# 查找数组中对的数量，以使它们的 XOR 为 0 > 原文： [https://www.geeksforgeeks.org/find-number-pairs-array-xor-0/](https://www.geeksforgeeks.org/find-number-pairs-array-xor-0/) 给定大小为 N 的数组![A[ ] ](https://img.kancloud.cn/f4/a5/f4a5c6b60e70dcecb9d2810436055aa9_32x28.png "Rendered by QuickLaTeX.com")。求对（i，j）的对数，以使 XOR  = 0，并且 1 < = i < j < =N。 **示例**： ``` Input : A[] = {1, 3, 4, 1, 4} Output : 2 Explanation : Index (0, 3) and (2, 4) Input : A[] = {2, 2, 2} Output : 3 ``` **第一种方法**： <font size="3">**排序**</font> 仅当时才满足 XOR  = 0 的要求。 因此，我们将首先对数组进行排序，然后计算每个元素的频率。 通过组合，我们可以观察到，如果某个元素的频率为，那么它将为答案提供。 以下是上述方法的实现： ## C++ ```cpp // C++ program to find number  // of pairs in an array such // that their XOR is 0 #include <bits/stdc++.h> using namespace std; // Function to calculate the // count int calculate(int a[], int n) {     // Sorting the list using     // built in function     sort(a, a + n);     int count = 1;     int answer = 0;     // Traversing through the     // elements     for (int i = 1; i < n; i++) {         if (a[i] == a[i - 1]){             // Counting frequency of each              // elements             count += 1;         }          else          {             // Adding the contribution of             // the frequency to the answer             answer = answer + (count * (count - 1)) / 2;             count = 1;         }     }     answer = answer + (count * (count - 1)) / 2;     return answer; } // Driver Code int main() {     int a[] = { 1, 2, 1, 2, 4 };     int n = sizeof(a) / sizeof(a[0]);     // Print the count     cout << calculate(a, n);     return 0; } // This article is contributed by Sahil_Bansall. ``` ## Java ```java // Java program to find number  // of pairs in an array such // that their XOR is 0 import java.util.*; class GFG  {     // Function to calculate      // the count     static int calculate(int a[], int n)     {         // Sorting the list using         // built in function         Arrays.sort(a);         int count = 1;         int answer = 0;         // Traversing through the         // elements         for (int i = 1; i < n; i++)          {             if (a[i] == a[i - 1])             {                 // Counting frequency of each                  // elements                 count += 1;             }              else             {                 // Adding the contribution of                 // the frequency to the answer                 answer = answer + (count * (count - 1)) / 2;                 count = 1;             }         }         answer = answer + (count * (count - 1)) / 2;         return answer;     }     // Driver Code     public static void main (String[] args)      {         int a[] = { 1, 2, 1, 2, 4 };         int n = a.length;         // Print the count         System.out.println(calculate(a, n));     } } // This code is contributed by Ansu Kumari. ``` ## Python3 ```py # Python3 program to find number of pairs # in an array such that their XOR is 0 # Function to calculate the count def calculate(a) :     # Sorting the list using     # built in function     a.sort()     count = 1     answer = 0     # Traversing through the elements     for i in range(1, len(a)) :         if a[i] == a[i - 1] :             # Counting frequncy of each elements             count += 1         else :             # Adding the contribution of             # the frequency to the answer             answer = answer + count * (count - 1) // 2             count = 1     answer = answer + count * (count - 1) // 2     return answer # Driver Code if __name__ == '__main__':     a = [1, 2, 1, 2, 4]     # Print the count     print(calculate(a)) ``` ## C# ```cs // C# program to find number  // of pairs in an array such // that their XOR is 0 using System; class GFG  {     // Function to calculate      // the count     static int calculate(int []a, int n)     {         // Sorting the list using         // built in function         Array.Sort(a);         int count = 1;         int answer = 0;         // Traversing through the         // elements         for (int i = 1; i < n; i++)          {             if (a[i] == a[i - 1])             {                 // Counting frequency of each                  // elements                 count += 1;             }              else             {                 // Adding the contribution of                 // the frequency to the answer                 answer = answer + (count * (count - 1)) / 2;                 count = 1;             }         }         answer = answer + (count * (count - 1)) / 2;         return answer;     }     // Driver Code     public static void Main ()      {         int []a = { 1, 2, 1, 2, 4 };         int n = a.Length;         // Print the count         Console.WriteLine(calculate(a, n));     } } // This code is contributed by vt_m. ``` ## PHP ```php <?php // PHP program to find number  // of pairs in an array such // that their XOR is 0 // Function to calculate  // the count function calculate($a, $n) {     // Sorting the list using     // built in function     sort($a);     $count = 1;     $answer = 0;     // Traversing through the     // elements     for ($i = 1; $i < $n; $i++)     {         if ($a[$i] == $a[$i - 1])         {             // Counting frequency of              // each elements             $count += 1;         }          else         {             // Adding the contribution of             // the frequency to the answer             $answer = $answer + ($count *                         ($count - 1)) / 2;             $count = 1;         }     }     $answer = $answer + ($count *                 ($count - 1)) / 2;     return $answer; }     // Driver Code     $a = array(1, 2, 1, 2, 4);     $n = count($a);     // Print the count     echo calculate($a, $n); // This code is contributed by anuj_67\. ?> ``` **输出**： ``` 2 ``` **时间复杂度**：`O(N log N)` ， **第二种方法**： <font size="3">**散列（索引映射）**</font> 如果我们可以计算数组中每个元素的频率，则解决方案很方便。 索引映射技术可用于计算每个元素的频率。 下面是上述方法的实现： ## C++ ``` // C++ program to find number of pairs // in an array such that their XOR is 0 #include <bits/stdc++.h> using namespace std; // Function to calculate the answer int calculate(int a[], int n){     // Finding the maximum of the array     int *maximum = max_element(a, a + 5);     // Creating frequency array     // With initial value 0     int frequency[*maximum + 1] = {0};     // Traversing through the array      for(int i = 0; i < n; i++)     {          // Counting frequency         frequency[a[i]] += 1;     }     int answer = 0;     // Traversing through the frequency array     for(int i = 0; i < (*maximum)+1; i++)     {          // Calculating answer         answer = answer + frequency[i] * (frequency[i] - 1) ;     }     return answer/2; } // Driver Code int main() {    int a[] = {1, 2, 1, 2, 4};    int n = sizeof(a) / sizeof(a[0]);     // Function calling    cout << (calculate(a,n)); } // This code is contributed by Smitha ```