子数组的异或（元素范围）
给定一个包含 n 个整数的数组 arr[] 和一些查询。每个查询的形式为 (L, R)，其中 L 和 R 是数组的索引。求子数组arr[L…R]的异或值，即[L,R]范围内的所有元素异或得到的值。假设基于 0 的索引并且每个数组元素是一个 32 位无符号整数。
例子：
Input : arr[] = {2, 5, 1, 6, 1, 2, 5}
L = 1, R = 4
Output : 3
The XOR value of arr[1...4] is 3.
Input : arr[] = {2, 5, 1, 6, 1, 2, 5}
L = 0, R = 6
Output : 6
The XOR value of arr[0...6] is 6.方法：
一个简单的解决方案是对每个查询从索引 L 到索引 R 遍历给定数组。这导致处理每个查询的时间复杂度为 O(n)。如果有 q 个查询，则所需的总时间将为 O(q*n)。对于大量查询和大型数组，此解决方案不是最佳的。
一个有效的解决方案是首先对数组进行预处理。有两个观察会有所帮助：
- 问题陈述中提到每个数组元素都是一个 32 位无符号数。因此，结果也将是一个 32 位无符号数。
- 当多个位异或时，如果有奇数个 1，则结果为 1，否则结果为 0。
使用观察 1，创建一个大小为 32*n 的二维数组 cnt。该数组将用于存储 1 的计数。元素cnt[i][j] 表示第i 位到索引j 的1 的计数，即在第i 位位置的子数组arr[0..j] 中存在多少个1。
根据观察 2 获得 XOR 值，需要找到子数组 arr[L…R] 中所有 32 位位置的 1 的个数。这可以在cnt数组的帮助下完成。子数组 arr[L…R] 中第 i 个位的个数为 cnt[i][R] – cnt[i][L-1]。如果第 i 位 1 的个数是奇数，则结果中将设置第 i 位。如果设置了第 i 位，则可以通过添加对应于第 i 位的 2 的幂来获得结果。
根据该算法处理每个查询将花费 O(32) 即恒定时间。创建 cnt 数组的预处理阶段将花费 O(n) 时间。因此，该解决方案对于 q 个查询的总时间复杂度为 O(n+q)。
执行：
C++
#include
using namespace std;
// Function to preprocess the array and find count of
// number of ones upto jth index for ith bit.
void preprocess(int arr[], int n, vector >& cnt)
{
int i, j;
// Run a loop for each bit position from
// 0 to 32.
for (i = 0; i < 32; i++) {
cnt[i][0] = 0;
for (j = 0; j < n; j++) {
if (j > 0) {
// store previous count of 1s
// for ith bit position.
cnt[i][j] = cnt[i][j - 1];
}
// Check if ith bit for jth element
// of array is set or not. If it is
// set then increase count of 1 for
// ith bit by 1.
if (arr[j] & (1 << i))
cnt[i][j]++;
}
}
}
// Function to find XOR value for a range of array elements.
int findXOR(int L, int R, const vector > cnt)
{
// variable to store final answer.
int ans = 0;
// variable to store number of 1s for ith bit
// in the range L to R.
int noOfOnes;
int i, j;
// Find number of 1s for each bit position from 0
// to 32.
for (i = 0; i < 32; i++) {
noOfOnes = cnt[i][R] - ((L > 0) ? cnt[i][L - 1] : 0);
// If number of 1s are odd then in the result
// ith bit will be set, i.e., 2^i will be present in
// the result. Add 2^i in ans variable.
if (noOfOnes & 1) {
ans += (1 << i);
}
}
return ans;
}
int main()
{
int arr[] = { 2, 5, 1, 6, 1, 2, 5 };
int n = sizeof(arr) / sizeof(arr[0]);
vector > cnt(32, vector(n));
preprocess(arr, n, cnt);
int L = 1;
int R = 4;
cout << findXOR(L, R, cnt);
return 0;
} Java
class GFG
{
// Function to preprocess the array and find count of
// number of ones upto jth index for ith bit.
static void preprocess(int arr[], int n, int[][] cnt)
{
int i, j;
// Run a loop for each bit position from
// 0 to 32.
for (i = 0; i < 32; i++)
{
cnt[i][0] = 0;
for (j = 0; j < n; j++)
{
if (j > 0)
{
// store previous count of 1s
// for ith bit position.
cnt[i][j] = cnt[i][j - 1];
}
// Check if ith bit for jth element
// of array is set or not. If it is
// set then increase count of 1 for
// ith bit by 1.
if ((arr[j] & (1 << i)) >= 1)
cnt[i][j]++;
}
}
}
// Function to find XOR value for a range of array elements.
static int findXOR(int L, int R, int[][] cnt)
{
// variable to store final answer.
int ans = 0;
// variable to store number of 1s for ith bit
// in the range L to R.
int noOfOnes;
int i, j;
// Find number of 1s for each bit position from 0
// to 32.
for (i = 0; i < 32; i++)
{
noOfOnes = cnt[i][R] - ((L > 0) ? cnt[i][L - 1] : 0);
// If number of 1s are odd then in the result
// ith bit will be set, i.e., 2^i will be present in
// the result. Add 2^i in ans variable.
if (noOfOnes % 2 == 1)
{
ans += (1 << i);
}
}
return ans;
}
// Driver code
public static void main(String[] args)
{
int arr[] = { 2, 5, 1, 6, 1, 2, 5 };
int n = arr.length;
int[][] cnt = new int[32][n];
preprocess(arr, n, cnt);
int L = 1;
int R = 4;
System.out.print(findXOR(L, R, cnt));
}
}
// This code is contributed by Rajput-JiPython3
# Function to preprocess the array and
# find count of number of ones upto
# jth index for ith bit.
def preprocess(arr, n, cnt):
# Run a loop for each bit position
# from 0 to 32.
for i in range(32):
cnt[i][0] = 0
for j in range(n):
if (j > 0):
# store previous count of 1s
# for ith bit position.
cnt[i][j] = cnt[i][j - 1]
# Check if ith bit for jth element
# of array is set or not. If it is
# set then increase count of 1 for
# ith bit by 1.
if (arr[j] & (1 << i)):
cnt[i][j] += 1
# Function to find XOR value
# for a range of array elements.
def findXOR(L, R, cnt):
# variable to store final answer.
ans = 0
# variable to store number of 1s
# for ith bit in the range L to R.
noOfOnes = 0
# Find number of 1s for each
# bit position from 0 to 32.
for i in range(32):
if L > 0:
noOfOnes = cnt[i][R] - cnt[i][L - 1]
else:
noOfOnes = cnt[i][R]
# If number of 1s are odd then in the result
# ith bit will be set, i.e., 2^i will be
# present in the result. Add 2^i in ans variable.
if (noOfOnes & 1):
ans += (1 << i)
return ans
# Driver Code
arr = [2, 5, 1, 6, 1, 2, 5]
n = len(arr)
cnt = [[0 for i in range(n)]
for i in range(32)]
preprocess(arr, n, cnt)
L = 1
R = 4
print(findXOR(L, R, cnt))
# This code is contributed by Mohit KumarC#
using System;
class GFG
{
// Function to preprocess the array and find count of
// number of ones upto jth index for ith bit.
static void preprocess(int []arr, int n, int[,] cnt)
{
int i, j;
// Run a loop for each bit position from
// 0 to 32.
for (i = 0; i < 32; i++)
{
cnt[i, 0] = 0;
for (j = 0; j < n; j++)
{
if (j > 0)
{
// store previous count of 1s
// for ith bit position.
cnt[i, j] = cnt[i, j - 1];
}
// Check if ith bit for jth element
// of array is set or not. If it is
// set then increase count of 1 for
// ith bit by 1.
if ((arr[j] & (1 << i)) >= 1)
cnt[i, j]++;
}
}
}
// Function to find XOR value for a range of array elements.
static int findXOR(int L, int R, int[,] cnt)
{
// variable to store readonly answer.
int ans = 0;
// variable to store number of 1s for ith bit
// in the range L to R.
int noOfOnes;
int i;
// Find number of 1s for each bit position from 0
// to 32.
for (i = 0; i < 32; i++)
{
noOfOnes = cnt[i, R] - ((L > 0) ? cnt[i, L - 1] : 0);
// If number of 1s are odd then in the result
// ith bit will be set, i.e., 2^i will be present in
// the result. Add 2^i in ans variable.
if (noOfOnes % 2 == 1)
{
ans += (1 << i);
}
}
return ans;
}
// Driver code
public static void Main(String[] args)
{
int []arr = { 2, 5, 1, 6, 1, 2, 5 };
int n = arr.Length;
int[,] cnt = new int[32, n];
preprocess(arr, n, cnt);
int L = 1;
int R = 4;
Console.Write(findXOR(L, R, cnt));
}
}
// This code is contributed by Rajput-JiJavascript
输出：
3时间复杂度： O(n+q)
辅助空间： O(n)