两个数组 XOR 之间偶数设置位的计数

给定两个数组A[] 和 B[] 分别具有N和M 个正元素。任务是为数组 B 的每个元素计算数组 A 中具有偶数位异或中设置位的元素的数量。
例子：

Input: A[] = { 4, 2, 15, 9, 8, 8 }, B[] = { 3, 4, 22 } 
Output: 2 4 4 
Explanation: 
Binary representation of elements of A are : 100, 10, 1111, 1001, 1000, 1000 
Binary representation of elements of B are : 11, 101, 10110 
Now for element 3(11), 
3^4 = 11^100 = 111 
3^2 = 11^10 = 01 
3^15 = 11^1111 = 1100 
3^9 = 11^1001 = 1111 
3^8 = 11^1000 = 1011 
3^8 = 11^1000 = 1011 
Only 2 elements {15, 9} in A[] are there for element 3 such that count of set bit after XOR is even. So the count is 2. 
Similarly, Count for element 4 and 22 is 4. 
Input: A[] = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 }, B[] = { 4 } 
Output: 5 
Explanation: 
The element in A[] such that count of set bit after XOR is even is {1, 2, 4, 7, 8}. So the count is 5.
 

朴素的方法：这个想法是计算数组B[]中每个元素与数组A[] 中每个元素的异或，并计算具有偶数设置位的数字。
时间复杂度： O(N*M)，其中 N 和 M 分别是数组A[]和B[]的长度。
高效的方法：这个想法是使用 XOR 的属性。对于任何两个数字，如果两个数字的设置位计数是偶数或奇数，则两个数字异或后的设置位计数是偶数。
以下是基于上述属性的步骤：

1. 计算数组A[] 中具有偶数(比如a )和奇数(比如b )设置位的元素的数量。
2. 对于数组B[]中的每个元素：
   • 如果当前元素有偶数设置位，则数组A[] 中与当前元素异或设置偶数为偶数的元素为a 。
   • 如果当前元素的设置位为奇数，则数组A[] 中与当前元素异或设置位为偶数的元素为b 。

下面是上述方法的实现：

CPP

// C++ program for the above approach
#include 
using namespace std;
 
// Function that count the XOR of B[]
// with all the element in A[] having
// even set bit
void countEvenBit(int A[], int B[], int n, int m)
{
    int i, j, cntOdd = 0, cntEven = 0;
    for (i = 0; i < n; i++) {
 
        // Count the set bits in A[i]
        int x = __builtin_popcount(A[i]);
 
        // check for even or Odd
        if (x & 1) {
            cntEven++;
        }
        else {
            cntOdd++;
        }
    }
 
    // To store the count of element for
    // B[] such that XOR with all the
    // element in A[] having even set bit
    int CountB[m];
 
    for (i = 0; i < m; i++) {
 
        // Count set bit for B[i]
        int x = __builtin_popcount(B[i]);
 
        // check for Even or Odd
        if (x & 1) {
            CountB[i] = cntEven;
        }
        else {
            CountB[i] = cntOdd;
        }
    }
 
    for (i = 0; i < m; i++) {
        cout << CountB[i] << ' ';
    }
}
 
// Driver Code
int main()
{
    int A[] = { 4, 2, 15, 9, 8, 8 };
    int B[] = { 3, 4, 22 };
 
    countEvenBit(A, B, 6, 3);
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
 
class GFG{
  
// Function that count the XOR of B[]
// with all the element in A[] having
// even set bit
static void countEvenBit(int A[], int B[], int n, int m)
{
    int i, j, cntOdd = 0, cntEven = 0;
    for (i = 0; i < n; i++) {
  
        // Count the set bits in A[i]
        int x = Integer.bitCount(A[i]);
  
        // check for even or Odd
        if (x % 2 == 1) {
            cntEven++;
        }
        else {
            cntOdd++;
        }
    }
  
    // To store the count of element for
    // B[] such that XOR with all the
    // element in A[] having even set bit
    int []CountB = new int[m];
  
    for (i = 0; i < m; i++) {
  
        // Count set bit for B[i]
        int x = Integer.bitCount(B[i]);
  
        // check for Even or Odd
        if (x%2 == 1) {
            CountB[i] = cntEven;
        }
        else {
            CountB[i] = cntOdd;
        }
    }
  
    for (i = 0; i < m; i++) {
        System.out.print(CountB[i] +" ");
    }
}
  
// Driver Code
public static void main(String[] args)
{
    int A[] = { 4, 2, 15, 9, 8, 8 };
    int B[] = { 3, 4, 22 };
  
    countEvenBit(A, B, 6, 3);
}
}
 
// This code is contributed by sapnasingh4991

Python3

# Python3 program for the above approach
 
# Function that count the XOR of B
# with all the element in A having
# even set bit
def countEvenBit(A, B, n, m):
 
    i, j, cntOdd = 0, 0, 0
    cntEven = 0
    for i in range(n):
 
        # Count the set bits in A[i]
        x = bin(A[i])[2:].count('1')
 
        # check for even or Odd
        if (x & 1):
            cntEven += 1
 
        else :
            cntOdd += 1
 
    # To store the count of element for
    # B such that XOR with all the
    # element in A having even set bit
    CountB = [0]*m
 
    for i in range(m):
 
        # Count set bit for B[i]
        x = bin(B[i])[2:].count('1')
 
        # check for Even or Odd
        if (x & 1):
            CountB[i] = cntEven
 
        else:
            CountB[i] = cntOdd
 
    for i in range(m):
        print(CountB[i], end=" ")
 
# Driver Code
if __name__ == '__main__':
 
    A = [ 4, 2, 15, 9, 8, 8]
    B = [ 3, 4, 22 ]
 
    countEvenBit(A, B, 6, 3)
 
# This code is contributed by mohit kumar 29

C#

// C# program for the above approach
using System;
 
class GFG{
 
// Function that count the XOR of []B
// with all the element in []A having
// even set bit
static void countEvenBit(int []A, int []B, int n, int m)
{
    int i, cntOdd = 0, cntEven = 0;
    for (i = 0; i < n; i++)
    {
 
        // Count the set bits in A[i]
        int x = bitCount(A[i]);
 
        // check for even or Odd
        if (x % 2 == 1) {
            cntEven++;
        }
        else {
            cntOdd++;
        }
    }
 
    // To store the count of element for
    // []B such that XOR with all the
    // element in []A having even set bit
    int []CountB = new int[m];
 
    for (i = 0; i < m; i++) {
 
        // Count set bit for B[i]
        int x = bitCount(B[i]);
 
        // check for Even or Odd
        if (x % 2 == 1) {
            CountB[i] = cntEven;
        }
        else {
            CountB[i] = cntOdd;
        }
    }
 
    for (i = 0; i < m; i++) {
        Console.Write(CountB[i] +" ");
    }
}
static int bitCount(int x)
{
    int setBits = 0;
    while (x != 0) {
        x = x & (x - 1);
        setBits++;
    }
    return setBits;
}
 
// Driver Code
public static void Main(String[] args)
{
    int []A = { 4, 2, 15, 9, 8, 8 };
    int []B = { 3, 4, 22 };
 
    countEvenBit(A, B, 6, 3);
}
}
 
// This code is contributed by Rajput-Ji

Javascript

 
// Javascript program for the above approach
 
// Function that count the XOR of B[]
// with all the element in A[] having
// even set bit
function countEvenBit(A, B, n, m)
{
    let i, j, cntOdd = 0, cntEven = 0;
    for (i = 0; i < n; i++) {
 
        // Count the set bits in A[i]
        let x = bitCount(A[i]);
 
        // check for even or Odd
        if (x & 1) {
            cntEven++;
        }
        else {
            cntOdd++;
        }
    }
 
    // To store the count of element for
    // B[] such that XOR with all the
    // element in A[] having even set bit
    let CountB = new Array(m);
 
    for (i = 0; i < m; i++) {
 
        // Count set bit for B[i]
        let x = bitCount(B[i]);
 
        // check for Even or Odd
        if (x & 1) {
            CountB[i] = cntEven;
        }
        else {
            CountB[i] = cntOdd;
        }
    }
 
    for (i = 0; i < m; i++) {
        document.write(CountB[i] + " ");
    }
}
 
function bitCount(x)
{
    let setBits = 0;
    while (x != 0) {
        x = x & (x - 1);
        setBits++;
    }
    return setBits;
}
 
// Driver Code
    let A = [ 4, 2, 15, 9, 8, 8 ];
    let B = [ 3, 4, 22 ];
 
    countEvenBit(A, B, 6, 3);
 

输出：

2 4 4

时间复杂度： O(N + M)，其中 N 和 M 分别是给定的两个数组的长度。

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