给定一个数组arr[]和两个整数X和K 。任务是对数组K次执行以下操作：
- 对数组进行排序。
- 将排序数组的每个替代元素与X异或，即arr[0], arr[2], arr[4], …
重复上述步骤K次后，打印修改后的数组中的最大和最小元素。
例子：
Input: arr[] = {9, 7, 11, 15, 5}, K = 1, X = 2
Output: 7 13
Since the operations has to be performed only once,
the sorted array will be {5, 7, 9, 11, 15}
Now, apply xor with 2 on alternate elements i.e. 5, 9 and 15.
{5 ^ 2, 7, 9 ^ 2, 11, 15 ^ 2} which is equal to
{7, 7, 11, 11, 13}
Input: arr[] = {605, 986}, K = 548, X = 569
Output: 605 986
方法：不是在每次迭代中对数组进行排序，而是可以维护一个频率数组，该数组将存储数组中每个元素的频率。遍历数组中从 1 到最大元素，可以按排序顺序处理元素，每次操作后，当交替元素与给定整数进行异或时，可以调整相同元素的频率。检查编程实现以获取更多详细信息。
下面是上述方法的实现：
C++
// C++ implementation of the approach
#include
using namespace std;
#define MAX 100000
// Function to find the maximum and the
// minimum elements from the array after
// performing the given operation k times
void xorOnSortedArray(int arr[], int n, int k, int x)
{
// To store the current sequence of elements
int arr1[MAX + 1] = { 0 };
// To store the next sequence of elements
// after xoring with current elements
int arr2[MAX + 1] = { 0 };
int xor_val[MAX + 1];
// Store the frequency of elements of arr[] in arr1[]
for (int i = 0; i < n; i++)
arr1[arr[i]]++;
// Storing all precomputed XOR values so that
// we don't have to do it again and again
// as XOR is a costly operation
for (int i = 0; i <= MAX; i++)
xor_val[i] = i ^ x;
// Perform the operations k times
while (k--) {
// The value of count decide on how many elements
// we have to apply XOR operation
int count = 0;
for (int i = 0; i <= MAX; i++) {
int store = arr1[i];
// If current element is present in
// the array to be modified
if (arr1[i] > 0) {
// Suppose i = m and arr1[i] = num, it means
// 'm' appears 'num' times
// If the count is even we have to perform
// XOR operation on alternate 'm' starting
// from the 0th index because count is even
// and we have to perform XOR operations
// starting with initial 'm'
// Hence there will be ceil(num/2) operations on
// 'm' that will change 'm' to xor_val[m] i.e. m^x
if (count % 2 == 0) {
int div = ceil((float)arr1[i] / 2);
// Decrease the frequency of 'm' from arr1[]
arr1[i] = arr1[i] - div;
// Increase the frequency of 'm^x' in arr2[]
arr2[xor_val[i]] += div;
}
// If the count is odd we have to perform
// XOR operation on alternate 'm' starting
// from the 1st index because count is odd
// and we have to leave the 0th 'm'
// Hence there will be (num/2) XOR operations on
// 'm' that will change 'm' to xor_val[m] i.e. m^x
else if (count % 2 != 0) {
int div = arr1[i] / 2;
arr1[i] = arr1[i] - div;
arr2[xor_val[i]] += div;
}
}
// Updating the count by frequency of
// the current elements as we have
// processed that many elements
count = count + store;
}
// Updating arr1[] which will now store the
// next sequence of elements
// At this time, arr1[] stores the remaining
// 'm' on which XOR was not performed and
// arr2[] stores the frequency of 'm^x' i.e.
// those 'm' on which operation was performed
// Updating arr1[] with frequency of remaining
// 'm' & frequency of 'm^x' from arr2[]
// With help of arr2[], we prevent sorting of
// the array again and again
for (int i = 0; i <= MAX; i++) {
arr1[i] = arr1[i] + arr2[i];
// Resetting arr2[] for next iteration
arr2[i] = 0;
}
}
// Finding the maximum and the minimum element
// from the modified array after the operations
int min = INT_MAX;
int max = INT_MIN;
for (int i = 0; i <= MAX; i++) {
if (arr1[i] > 0) {
if (min > i)
min = i;
if (max < i)
max = i;
}
}
// Printing the max and the min element
cout << min << " " << max << endl;
}
// Driver code
int main()
{
int arr[] = { 605, 986 };
int n = sizeof(arr) / sizeof(arr[0]);
int k = 548, x = 569;
xorOnSortedArray(arr, n, k, x);
return 0;
} Java
// Java implementation of the approach
class GFG
{
static int MAX = 100000;
// Function to find the maximum and the
// minimum elements from the array after
// performing the given operation k times
public static void xorOnSortedArray(int[] arr, int n,
int k, int x)
{
// To store the current sequence of elements
int[] arr1 = new int[MAX + 1];
// To store the next sequence of elements
// after xoring with current elements
int[] arr2 = new int[MAX + 1];
int[] xor_val = new int[MAX + 1];
// Store the frequency of elements
// of arr[] in arr1[]
for (int i = 0; i < n; i++)
arr1[arr[i]]++;
// Storing all precomputed XOR values so that
// we don't have to do it again and again
// as XOR is a costly operation
for (int i = 0; i <= MAX; i++)
xor_val[i] = i ^ x;
// Perform the operations k times
while (k-- > 0)
{
// The value of count decide on how many elements
// we have to apply XOR operation
int count = 0;
for (int i = 0; i <= MAX; i++)
{
int store = arr1[i];
// If current element is present in
// the array to be modified
if (arr1[i] > 0)
{
// Suppose i = m and arr1[i] = num, it means
// 'm' appears 'num' times
// If the count is even we have to perform
// XOR operation on alternate 'm' starting
// from the 0th index because count is even
// and we have to perform XOR operations
// starting with initial 'm'
// Hence there will be ceil(num/2) operations on
// 'm' that will change 'm' to xor_val[m] i.e. m^x
if (count % 2 == 0)
{
int div = (int) Math.ceil(arr1[i] / 2);
// Decrease the frequency of 'm' from arr1[]
arr1[i] = arr1[i] - div;
// Increase the frequency of 'm^x' in arr2[]
arr2[xor_val[i]] += div;
}
// If the count is odd we have to perform
// XOR operation on alternate 'm' starting
// from the 1st index because count is odd
// and we have to leave the 0th 'm'
// Hence there will be (num/2) XOR operations on
// 'm' that will change 'm' to xor_val[m] i.e. m^x
else if (count % 2 != 0)
{
int div = arr1[i] / 2;
arr1[i] = arr1[i] - div;
arr2[xor_val[i]] += div;
}
}
// Updating the count by frequency of
// the current elements as we have
// processed that many elements
count = count + store;
}
// Updating arr1[] which will now store the
// next sequence of elements
// At this time, arr1[] stores the remaining
// 'm' on which XOR was not performed and
// arr2[] stores the frequency of 'm^x' i.e.
// those 'm' on which operation was performed
// Updating arr1[] with frequency of remaining
// 'm' & frequency of 'm^x' from arr2[]
// With help of arr2[], we prevent sorting of
// the array again and again
for (int i = 0; i <= MAX; i++)
{
arr1[i] = arr1[i] + arr2[i];
// Resetting arr2[] for next iteration
arr2[i] = 0;
}
}
// Finding the maximum and the minimum element
// from the modified array after the operations
int min = Integer.MAX_VALUE;
int max = Integer.MIN_VALUE;
for (int i = 0; i <= MAX; i++)
{
if (arr1[i] > 0)
{
if (min > i)
min = i;
if (max < i)
max = i;
}
}
// Printing the max and the min element
System.out.println(min + " " + max);
}
// Driver code
public static void main(String[] args)
{
int[] arr = { 605, 986 };
int n = arr.length;
int k = 548, x = 569;
xorOnSortedArray(arr, n, k, x);
}
}
// This code is contributed by
// sanjeev2552Python 3
# Python 3 implementation of the approach
MAX = 10000
import sys
from math import ceil, floor
# Function to find the maximum and the
# minimum elements from the array after
# performing the given operation k times
def xorOnSortedArray(arr, n, k, x):
# To store the current sequence of elements
arr1 = [0 for i in range(MAX + 1)]
# To store the next sequence of elements
# after xoring with current elements
arr2 = [0 for i in range(MAX + 1)]
xor_val = [0 for i in range(MAX + 1)]
# Store the frequency of elements
# of arr[] in arr1[]
for i in range(n):
arr1[arr[i]] += 1
# Storing all precomputed XOR values
# so that we don't have to do it
# again and again as XOR is a costly operation
for i in range(MAX + 1):
xor_val[i] = i ^ x
# Perform the operations k times
while (k > 0):
k -= 1
# The value of count decide on
# how many elements we have to
# apply XOR operation
count = 0
for i in range(MAX + 1):
store = arr1[i]
# If current element is present in
# the array to be modified
if (arr1[i] > 0):
# Suppose i = m and arr1[i] = num,
# it means 'm' appears 'num' times
# If the count is even we have to
# perform XOR operation on alternate
# 'm' starting from the 0th index because
# count is even and we have to perform
# XOR operations starting with initial 'm'
# Hence there will be ceil(num/2)
# operations on 'm' that will change
# 'm' to xor_val[m] i.e. m^x
if (count % 2 == 0):
div = arr1[i] // 2 + 1
# Decrease the frequency of
# 'm' from arr1[]
arr1[i] = arr1[i] - div
# Increase the frequency of
# 'm^x' in arr2[]
arr2[xor_val[i]] += div
# If the count is odd we have to perform
# XOR operation on alternate 'm' starting
# from the 1st index because count is odd
# and we have to leave the 0th 'm'
# Hence there will be (num/2) XOR operations on
# 'm' that will change 'm' to xor_val[m] i.e. m^x
elif (count % 2 != 0):
div = arr1[i] // 2
arr1[i] = arr1[i] - div
arr2[xor_val[i]] += div
# Updating the count by frequency of
# the current elements as we have
# processed that many elements
count = count + store
# Updating arr1[] which will now store the
# next sequence of elements
# At this time, arr1[] stores the remaining
# 'm' on which XOR was not performed and
# arr2[] stores the frequency of 'm^x' i.e.
# those 'm' on which operation was performed
# Updating arr1[] with frequency of remaining
# 'm' & frequency of 'm^x' from arr2[]
# With help of arr2[], we prevent sorting of
# the array again and again
for i in range(MAX+1):
arr1[i] = arr1[i] + arr2[i]
# Resetting arr2[] for next iteration
arr2[i] = 0
# Finding the maximum and the minimum element
# from the modified array after the operations
mn = sys.maxsize
mx = -sys.maxsize-1
for i in range(MAX + 1):
if (arr1[i] > 0):
if (mn > i):
mn = i
if (mx < i):
mx = i
# Printing the max and the min element
print(mn,mx)
# Driver code
if __name__ == '__main__':
arr = [605, 986]
n = len(arr)
k = 548
x = 569
xorOnSortedArray(arr, n, k, x)
# This code is contributed by Surendra_GangwarC#
// C# implementation of the approach
using System;
class GFG
{
static int MAX = 100000;
// Function to find the maximum and the
// minimum elements from the array after
// performing the given operation k times
public static void xorOnSortedArray(int[] arr, int n,
int k, int x)
{
// To store the current sequence of elements
int[] arr1 = new int[MAX + 1];
// To store the next sequence of elements
// after xoring with current elements
int[] arr2 = new int[MAX + 1];
int[] xor_val = new int[MAX + 1];
// Store the frequency of elements
// of arr[] in arr1[]
for (int i = 0; i < n; i++)
arr1[arr[i]]++;
// Storing all precomputed XOR values so that
// we don't have to do it again and again
// as XOR is a costly operation
for (int i = 0; i <= MAX; i++)
xor_val[i] = i ^ x;
// Perform the operations k times
while (k-- > 0)
{
// The value of count decides on
// how many elements we have to
// apply XOR operation
int count = 0;
for (int i = 0; i <= MAX; i++)
{
int store = arr1[i];
// If current element is present in
// the array to be modified
if (arr1[i] > 0)
{
// Suppose i = m and arr1[i] = num,
// it means 'm' appears 'num' times
// If the count is even we have to perform
// XOR operation on alternate 'm' starting
// from the 0th index because count is even
// and we have to perform XOR operations
// starting with initial 'm'
// Hence there will be ceil(num/2) operations on
// 'm' that will change 'm' to xor_val[m] i.e. m^x
if (count % 2 == 0)
{
int div = (int) Math.Ceiling((double)(arr1[i] / 2));
// Decrease the frequency of 'm' from arr1[]
arr1[i] = arr1[i] - div;
// Increase the frequency of 'm^x' in arr2[]
arr2[xor_val[i]] += div;
}
// If the count is odd we have to perform
// XOR operation on alternate 'm' starting
// from the 1st index because count is odd
// and we have to leave the 0th 'm'
// Hence there will be (num/2) XOR operations on
// 'm' that will change 'm' to xor_val[m] i.e. m^x
else if (count % 2 != 0)
{
int div = arr1[i] / 2;
arr1[i] = arr1[i] - div;
arr2[xor_val[i]] += div;
}
}
// Updating the count by frequency of
// the current elements as we have
// processed that many elements
count = count + store;
}
// Updating arr1[] which will now store the
// next sequence of elements
// At this time, arr1[] stores the remaining
// 'm' on which XOR was not performed and
// arr2[] stores the frequency of 'm^x' i.e.
// those 'm' on which operation was performed
// Updating arr1[] with frequency of remaining
// 'm' & frequency of 'm^x' from arr2[]
// With help of arr2[], we prevent sorting of
// the array again and again
for (int i = 0; i <= MAX; i++)
{
arr1[i] = arr1[i] + arr2[i];
// Resetting arr2[] for next iteration
arr2[i] = 0;
}
}
// Finding the maximum and the minimum element
// from the modified array after the operations
int min = int.MaxValue;
int max = int.MinValue;
for (int i = 0; i <= MAX; i++)
{
if (arr1[i] > 0)
{
if (min > i)
min = i;
if (max < i)
max = i;
}
}
// Printing the max and the min element
Console.WriteLine(min + " " + max);
}
// Driver code
public static void Main(String[] args)
{
int[] arr = { 605, 986 };
int n = arr.Length;
int k = 548, x = 569;
xorOnSortedArray(arr, n, k, x);
}
}
// This code is contributed by Rajput-JiJavascript
输出：
605 986如果您希望与专家一起参加现场课程，请参阅DSA 现场工作专业课程和学生竞争性编程现场课程。