Dada una array a[] de tamaño N y un número entero K, la tarea es encontrar un valor X en el rango [0, K] que pueda maximizar el valor de la función dada
- Xor-sum(X) = (X XOR A[0]) + (X Xor A[1]) + (X Xor A[2]) + __________+ (X Xor A[N-1]).
Ejemplos:
Entrada: a[] = {1, 2, 3, 4, 5, 6}, N=6, K=10
Salida: 8
Explicación: El valor de X es 1 para el cual la suma requerida se vuelve máxima.Entrada: a[] = {1, 6}, N=2, K=7
Salida: 0
Enfoque: La idea es considerar todos y cada uno de los valores de K y luego encontrar el valor de X que satisfaga la condición anterior. Siga los pasos a continuación para resolver el problema:
- Inicialice las variables max y X como 0 y -1 para almacenar la suma máxima definida y el valor de X para el que está sucediendo.
- Itere sobre el rango [0, K] usando la variable j y realice las siguientes tareas:
- Inicialice la suma variable como 0 para almacenar la suma definida.
- Itere sobre el rango [0, N) usando la variable i y realice las siguientes tareas:
- Establece el valor de sum como sum + j^a[i].
- Si sum es mayor que max, establezca el valor de max como sum y X como j.
- Después de realizar los pasos anteriores, imprima el valor de X como respuesta.
A continuación se muestra la implementación del enfoque anterior.
C++14
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to find the value of X for which
// the given sum maximises
void findValue(vector<int>& a, int N, int K)
{
// Variables to store the maximum
// sum and that particular value
// of X.
long long int max = 0, X = -1;
// Check every value of K
for (int j = 0; j <= K; j++) {
// Variable to store the desired sum
long long int sum = 0;
for (int i = 0; i < N; i++) {
(sum = sum + (j ^ a[i]));
}
// Check the condition
if (sum > max) {
max = sum;
X = j;
}
}
// Print the result
cout << X;
}
// Driver Code
int main()
{
vector<int> a = { 1, 2, 3, 4, 5, 6 };
int N = 6, K = 10;
findValue(a, N, K);
return 0;
}
C
// C program for the above approach
#include <stdio.h>
// Function to find the value of X for which
// the given sum maximises
void findValue(int *a, int N, int K)
{
// Variables to store the maximum
// sum and that particular value
// of X.
long long int max = 0, X = -1;
// Check every value of K
for (int j = 0; j <= K; j++) {
// Variable to store the desired sum
long long int sum = 0;
for (int i = 0; i < N; i++) {
(sum = sum + (j ^ a[i]));
}
// Check the condition
if (sum > max) {
max = sum;
X = j;
}
}
// Print the result
printf("%lli", X);
}
// Driver Code
int main()
{
int a[] = { 1, 2, 3, 4, 5, 6 };
int N = 6, K = 10;
findValue(a, N, K);
return 0;
}
// This code is contributed by palashi.
Java
// Java program for the above approach
class GFG {
// Function to find the value of X for which
// the given sum maximises
public static void findValue(int[] a, int N, int K) {
// Variables to store the maximum
// sum and that particular value
// of X.
int max = 0, X = -1;
// Check every value of K
for (int j = 0; j <= K; j++) {
// Variable to store the desired sum
int sum = 0;
for (int i = 0; i < N; i++) {
sum = sum + (j ^ a[i]);
}
// Check the condition
if (sum > max) {
max = sum;
X = j;
}
}
// Print the result
System.out.println(X);
}
// Driver Code
public static void main(String args[])
{
int[] a = { 1, 2, 3, 4, 5, 6 };
int N = 6, K = 10;
findValue(a, N, K);
}
}
// This code is contributed by saurabh_jaiswal.
Python3
# Python3 program for the above approach # Function to find the value of X for which # the given sum maximises def findValue(a, N, K) : # Variables to store the maximum # sum and that particular value # of X. max = 0; X = -1; # Check every value of K for j in range( K + 1) : # Variable to store the desired sum sum = 0; for i in range(N) : sum = sum + (j ^ a[i]); # Check the condition if (sum > max) : max = sum; X = j; # Print the result print(X); # Driver Code if __name__ == "__main__" : a = [ 1, 2, 3, 4, 5, 6 ]; N = 6; K = 10; findValue(a, N, K); # This code is contributed by AnkThon
C#
// C# program for the above approach
using System;
public class GFG {
// Function to find the value of X for which
// the given sum maximises
public static void findValue(int[] a, int N, int K) {
// Variables to store the maximum
// sum and that particular value
// of X.
int max = 0, X = -1;
// Check every value of K
for (int j = 0; j <= K; j++) {
// Variable to store the desired sum
int sum = 0;
for (int i = 0; i < N; i++) {
sum = sum + (j ^ a[i]);
}
// Check the condition
if (sum > max) {
max = sum;
X = j;
}
}
// Print the result
Console.WriteLine(X);
}
// Driver Code
public static void Main(string []args)
{
int[] a = { 1, 2, 3, 4, 5, 6 };
int N = 6, K = 10;
findValue(a, N, K);
}
}
// This code is contributed by AnkThon
Javascript
<script>
// JavaScript Program to implement
// the above approach
// Function to find the value of X for which
// the given sum maximises
function findValue(a, N, K) {
// Variables to store the maximum
// sum and that particular value
// of X.
let max = 0, X = -1;
// Check every value of K
for (let j = 0; j <= K; j++) {
// Variable to store the desired sum
let sum = 0;
for (let i = 0; i < N; i++) {
(sum = sum + (j ^ a[i]));
}
// Check the condition
if (sum > max) {
max = sum;
X = j;
}
}
// Print the result
document.write(X);
}
// Driver Code
let a = [1, 2, 3, 4, 5, 6];
let N = 6, K = 10;
findValue(a, N, K);
// This code is contributed by Potta Lokesh
</script>
Producción
8
Complejidad temporal: O(K*N)
Espacio auxiliar: O(1)
Publicación traducida automáticamente
Artículo escrito por pavankumarb289 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA