在 [0, K] 范围内找到一个值 X，它可以在给定数组上最大化 X XOR 和
给定一个大小为N的数组a[]和一个整数K，任务是在[0, K]范围内找到一个值X ，它可以最大化给定函数的值
- Xor-sum(X) = (X XOR A[0]) + (X Xor A[1]) + (X Xor A[2]) + __________+ (X Xor A[N-1]).
例子：
Input: a[] = {1, 2, 3, 4, 5, 6}, N=6, K=10
Output: 8
Explanation: The value of X is 1 for which the required sum becomes maximum.
Input: a[] = {1, 6}, N=2, K=7
Output: 0
方法：想法是考虑K的每一个值，然后找到满足上述条件的X值。请按照以下步骤解决问题：
- 将变量max和X初始化为0和-1以存储定义的最大总和以及发生它的X的值。
- 使用变量j迭代范围[0, K]并执行以下任务：
- 将变量sum初始化为0以存储定义的总和。
- 使用变量i遍历范围[0, N)并执行以下任务：
- 将 sum 的值设置为sum + j^a[i]。
- 如果sum大于max，则将max的值设置为sum ，将X设置为j。
- 执行上述步骤后，打印X的值作为答案。
下面是上述方法的实现
C++14
// C++ program for the above approach
#include
using namespace std;
// Function to find the value of X for which
// the given sum maximises
void findValue(vector& 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 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
// 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 AnkThonC#
// 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 AnkThonJavascript
输出
8时间复杂度： O(K*N)
辅助空间： O(1)