Dada una array de enteros. La tarea es calcular el producto de todos los números compuestos en una array.
Nota: 1 no es ni primo ni compuesto.
Ejemplos:
Input: arr[] = {2, 3, 4, 5, 6, 7}
Output: 24
Composite numbers are 4 and 6.
So, product = 24
Input: arr[] = {11, 13, 17, 20, 19}
Output: 20
Enfoque ingenuo: una solución simple es atravesar la array y hacer una prueba de primalidad en cada elemento. Si el elemento no es primo ni 1, multiplícalo al producto corriente.
Complejidad de tiempo: O (Nsqrt (N))
Enfoque eficiente: el uso de Sieve of Eratosthenes genera un vector booleano hasta el tamaño del elemento máximo de la array que se puede usar para verificar si un número es primo o no. También agregue 0 y 1 como números primos para que no se cuenten como números compuestos. Ahora recorra la array y encuentre el producto de esos elementos que son compuestos usando el vector booleano generado.
C++
// C++ program to find the product
// of all the composite numbers
// in an array
#include <bits/stdc++.h>
using namespace std;
// Function that returns the
// the product of all composite numbers
int compositeProduct(int arr[], int n)
{
// Find maximum value in the array
int max_val = *max_element(arr, arr + n);
// Use sieve to find all prime numbers
// less than or equal to max_val
// Create a boolean array "prime[0..n]". A
// value in prime[i] will finally be false
// if i is Not a prime, else true.
vector<bool> prime(max_val + 1, true);
// Set 0 and 1 as primes as
// they don't need to be
// counted as composite numbers
prime[0] = true;
prime[1] = true;
for (int p = 2; p * p <= max_val; p++) {
// If prime[p] is not changed, then
// it is a prime
if (prime[p] == true) {
// Update all multiples of p
for (int i = p * 2; i <= max_val; i += p)
prime[i] = false;
}
}
// Find the product of all
// composite numbers in the arr[]
int product = 1;
for (int i = 0; i < n; i++)
if (!prime[arr[i]]) {
product *= arr[i];
}
return product;
}
// Driver code
int main()
{
int arr[] = { 2, 3, 4, 5, 6, 7 };
int n = sizeof(arr) / sizeof(arr[0]);
cout << compositeProduct(arr, n);
return 0;
}
Java
// Java program to find the product
// of all the composite numbers
// in an array
import java.util.*;
class GFG {
// Function that returns the
// the product of all composite numbers
static int compositeProduct(int arr[], int n)
{
// Find maximum value in the array
int max_val = Arrays.stream(arr).max().getAsInt();
// Use sieve to find all prime numbers
// less than or equal to max_val
// Create a boolean array "prime[0..n]". A
// value in prime[i] will finally be false
// if i is Not a prime, else true.
boolean[] prime = new boolean[max_val + 1];
Arrays.fill(prime, true);
// Set 0 and 1 as primes as
// they don't need to be
// counted as composite numbers
prime[0] = true;
prime[1] = true;
for (int p = 2; p * p <= max_val; p++) {
// If prime[p] is not changed, then
// it is a prime
if (prime[p] == true) {
// Update all multiples of p
for (int i = p * 2; i <= max_val; i += p) {
prime[i] = false;
}
}
}
// Find the product of all
// composite numbers in the arr[]
int product = 1;
for (int i = 0; i < n; i++) {
if (!prime[arr[i]]) {
product *= arr[i];
}
}
return product;
}
// Driver code
public static void main(String[] args)
{
int arr[] = { 2, 3, 4, 5, 6, 7 };
int n = arr.length;
System.out.println(compositeProduct(arr, n));
}
}
// This code has been contributed by 29AjayKumar
Python3
''' Python3 program to find product of all the composite numbers in given array''' import math as mt ''' function to find the product of all composite numbers in the given array ''' def compositeProduct(arr, n): # find the maximum value in the array max_val = max(arr) ''' USE SIEVE TO FIND ALL PRIME NUMBERS LESS THAN OR EQUAL TO max_val Create a boolean array "prime[0..n]". A value in prime[i] will finally be false if i is Not a prime, else true. ''' prime =[True for i in range(max_val + 1)] ''' Set 0 and 1 as primes as they don't need to be counted as composite numbers ''' prime[0]= True prime[1]= True for p in range(2, mt.ceil(mt.sqrt(max_val))): # Remaining part of SIEVE ''' if prime[p] is not changed, than it is prime ''' if prime[p]: # update all multiples of p for i in range(p * 2, max_val + 1, p): prime[i]= False # find the product of all composite numbers in the arr[] product = 1 for i in range(n): if prime[arr[i]]== False: product*= arr[i] return product # Driver code arr =[2, 3, 4, 5, 6, 7] n = len(arr) print(compositeProduct(arr, n)) # contributed by Mohit kumar 29
C#
// C# program to find the product
// of all the composite numbers
// in an array
using System;
using System.Linq;
public class GFG {
// Function that returns the
// the product of all composite numbers
static int compositeProduct(int[] arr, int n)
{
// Find maximum value in the array
int max_val = arr.Max();
// Use sieve to find all prime numbers
// less than or equal to max_val
// Create a boolean array "prime[0..n]". A
// value in prime[i] will finally be false
// if i is Not a prime, else true.
bool[] prime = new bool[max_val + 1];
for (int i = 0; i < max_val + 1; i++)
prime[i] = true;
// Set 0 and 1 as primes as
// they don't need to be
// counted as composite numbers
prime[0] = true;
prime[1] = true;
for (int p = 2; p * p <= max_val; p++) {
// If prime[p] is not changed, then
// it is a prime
if (prime[p] == true) {
// Update all multiples of p
for (int i = p * 2; i <= max_val; i += p) {
prime[i] = false;
}
}
}
// Find the product of all
// composite numbers in the arr[]
int product = 1;
for (int i = 0; i < n; i++) {
if (!prime[arr[i]]) {
product *= arr[i];
}
}
return product;
}
// Driver code
public static void Main()
{
int[] arr = { 2, 3, 4, 5, 6, 7 };
int n = arr.Length;
Console.WriteLine(compositeProduct(arr, n));
}
}
/* This code contributed by PrinciRaj1992 */
PHP
<?php
// PHP program to find the product
// of all the composite numbers
// in an array
// Function that returns the
// the product of all composite numbers
function compositeProduct($arr, $n)
{
// Find maximum value in the array
$max_val = max($arr);
// Use sieve to find all prime numbers
// less than or equal to max_val
// Create a boolean array "prime[0..n]". A
// value in prime[i] will finally be false
// if i is Not a prime, else true.
$prime = array_fill(0, $max_val + 1, true);
// Set 0 and 1 as primes as
// they don't need to be
// counted as composite numbers
$prime[0] = true;
$prime[1] = true;
for ($p = 2; $p * $p <= $max_val; $p++)
{
// If prime[p] is not changed,
// then it is a prime
if ($prime[$p] == true)
{
// Update all multiples of p
for ($i = $p * 2;
$i <= $max_val; $i += $p)
$prime[$i] = false;
}
}
// Find the product of all
// composite numbers in the arr[]
$product = 1;
for ($i = 0; $i < $n; $i++)
if (!$prime[$arr[$i]])
{
$product *= $arr[$i];
}
return $product;
}
// Driver code
$arr = array( 2, 3, 4, 5, 6, 7 );
$n = count($arr);
echo compositeProduct($arr, $n);
// This code is contributed by mits
?>
Javascript
<script>
// Javascript program to find the product
// of all the composite numbers
// in an array
// Function that returns the
// the product of all composite numbers
function compositeProduct(arr, n)
{
// Find maximum value in the array
let max_val = arr.sort((A, B) => B - A)[0];
// Use sieve to find all prime numbers
// less than or equal to max_val
// Create a boolean array "prime[0..n]". A
// value in prime[i] will finally be false
// if i is Not a prime, else true.
let prime = new Array(max_val + 1).fill(true);
// Set 0 and 1 as primes as
// they don't need to be
// counted as composite numbers
prime[0] = true;
prime[1] = true;
for (let p = 2; p * p <= max_val; p++)
{
// If prime[p] is not changed,
// then it is a prime
if (prime[p] == true)
{
// Update all multiples of p
for (let i = p * 2;
i <= max_val; i += p)
prime[i] = false;
}
}
// Find the product of all
// composite numbers in the arr[]
let product = 1;
for (let i = 0; i < n; i++)
if (!prime[arr[i]])
{
product *= arr[i];
}
return product;
}
// Driver code
let arr = new Array( 2, 3, 4, 5, 6, 7 );
let n = arr.length;
document.write(compositeProduct(arr, n));
// This code is contributed by gfgking
</script>
Producción:
24
Complejidad de tiempo: O(n + max_val 2 )
Espacio Auxiliar: O(max_val)
Publicación traducida automáticamente
Artículo escrito por imdhruvgupta y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA