对数组的范围查询，其每个元素都是索引值与前一个元素的 XOR · GeeksForGeeks 中文系列教程

# 对数组的范围查询，其每个元素都是索引值与前一个元素的 XOR > 原文： [https://www.geeksforgeeks.org/range-query-array-whose-element-xor-index-value-previous-element/](https://www.geeksforgeeks.org/range-query-array-whose-element-xor-index-value-previous-element/) 考虑一个 **arr []** ，它可以定义为： ![](https://img.kancloud.cn/8e/a0/8ea0eb51420968191e07fc58f1bc7ccc_300x123.png) 给您 **Q** 查询，形式为 **[l，r]** 。任务是为每个查询输出 arr [l]⊕arr [l + 1]……..⊕arr [r-1]⊕arr [r]的值。 **示例**： ``` 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，例如 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。因此，arr [l] = arr [l + 1] ..... arr [r] = brr [l-1] = brr [r]。现在，让我们观察一下 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，在奇数索引中 brr [i] = 1⊕3⊕5⊕…。对于偶数索引 brr [i] = 2 4 6 6…。 ⊕我对于偶数索引，有从 1 到 i / 2 乘以 2 的数字，这意味着位将向左移动 1，因此 brr [i] = 2 4 4 6…。 ⊕i =（1⊕2⊕3…..⊕i / 2）*2。对于奇数索引，从 0 到（i – 1）/ 2 的数字乘以 2 加 1。位左移 1，最后一位为 1。因此，brr [i] = 1 3 3 5…。 ⊕i =（0⊕1⊕2⊕…。⊕（i – 1）/ 2）* 2 + x。 x 是 1⊕1。…..⊕1“ 1”重复（i – 1）/ 2 + 1 次。因此，如果（i-1）/ 2 +1 为奇数，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，< = k。 ``` bitmask(K)00=4K xorsum bitmask(K)01=4K+1 bitmask(K)10=4K+2 bitmask(K)11=4k+3 --------------------- 000000000000=0 ``` 因此，当 0⊕Y = Y 时，则 1⊕2⊕3⊕…x =（floor（x / 4）x 4）…⊕x 这里最多有 3 个数字，因此我们可以在`O(1)`中进行计算。下面是此方法的实现： ## C++ ```cpp // CPP Program to solve range query on array // whose each element is XOR of index value // and previous element. #include <bits/stdc++.h> 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 // 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 ```py # 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 sahishelangia ``` ## C# ```cs // 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 ```php <?php // PHP Program to solve range // query on array whose each // element is XOR of index // value and previous element. // function return derived // formula value. function fun($x) { $y = ((int)($x / 4) * 4); // finding xor value // of range [y...x] $ans = 0; for ($i = $y; $i <= $x; $i++) $ans ^= $i; return $ans; } // function to solve // query for l and r. function query($x) { // if l or r is 0\. if ($x == 0) return 0; $k = (int)(($x + 1) / 2); // finding x is divisible // by 2 or not. return ($x %= 2) ? 2 * fun($k) : ((fun($k - 1) * 2) ^ ($k & 1)); } function allQueries($q, $l, $r) { for ($i = 0; $i < $q; $i++) echo (query($r[$i]) ^ query($l[$i] - 1)) , "\n"; } // Driver Code $q = 3; $l = array( 2, 2, 5 ); $r = array ( 4, 8, 9 ); allQueries($q, $l, $r); // This code is contributed by ajit ?> ``` **输出**： ``` 0 2 0 ``` * * * * * *