Dada una array de números positivos, la tarea es encontrar el AND bit a bit de la suma de números no primos y la suma de números primos. Tenga en cuenta que 1 no es ni primo ni compuesto.
Ejemplos :
Entrada: arr[] = {1, 3, 5, 10, 15, 7}
Salida: 9
Suma de números no primos = 10 + 15 = 25
Suma de números primos = 3 + 5 + 7 = 15
25 y 15 = 9
Entrada : arr[] = {3, 4, 6, 7}
Salida: 10
Enfoque ingenuo: una solución simple es atravesar la array y seguir verificando cada elemento si es primo o no. Si el número es primo, agréguelo a S1, que almacena la suma de los números primos de la array; de lo contrario, agréguelo a S2, que almacena la suma de los números no primos. Finalmente, imprima S1 y S2.
Complejidad de tiempo: O(N * sqrt(N))
Enfoque eficiente: Genere todos los números primos hasta el elemento máximo de la array utilizando el Tamiz de Eratóstenes y almacénelos en un hash. Ahora, recorra la array y verifique si el número es primo o no. Al final, calcule e imprima el AND bit a bit de la suma de números primos y la suma de números compuestos.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function to return the bitwise AND of the
// sum of primes and the sum of non-primes
int calculateAND(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);
// Remaining part of SIEVE
prime[0] = false;
prime[1] = false;
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;
}
}
// Store the sum of primes in S1 and
// the sum of non-primes in S2
int S1 = 0, S2 = 0;
for (int i = 0; i < n; i++) {
if (prime[arr[i]]) {
// The number is prime
S1 += arr[i];
}
else if (arr[i] != 1) {
// The number is not prime
S2 += arr[i];
}
}
// Return the bitwise AND of the sums
return (S1 & S2);
}
// Driver code
int main()
{
int arr[] = { 3, 4, 6, 7 };
int n = sizeof(arr) / sizeof(arr[0]);
cout << calculateAND(arr, n);
return 0;
}
Java
// Java implementation of the approach
import java.util.Arrays;
import java.util.Collections;
import java.util.List;
class GFG
{
static int getMax(int[] A)
{
int max = Integer.MIN_VALUE;
for (int i: A)
{
max = Math.max(max, i);
}
return max;
}
// Function to return the bitwise AND of the
// sum of primes and the sum of non-primes
static int calculateAND(int arr[], int n)
{
// using Collections.max() to find
// maximum element using only 1 line.
// Find maximum value in the array
int max_val = getMax(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.
boolean prime[] = new boolean [max_val + 1];
int i;
for (i = 0; i < max_val + 1; i++)
prime[i] = true;
// Remaining part of SIEVE
prime[0] = false;
prime[1] = false;
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 ( i = p * 2; i <= max_val; i += p)
prime[i] = false;
}
}
// Store the sum of primes in S1 and
// the sum of non-primes in S2
int S1 = 0, S2 = 0;
for (i = 0; i < n; i++)
{
if (prime[arr[i]])
{
// The number is prime
S1 += arr[i];
}
else if (arr[i] != 1)
{
// The number is not prime
S2 += arr[i];
}
}
// Return the bitwise AND of the sums
return (S1 & S2);
}
// Driver code
public static void main (String[] args)
{
int arr[] = { 3, 4, 6, 7 };
int n = arr.length;
System.out.println(calculateAND(arr, n));
}
}
// This code is contributed by AnkitRai01
Python3
# Python3 implementation of the approach # Function to return the bitwise AND of the # sum of primes and the sum of non-primes def calculateAND(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 = [True for i in range(max_val + 1)] # Remaining part of SIEVE prime[0] = False prime[1] = False for p in range(2, max_val + 1): if p * p >= max_val: break # If prime[p] is not changed, # then it is a prime if (prime[p]): # Update all multiples of p for i in range(2 * p, max_val + 1, p): prime[i] = False # Store the sum of primes in S1 and # the sum of non-primes in S2 S1 = 0 S2 = 0 for i in range(n): if (prime[arr[i]]): # The number is prime S1 += arr[i] elif (arr[i] != 1): # The number is not prime S2 += arr[i] # Return the bitwise AND of the sums return (S1 & S2) # Driver code arr = [3, 4, 6, 7] n = len(arr) print(calculateAND(arr, n)) # This code is contributed by Mohit Kumar
C#
// C# implementation of the approach
using System;
using System.Collections.Generic;
class GFG
{
static int getMax(int[] A)
{
int max = int.MinValue;
foreach (int i in A)
{
max = Math.Max(max, i);
}
return max;
}
// Function to return the bitwise AND of the
// sum of primes and the sum of non-primes
static int calculateAND(int []arr, int n)
{
// using Collections.max() to find
// maximum element using only 1 line.
// Find maximum value in the array
int max_val = getMax(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.
bool []prime = new bool [max_val + 1];
int i;
for (i = 0; i < max_val + 1; i++)
prime[i] = true;
// Remaining part of SIEVE
prime[0] = false;
prime[1] = false;
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 (i = p * 2; i <= max_val; i += p)
prime[i] = false;
}
}
// Store the sum of primes in S1 and
// the sum of non-primes in S2
int S1 = 0, S2 = 0;
for (i = 0; i < n; i++)
{
if (prime[arr[i]])
{
// The number is prime
S1 += arr[i];
}
else if (arr[i] != 1)
{
// The number is not prime
S2 += arr[i];
}
}
// Return the bitwise AND of the sums
return (S1 & S2);
}
// Driver code
public static void Main (String[] args)
{
int []arr = { 3, 4, 6, 7 };
int n = arr.Length;
Console.WriteLine(calculateAND(arr, n));
}
}
// This code is contributed by Rajput-Ji
Javascript
<script>
// Javascript implementation of the approach
// Function to return the bitwise AND of the
// sum of primes and the sum of non-primes
function calculateAND(arr, n)
{
// Find maximum value in the array
var max_val = arr.reduce((a,b)=>Math.max(a,b));
// 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.
var prime = Array(max_val + 1).fill(true);
// Remaining part of SIEVE
prime[0] = false;
prime[1] = false;
for (var 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 (var i = p * 2; i <= max_val; i += p)
prime[i] = false;
}
}
// Store the sum of primes in S1 and
// the sum of non-primes in S2
var S1 = 0, S2 = 0;
for (var i = 0; i < n; i++) {
if (prime[arr[i]]) {
// The number is prime
S1 += arr[i];
}
else if (arr[i] != 1) {
// The number is not prime
S2 += arr[i];
}
}
// Return the bitwise AND of the sums
return (S1 & S2);
}
// Driver code
var arr = [ 3, 4, 6, 7 ];
var n = arr.length;
document.write( calculateAND(arr, n));
</script>
Producción:
10
Complejidad de tiempo: O(N * log(log(N))
Complejidad de espacio: O(max_val) donde max_val es el valor máximo de un elemento en la array dada.
Publicación traducida automáticamente
Artículo escrito por NikhilRathor y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA