范围查询数组，每个元素都是索引值和前一个元素的异或
考虑一个arr[]可以定义为：
您将获得[l, r]形式的Q查询。任务是为每个查询输出 arr[l] ⊕ arr[l+1] ⊕ ….. ⊕ arr[r-1] ⊕ arr[r] 的值。
例子 ：
Input : q = 3
q1 = { 2, 4 }
q2 = { 2, 8 }
q3 = { 5, 9 }
Output : 7
9
15
The beginning of the array with constraint look like:
arr[] = { 0, 1, 3, 0, 4, 1, 7, 0, 8, 1, 11, .... }
For q1, 3 ⊕ 0 ⊕ 4 = 7
For q2, 3 ⊕ 0 ⊕ 4 ⊕ 1 ⊕ 7 ⊕ 0 ⊕ 8 = 9
For q3, 1 ⊕ 7 ⊕ 0 ⊕ 8 ⊕ 1 = 15让我们观察arr[]
arr[0] = 0
arr[1] = 1
arr[2] = 1 ⊕ 2
arr[3] = 1 ⊕ 2 ⊕ 3
arr[4] = 1 ⊕ 2 ⊕ 3 ⊕ 4
arr[5] = 1 ⊕ 2 ⊕ 3 ⊕ 4 ⊕ 5
....让我们创建另一个数组，比如 brr[]，其中 brr[i] = arr[0] ⊕ arr[1] ⊕ arr[2] ⊕ ..... ⊕ arr[i]。
brr[i] = arr[0] ⊕ arr[1] ⊕ arr[2] ⊕ … ⊕ arr[i-1] ⊕ arr[i] = brr[j] ⊕ arr[j+1] ⊕ arr[j+ 2] ⊕ ….. ⊕ arr[i]，对于任何 0 <= j <= i。
所以，arr[l] ⊕ arr[l+1] ⊕ ..... ⊕ arr[r] = brr[l-1] ⊕ brr[r]。
现在，让我们观察 brr[]：
brr[1] = 1
brr[2] = 2
brr[3] = 1 ⊕ 3
brr[4] = 2 ⊕ 4
brr[5] = 1 ⊕ 3 ⊕ 5
brr[6] = 2 ⊕ 4 ⊕ 6
brr[7] = 1 ⊕ 3 ⊕ 5 ⊕ 7
brr[8] = 2 ⊕ 4 ⊕ 6 ⊕ 8很容易观察到奇数索引 brr[i] = 1 ⊕ 3 ⊕ 5 ⊕ ...。 ⊕ i 和偶数索引 brr[i] = 2 ⊕ 4 ⊕ 6 ⊕ ...。 ⊕ 一世。
对于偶数索引，有从 1 到 i/2 乘以 2 的数字，这意味着位向左移动 1，因此，brr[i] = 2 ⊕ 4 ⊕ 6 ⊕ ...。 ⊕ i = (1 ⊕ 2 ⊕ 3 ⊕ ….. ⊕ i/2) * 2。
对于奇数索引，有从 0 到 (i – 1)/2 的数字乘以 2 并加 1。这意味着位向左移动 1，最后一位变为 1。所以，brr[i] = 1 ⊕ 3 ⊕ 5 ⊕ …. ⊕ i = (0 ⊕ 1 ⊕ 2 ⊕ .... ⊕ (i – 1)/2) * 2 + x。
x 是 1 ⊕ 1 ⊕ 1 ⊕ ….. ⊕ 1 个“一”重复 (i – 1)/2 + 1 次。因此，如果 (i-1)/2 + 1 是奇数，则 x = 1，否则 x = 0。
现在，计算 1 ⊕ 2 ⊕ 3 ⊕ ...。 ⊕ x。
让我们证明 (4K) ⊕ (4K + 1) ⊕ (4K + 2) ⊕ (4K + 3) = 0 对于 0 <= k。
bitmask(K)00=4K
xorsum bitmask(K)01=4K+1
bitmask(K)10=4K+2
bitmask(K)11=4k+3
---------------------
000000000000=0所以 0 ⊕ Y = Y 然后 1 ⊕ 2 ⊕ 3 ⊕ … ⊕ x = (floor(x/4) x 4) ⊕ … ⊕ x 这里最多有 3 个数字，所以我们可以在 O(1) 中计算。
下面是这种方法的实现：
C++
// CPP Program to solve range query on array
// whose each element is XOR of index value
// and previous element.
#include
using namespace std;
// function return derived formula value.
int fun(int x)
{
int y = (x / 4) * 4;
// finding xor value of range [y...x]
int ans = 0;
for (int i = y; i <= x; i++)
ans ^= i;
return ans;
}
// function to solve query for l and r.
int query(int x)
{
// if l or r is 0.
if (x == 0)
return 0;
int k = (x + 1) / 2;
// finding x is divisible by 2 or not.
return (x %= 2) ? 2 * fun(k) : ((fun(k - 1) * 2) ^ (k & 1));
}
void allQueries(int q, int l[], int r[])
{
for (int i = 0; i < q; i++)
cout << (query(r[i]) ^ query(l[i] - 1)) << endl;
}
// Driven Program
int main()
{
int q = 3;
int l[] = { 2, 2, 5 };
int r[] = { 4, 8, 9 };
allQueries(q, l, r);
return 0;
} Java
// Java Program to solve range query on array
// whose each element is XOR of index value
// and previous element.
import java.io.*;
class GFG {
// function return derived formula value.
static int fun(int x)
{
int y = (x / 4) * 4;
// finding xor value of range [y...x]
int ans = 0;
for (int i = y; i <= x; i++)
ans ^= i;
return ans;
}
// function to solve query for l and r.
static int query(int x)
{
// if l or r is 0.
if (x == 0)
return 0;
int k = (x + 1) / 2;
// finding x is divisible by 2 or not.
return ((x %= 2) != 0) ? 2 * fun(k) :
((fun(k - 1) * 2) ^ (k & 1));
}
static void allQueries(int q, int l[], int r[])
{
for (int i = 0; i < q; i++)
System.out.println((query(r[i]) ^
query(l[i] - 1))) ;
}
// Driven Program
public static void main (String[] args) {
int q = 3;
int []l = { 2, 2, 5 };
int []r = { 4, 8, 9 };
allQueries(q, l, r);
}
}
// This code is contributed by vt_m.Python3
# Python3 Program to solve range query
# on array whose each element is XOR of
# index value and previous element.
# function return derived formula value.
def fun(x):
y = (x // 4) * 4
# finding xor value of range [y...x]
ans = 0
for i in range(y, x + 1):
ans ^= i
return ans
# function to solve query for l and r.
def query(x):
# if l or r is 0.
if (x == 0):
return 0
k = (x + 1) // 2
# finding x is divisible by 2 or not.
if x % 2 == 0:
return((fun(k - 1) * 2) ^ (k & 1))
else:
return(2 * fun(k))
def allQueries(q, l, r):
for i in range(q):
print(query(r[i]) ^ query(l[i] - 1))
# Driver Code
q = 3
l = [ 2, 2, 5 ]
r = [ 4, 8, 9 ]
allQueries(q, l, r)
# This code is contributed
# by sahishelangiaC#
// C# Program to solve range query on array
// whose each element is XOR of index value
// and previous element.
using System;
class GFG {
// function return derived formula value.
static int fun(int x)
{
int y = (x / 4) * 4;
// finding xor value of range [y...x]
int ans = 0;
for (int i = y; i <= x; i++)
ans ^= i;
return ans;
}
// function to solve query for l and r.
static int query(int x)
{
// if l or r is 0.
if (x == 0)
return 0;
int k = (x + 1) / 2;
// finding x is divisible by 2 or not.
return ((x %= 2)!=0) ? 2 * fun(k) :
((fun(k - 1) * 2) ^ (k & 1));
}
static void allQueries(int q, int []l, int []r)
{
for (int i = 0; i < q; i++)
Console.WriteLine((query(r[i])
^ query(l[i] - 1))) ;
}
// Driven Program
public static void Main ()
{
int q = 3;
int []l = { 2, 2, 5 };
int []r = { 4, 8, 9 };
allQueries(q, l, r);
}
}
// This code is contributed by vt_m.PHP
Javascript
输出 ：
0
2
0