Dada una secuencia de enteros positivos a1, a2, …, an. Encuentre todos esos índices i tales que el i-ésimo elemento sea igual a la media aritmética de todos los demás elementos (es decir, todos los elementos excepto este).
Ejemplos:
Input : 5
1 2 3 4 5
Output : 1 no. of elements
2 index of element
Average of 1, 2, 4 & 5 is 3 so the
output is single index i.e. 3.
Input : 4
50 50 50 50
Output : 4 no. of elements
0 1 2 3 index of element
Average of 50, 50, 50 & 50 is 50 and
all the indexes has the same i.e. 50
so the output is indexes 1, 2, 3 & 4.
C++
// CPP program to print all such indices such
// that the i-th element equals the arithmetic
// mean of all other elements
#include <bits/stdc++.h>
using namespace std;
// function to find number of elements
// satisfying condition and their indexes
void averageNumbers(int arr[], int n, int sum)
{
int cnt = 0;
// calculating average
sum /= (double)n;
// counting how many elements
// satisfies the condition.
cout << count(arr, arr + n, sum)
<< endl;
for (int i = 0; i < n; i++) {
if ((double)arr[i] == sum) {
// output the indices.
cout << i << " ";
cnt++;
}
}
}
// Driver code
int main()
{
int n;
int arr[] = { 1, 2, 3, 4, 5 };
n = sizeof(arr) / sizeof(arr[0]);
double sum = 0;
int cnt = 0;
// sum of the elements of the array
for (int i = 0; i < n; i++) {
sum += (double)arr[i];
}
averageNumbers(arr, n, sum);
return 0;
}
C
// C program to print all such indices such
// that the i-th element equals the arithmetic
// mean of all other elements
#include <stdio.h>
// function to find number of elements
// satisfying condition and their indexes
void averageNumbers(int arr[], int n, int sum)
{
int cnt = 0;
// calculating average
sum /= (double)n;
for (int i = 0; i < n; i++) {
if ((double)arr[i] == sum) {
// counting how many elements
// satisfies the condition
cnt++;
}
}
printf("%d\n",cnt);
for (int i = 0; i < n; i++) {
if ((double)arr[i] == sum) {
// output the indices.
printf("%d ",i);
cnt++;
}
}
}
// Driver code
int main()
{
int n;
int arr[] = { 1, 2, 3, 4, 5 };
n = sizeof(arr) / sizeof(arr[0]);
double sum = 0;
int cnt = 0;
// sum of the elements of the array
for (int i = 0; i < n; i++) {
sum += (double)arr[i];
}
averageNumbers(arr, n, sum);
return 0;
}
// This code is contributed by kothvvsaakash.
Java
// Java program to print all such indices such
// that the i-th element equals the arithmetic
// mean of all other elements
public class GFG {
// function to find number of elements
// satisfying condition and their indexes
static void averageNumbers(int arr[], int n, int sum) {
int cnt = 0;
// calculating average
sum /= (double) n;
// counting how many elements
// satisfies the condition.
System.out.println(count(arr, sum));
for (int i = 0; i < n; i++) {
if ((double) arr[i] == sum) {
// output the indices.
System.out.print(i + " ");
cnt++;
}
}
}
static int count(int[] array, int sum) {
int count = 0;
for (int i = 0; i < array.length; i++) {
if (array[i] == sum) {
count++;
}
}
return count;
}
// Driver code
public static void main(String[] args) {
int n;
int arr[] = {1, 2, 3, 4, 5};
n = arr.length;
int sum = 0;
int cnt = 0;
// sum of the elements of the array
for (int i = 0; i < n; i++) {
sum += (double) arr[i];
}
averageNumbers(arr, n, sum);
}
}
// This code is contributed by 29AjayKumar
Python3
# Python 3 program to print all such indices # such that the i-th element equals the # arithmetic mean of all other elements # Function to find number of elements # satisfying condition and their indexes def averageNumbers(arr, n, sum): cnt = 0 # calculating average sum /= n # counting how many elements # satisfies the condition. print(count(arr, sum)) for i in range(0, n): if (arr[i] == sum): # output the indices. print(i, " ") cnt += 1 def count(array, sum): count = 0 for i in range(0, len(array)): if (array[i] == sum): count += 1 return count # Driver code if __name__ == '__main__': n = 0 arr = [ 1, 2, 3, 4, 5 ] n = len(arr) sum = 0 cnt = 0 # sum of the elements of the array for i in range(0, n): sum += arr[i] averageNumbers(arr, n, sum) # This code contributed by 29AjayKumar
C#
// C# program to print all such indices such
// that the i-th element equals the arithmetic
// mean of all other elements
using System;
public class GFG {
// function to find number of elements
// satisfying condition and their indexes
static void averageNumbers(int []arr, int n, int sum) {
int cnt = 0;
// calculating average
sum /= n;
// counting how many elements
// satisfies the condition.
Console.WriteLine(count(arr, sum));
for (int i = 0; i < n; i++) {
if ((double) arr[i] == sum) {
// output the indices.
Console.Write(i + " ");
cnt++;
}
}
}
static int count(int[] array, int sum) {
int count = 0;
for (int i = 0; i < array.Length; i++) {
if (array[i] == sum) {
count++;
}
}
return count;
}
// Driver code
public static void Main() {
int n;
int []arr = {1, 2, 3, 4, 5};
n = arr.Length;
int sum = 0;
// sum of the elements of the array
for (int i = 0; i < n; i++) {
sum += arr[i];
}
averageNumbers(arr, n, sum);
}
}
// This code is contributed by 29AjayKumar
PHP
<?php
// PHP program to print all such indices
// such that the i-th element equals the
// arithmetic mean of all other elements
// counting how many elements
// satisfies the condition.
function coun_t($arr, $sum)
{
$cnt = 0;
for ( $i = 0; $i < count($arr); $i++)
{
if ($arr[$i] == $sum)
{
$cnt++;
}
}
return $cnt;
}
// function to find number of elements
// satisfying condition and their indexes
function averageNumbers($arr, $n, $sum)
{
$cnt = 0;
// calculating average
$sum /= $n;
// counting how many elements
// satisfies the condition.
echo coun_t($arr, $sum) . "\n";
for ( $i = 0; $i < $n; $i++)
{
if ($arr[$i] == $sum)
{
// output the indices.
echo $i . " ";
$cnt++;
}
}
}
// Driver Code
$n = 0;
$arr = array( 1, 2, 3, 4, 5 );
$n = count($arr);
$sum = 0;
$cnt = 0;
// sum of the elements of the array
for ($i = 0; $i < $n; $i++)
{
$sum += $arr[$i];
}
averageNumbers($arr, $n, $sum);
// This code is contributed by
// Rajput-Ji
?>
Javascript
<script>
// Javascript program to print all such indices such
// that the i-th element equals the arithmetic
// mean of all other elements
// function to find number of elements
// satisfying condition and their indexes
function averageNumbers(arr, n, sum) {
let cnt = 0;
// calculating average
sum /= n;
// counting how many elements
// satisfies the condition.
document.write(count(arr, sum) + "</br>");
for (let i = 0; i < n; i++) {
if (arr[i] == sum) {
// output the indices.
document.write(i + " ");
cnt++;
}
}
}
function count(array, sum) {
let count = 0;
for (let i = 0; i < array.length; i++) {
if (array[i] == sum) {
count++;
}
}
return count;
}
// Driver code
let n;
let arr = [1, 2, 3, 4, 5];
n = arr.length;
let sum = 0;
// sum of the elements of the array
for (let i = 0; i < n; i++) {
sum += arr[i];
}
averageNumbers(arr, n, sum);
// This code is contributed by mukesh07.
</script>
Producción
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Complejidad de Tiempo: O(n), Espacio Auxiliar: O(1)
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Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA