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📜  与K进行XOR运算后查询子数组中奇偶校验元素的计数

📅  最后修改于: 2021-04-27 21:05:34             🧑  作者: Mango

给定由N个元素组成的数组arr []和由LRK表示的Q个查询。任务是在对K进行按位异或后,打印子数组[L,R]中奇偶校验元素的计数。

例子:

方法:想法是使用MO的算法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。

  • 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。块中的所有查询均按R值的升序排序。
  • 计算奇数奇偶校验元素,然后将偶数奇偶校验元素计算为

(R - L + 1- odd parity elements)

  • XOR后使用奇偶校验元素进行观察:
    • 两个奇数奇偶校验元素的XOR是偶数奇偶校验元素。
    • 两个偶数奇偶校验元素的XOR是偶数奇偶校验元素。
    • 一个偶数奇偶校验元素和另一个奇数奇偶校验元素的XOR是奇数奇偶校验元素,反之亦然。
  • 一个接一个地处理所有查询并增加奇数奇偶校验元素的数量,现在我们将检查K的奇偶校验。如果K具有偶校验,然后偶校验的计数保持我们交换他们其他人。
    • count_oddP存储上一次查询中的奇数奇偶校验元素的计数。
    • 删除先前查询的多余元素,并为当前查询添加新元素。例如,如果上一个查询为[0,8],而当前查询为[3,9],则删除元素arr [0],arr [1]和arr [2]并添加arr [9]。
  • 为了显示结果,请按查询顺序对查询进行排序。

添加元素

  • 如果当前元素具有奇校验,则增加奇校验的计数。

移除元素

  • 如果当前元素具有奇校验,则减少奇校验的计数。

下面是上述方法的实现:

C++
// C++ program to count odd and
// even parity elements in subarray
// after XOR with K
 
#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, R, K, 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,
            k = q[i].K;
 
        // 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++;
        }
 
        // If parity of K is even
        // then the count of odd
        // and even parity remains
        // the same
        if (!__builtin_parity(k)) {
            q[i].odd = count_oddP;
            q[i].even
                = R - L + 1 - count_oddP;
        }
        // If parity of K is odd
        // we swap the count of
        // odd and even parity
        // elements
        else {
            q[i].odd
                = R - L + 1 - count_oddP;
            q[i].even = 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 };
    int n = sizeof(arr) / sizeof(arr[0]);
 
    Query q[] = { { 0, 5, 3, 0, 0, 0 },
                  { 1, 4, 8, 1, 0, 0 },
                  { 4, 6, 10, 2, 0, 0 } };
 
    int m = sizeof(q) / sizeof(q[0]);
 
    queryResults(arr, n, q, m);
 
    printResults(q, m);
 
    return 0;
}


Java
// Java program to count odd and
// even parity elements in subarray
// after XOR with K
import java.io.*;
import java.util.*;
 
// Class to represent
// a query range
class Query
{
    int L, R, K, index;
 
    // Count of odd
    // parity elements
    int odd;
 
    // Count of even
    // parity elements
    int even;
 
    Query(int L, int R, int K,
          int index, int odd, int even)
    {
        this.L = L;
        this.R = R;
        this.K = K;
        this.index = index;
        this.odd = odd;
        this.even = even;
    }
}
 
class GFG{
 
// Variable to represent block size.
static int block;
 
// To store the count of
// odd parity elements
static int count_oddP;
 
// Function to Add elements
// of current range
static void add(int currL, int a[])
{
     
    // _builtin_parity(x)returns true
    // if the number has odd parity else
    // it returns false for even parity.
    if (__builtin_parity(a[currL]))
        count_oddP++;
}
 
// Function to remove elements
// of previous range
static void remove(int currR, int a[])
{
     
    // _builtin_parity(x)returns true
    // if the number has odd parity else
    // it returns false for even parity.
    if (__builtin_parity(a[currR]))
        count_oddP--;
}
 
// Function to generate the result of queries
static 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)(Math.sqrt(n));
 
    // 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.
    Arrays.sort(q, (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;
    });
 
    // 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, k = q[i].K;
 
        // 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++;
        }
 
        // If parity of K is even
        // then the count of odd
        // and even parity remains
        // the same
        if (!__builtin_parity(k))
        {
            q[i].odd = count_oddP;
            q[i].even = R - L + 1 - count_oddP;
        }
         
        // If parity of K is odd
        // we swap the count of
        // odd and even parity
        // elements
        else
        {
            q[i].odd = R - L + 1 - count_oddP;
            q[i].even = count_oddP;
        }
    }
}
 
static boolean __builtin_parity(int K)
{
    return (Integer.bitCount(K) % 2) == 1;
}
 
// Function to display the results of
// queries in their initial order
static void printResults(Query q[], int m)
{
    Arrays.sort(q, (Query x, Query y) ->
 
                // sort all queries
                // in order of their
                // index value so that results
                // of queries can be printed
                // in same order as of input);
                x.index - y.index);
 
    for(int i = 0; i < m; i++)
    {
        System.out.println(q[i].odd + " " +
                           q[i].even);
    }
}
 
// Driver Code
public static void main(String[] args)
{
    int arr[] = { 5, 2, 3, 1, 4, 8, 10 };
    int n = arr.length;
 
    Query q[] = new Query[3];
    q[0] = new Query(0, 5, 3, 0, 0, 0);
    q[1] = new Query(1, 4, 8, 1, 0, 0);
    q[2] = new Query(4, 6, 10, 2, 0, 0);
 
    int m = q.length;
 
    queryResults(arr, n, q, m);
 
    printResults(q, m);
}
}
 
// This code is contributed by jithin


输出:
4 2
1 3
2 1