
{
int n = 8;
cout << avg_of_odd_num(n);
return 0;
}
Java
// java Program to find the average
// of sum of first n odd numbers
import java.io.*;
class GFG {
// Return the average of sum
// of first n odd numbers
static int avg_of_odd_num(int n)
{
return n;
}
// Driver Code
public static void main(String[] args)
{
int n = 8;
System.out.println(avg_of_odd_num(n));
}
}
// This code is contributed by vt_m
Python3
# Python 3 Program to # find the average # of sum of first n # odd numbers # Return the average of sum # of first n odd numbers def avg_of_odd_num(n) : return n # Driver Code n = 8 print(avg_of_odd_num(n)) # This code is contributed # by Nikita Tiwari.
C#
// C# Program to find the average
// of sum of first n odd numbers
using System;
class GFG {
// Return the average of sum
// of first n odd numbers
static int avg_of_odd_num(int n)
{
return n;
}
// Driver Code
public static void Main()
{
int n = 8;
Console.Write(avg_of_odd_num(n));
}
}
// This code is contributed by
// Smitha Dinesh Semwal
PHP
<?php
// PHP Program to find the average
// of sum of first n odd numbers
// Return the average of sum
// of first n odd numbers
function avg_of_odd_num($n)
{
return $n;
}
// Driver Code
$n = 8;
echo(avg_of_odd_num($n));
// This code is contributed by Ajit.
?>
Javascript
<script>
// javascript Program to find the average
// of sum of first n odd numbers
// Return the average of sum
// of first n odd numbers
function avg_of_odd_num(n)
{
return n;
}
// Driver Code
var n = 8;
document.write(avg_of_odd_num(n));
// This code is contributed by gauravrajput1
</script>
Producción :
8
Complejidad de tiempo : O(1)
Complejidad espacial: O(1) ya que usa variables constantes
Prueba
Sum of first n terms of an A.P.(Arithmetic Progression)
= (n/2) * [2*a + (n-1)*d].....(i)
where, a is the first term of the series
and d is the difference between the adjacent
terms of the series.
Here, a = 1, d = 2, applying these values to e. q.,
(i), we get
Sum = (n/2) * [2*1 + (n-1)*2]
= (n/2) * [2 + 2*n - 2]
= (n/2) * (2*n)
= n*n
= n2
Avg of first n odd numbers = n2/n
= n
Publicación traducida automáticamente
Artículo escrito por jaingyayak y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA