Dado un número impar n, encuentre el promedio de números impares de 1 a n.
Ejemplos:
Input : n = 9 Output : 5 Explanation (1 + 3 + 5 + 7 + 9)/5 = 25/5 = 5 Input : n = 221 Output : 111
Método 1 Podemos calcular el promedio sumando cada número impar hasta ny luego dividiendo la suma por el conteo.
A continuación se muestra la implementación del enfoque.
C++
// Program to find average of odd numbers
// till a given odd number.
#include <stdio.h>
// Function to calculate the average
// of odd numbers
int averageOdd(int n)
{
if (n % 2 == 0) {
printf("Invalid Input");
return -1;
}
int sum = 0, count = 0;
while (n >= 1) {
// count odd numbers
count++;
// store the sum of odd numbers
sum += n;
n = n - 2;
}
return sum / count;
}
// driver function
int main()
{
int n = 15;
printf("%d", averageOdd(n));
return 0;
}
Java
// Program to find average of odd numbers
// till a given odd number.
import java.io.*;
class GFG {
// Function to calculate the average
// of odd numbers
static int averageOdd(int n)
{
if (n % 2 == 0) {
System.out.println("Invalid Input");
return -1;
}
int sum = 0, count = 0;
while (n >= 1) {
// count odd numbers
count++;
// store the sum of odd numbers
sum += n;
n = n - 2;
}
return sum / count;
}
// driver function
public static void main(String args[])
{
int n = 15;
System.out.println(averageOdd(n));
}
}
/*This code is contributed by Nikita tiwari.*/
Python3
# Program to find average
# of odd numbers till a
# given odd number.
# Function to calculate
# the average of odd
# numbers
def averageOdd(n) :
if (n % 2 == 0) :
print("Invalid Input")
return -1
sm = 0
count = 0
while (n >= 1) :
# count odd numbers
count = count + 1
# store the sum of
# odd numbers
sm = sm + n
n = n - 2
return sm // count
# Driver function
n = 15
print(averageOdd(n))
# This code is contributed by Nikita Tiwari.
C#
// C# Program to find average
// of odd numbers till a given
// odd number.
using System;
class GFG {
// Function to calculate the
// average of odd numbers
static int averageOdd(int n)
{
if (n % 2 == 0) {
Console.Write("Invalid Input");
return -1;
}
int sum = 0, count = 0;
while (n >= 1) {
// count odd numbers
count++;
// store the sum of odd numbers
sum += n;
n = n - 2;
}
return sum / count;
}
// driver function
public static void Main()
{
int n = 15;
Console.Write(averageOdd(n));
}
}
/*This code is contributed by vt_m.*/
PHP
<?php
// Program to find average of odd
// numbers till a given odd number.
// Function to calculate the
// average of odd numbers
function averageOdd($n)
{
if ($n % 2 == 0)
{
echo("Invalid Input");
return -1;
}
$sum = 0;
$count = 0;
while ($n >= 1)
{
// count odd numbers
$count++;
// store the sum of
// odd numbers
$sum += $n;
$n = $n - 2;
}
return $sum / $count;
}
// Driver Code
$n = 15;
echo(averageOdd($n));
// This code is contributed by vt_m.
?>
Javascript
<script>
// Javascript Program to find average of odd numbers
// till a given odd number.
// Function to calculate the average
// of odd numbers
function averageOdd( n)
{
if (n % 2 == 0)
{
document.write("Invalid Input");
return -1;
}
let sum = 0, count = 0;
while (n >= 1) {
// count odd numbers
count++;
// store the sum of odd numbers
sum += n;
n = n - 2;
}
return sum / count;
}
// driver function
let n = 15;
document.write(averageOdd(n));
// This code is contributed by gauravrajput1
</script>
Producción:
8
Complejidad de tiempo: O(n)
Espacio auxiliar: O(1)
Método 2
El promedio de los números impares puede calcularse solo en pasos individuales
usando la siguiente fórmula
[n + 1] / 2
donde n es el último número impar.
¿Cómo funciona esta fórmula?
We know there are (n+1)/2 odd numbers till n. For example: There are two odd numbers till 3 and there are three odd numbers till 5. Sum of first k odd numbers is k*k Sum of odd numbers till n is ((n+1)/2)2 Average of odd numbers till n is (n + 1)/2
A continuación se muestra la implementación del enfoque.
C
// Program to find average of odd numbers
// till a given odd number.
#include <stdio.h>
// Function to calculate the average
// of odd numbers
int averageOdd(int n)
{
if (n % 2 == 0) {
printf("Invalid Input");
return -1;
}
return (n + 1) / 2;
}
// driver function
int main()
{
int n = 15;
printf("%d", averageOdd(n));
return 0;
}
Java
// Program to find average of odd
// numbers till a given odd number.
import java.io.*;
class GFG
{
// Function to calculate the
// average of odd numbers
static int averageOdd(int n)
{
if (n % 2 == 0)
{
System.out.println("Invalid Input");
return -1;
}
return (n + 1) / 2;
}
// driver function
public static void main(String args[])
{
int n = 15;
System.out.println(averageOdd(n));
}
}
// This code is contributed by Nikita tiwari.
Python3
# Program to find average of odd
# numbers till a given odd number.
# Function to calculate the
# average of odd numbers
def averageOdd(n) :
if (n % 2 == 0) :
print("Invalid Input")
return -1
return (n + 1) // 2
# driver function
n = 15
print(averageOdd(n))
# This code is contributed by Nikita tiwari.
C#
// C# Program to find average
// of odd numbers till a given
// odd number.
using System;
class GFG
{
// Function to calculate the
// average of odd numbers
static int averageOdd(int n)
{
if (n % 2 == 0)
{
Console.Write("Invalid Input");
return -1;
}
return (n + 1) / 2;
}
// driver function
public static void Main()
{
int n = 15;
Console.Write(averageOdd(n));
}
}
// This code is contributed by vt_m.
PHP
<?php
// Program to find average of odd
// numbers till a given odd number.
// Function to calculate the
// average of odd numbers
function averageOdd( $n)
{
if ($n % 2 == 0)
{
echo("Invalid Input");
return -1;
}
return ($n + 1) / 2;
}
// Driver Code
$n = 15;
echo(averageOdd($n));
// This code is contributed by vt_m.
?>
Javascript
<script>
// javascript Program to find average of odd numbers
// till a given odd number.
// Function to calculate the average
// of odd numbers
function averageOdd( n)
{
if (n % 2 == 0) {
document.write("Invalid Input");
return -1;
}
return (n + 1) / 2;
}
// driver function
let n = 15;
document.write( averageOdd(n));
// This code is contributed by todaysgaurav
</script>
Producción:
8
Complejidad temporal: O(1) desde que se realizan operaciones constantes
Complejidad espacial: O(1) ya que usa variables constantes