使用 MO 算法计算子阵列中奇偶校验元素的数量

给定一个由N 个元素和Q 个查询组成的数组 arr ，由 L和R表示一个范围，任务是打印子数组 [L, R] 中奇偶校验元素的计数。

例子：

Input:
arr[]=[5, 2, 3, 1, 4, 8, 10]
Q=2
1 3
0 4
Output:
2 1
3 2

Explanation:
In query 1, odd parity elements in subarray [1:3] are 2 and 1 and even parity element is 3.
In query 2, odd parity elements in subarray [0:4] are 2, 1 and 4 and even parity elements are 5 and 3.

Input:
arr[] = { 13, 17, 12, 10, 18, 19, 15, 7, 9, 6 }
Q=3
1 5
0 7
2 9
Output:
1 4
3 5
2 6
Explanation:
In query 1, odd parity element in subarray [1:4] is 19 and even parity elements are 17,12,10 and 18.
In query 2, odd parity elements in subarray [0:7] are 13, 19 and 7 and even parity elements are 17,12,10,18 and 15.
In query 3, odd parity elements in subarray [2:6] are 19 and 7 and even parity elements are 12,10,18, 15, 9 and 6.

方法：
MO 算法的思想是对所有查询进行预处理，以便一个查询的结果可以在下一个查询中使用。

1. 对所有查询进行排序，将L值从0 到 √n – 1的查询放在一起，然后是从√n 到 2×√n – 1 的查询，依此类推。块内的所有查询都按R值的升序排序。
2. 计算奇校验元素，然后计算偶校验元素为(R-L+1-奇校验元素)
3. 一一处理所有查询并增加奇校验元素的计数并将结果存储在结构中。
   • 让count_oddP存储先前查询中奇数奇偶校验元素的计数。
   • 删除先前查询的额外元素并为当前查询添加新元素。例如，如果之前的查询是 [0, 8] 而当前的查询是 [3, 9]，那么移除元素 arr[0]、arr[1] 和 arr[2] 并添加 arr[9]。
4. 为了显示结果，请按照提供的顺序对查询进行排序。

添加元素()

• 如果当前元素具有奇校验，则增加count_oddP的计数。

  删除元素()

  • 如果当前元素具有奇校验，则减少count_oddP的计数。

    下面的代码是上述方法的实现：

    C++

    // C++ program to count odd and
    // even parity elements in subarray
    // using MO's algorithm
      
    #include 
    using namespace std;
      
    #define MAX 100000
      
    // Variable to represent block size.
    // This is made global so compare()
    // of sort can use it.
    int block;
      
    // Structure to represent a query range 
    struct Query {
        // Starting index
        int L; 
        // Ending index
        int R;
        // Index of query
        int index;
        // Count of odd
        // parity elements
        int odd;
        // Count of even
        // parity elements
        int even;
    };
      
    // To store the count of
    // odd parity elements
    int count_oddP;
      
    // Function used to sort all queries so that
    // all queries of the same block are arranged
    // together and within a block, queries are
    // sorted in increasing order of R values.
    bool compare(Query x, Query y)
    {
        // Different blocks, sort by block.
        if (x.L / block != y.L / block)
            return x.L / block < y.L / block;
      
        // Same block, sort by R value
        return x.R < y.R;
    }
      
    // Function used to sort all queries in order of their
    // index value so that results of queries can be printed
    // in same order as of input
    bool compare1(Query x, Query y)
    {
        return x.index < y.index;
    }
      
    // Function to Add elements
    // of current range
    void add(int currL, int a[])
    {
        // _builtin_parity(x)returns true(1)
        // if the number has odd parity else
        // it returns false(0) for even parity.
        if (__builtin_parity(a[currL]))
            count_oddP++;
    }
      
    // Function to remove elements
    // of previous range
    void remove(int currR, int a[])
    {
        // _builtin_parity(x)returns true(1)
        // if the number has odd parity else
        // it returns false(0) for even parity.
        if (__builtin_parity(a[currR]))
            count_oddP--;
    }
      
    // Function to generate the result of queries
    void queryResults(int a[], int n, Query q[],
                    int m)
    {
      
        // Initialize number of odd parity
        // elements to 0
        count_oddP = 0;
      
        // Find block size
        block = (int)sqrt(n);
      
        // Sort all queries so that queries of
        // same blocks are arranged together.
        sort(q, q + m, compare);
      
        // Initialize current L, current R and
        // current result
        int currL = 0, currR = 0;
      
        for (int i = 0; i < m; i++) {
            // L and R values of current range
            int L = q[i].L, R = q[i].R;
      
            // Add Elements of current range
            while (currR <= R) {
                add(currR, a);
                currR++;
            }
            while (currL > L) {
                add(currL - 1, a);
                currL--;
            }
      
            // Remove element of previous range
            while (currR > R + 1)
      
            {
                remove(currR - 1, a);
                currR--;
            }
            while (currL < L) {
                remove(currL, a);
                currL++;
            }
      
            q[i].odd = count_oddP;
            q[i].even = R - L + 1 - count_oddP;
        }
    }
    // Function to display the results of
    // queries in their initial order
    void printResults(Query q[], int m)
    {
        sort(q, q + m, compare1);
        for (int i = 0; i < m; i++) {
            cout << q[i].odd << " " 
                   << q[i].even << endl;
        }
    }
      
    // Driver Code
    int main()
    {
      
        int arr[] = { 5, 2, 3, 1, 4, 8, 10, 12 };
        int n = sizeof(arr) / sizeof(arr[0]);
      
        Query q[] = { { 1, 3, 0, 0, 0 }, 
                      { 0, 4, 1, 0, 0 }, 
                      { 4, 7, 2, 0, 0 } };
      
        int m = sizeof(q) / sizeof(q[0]);
      
        queryResults(arr, n, q, m);
      
        printResults(q, m);
          
        return 0;
    }

    输出：

    2 1
    3 2
    2 2
    

    时间复杂度： O(Q × √n)

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