两个不同数组的子数组的最大OR和

给定两个正整数数组。从每个阵列中选择两个大小相等的子阵列，然后计算两个子阵列的最大可能OR值。
注意：令f(x，l，r)是数组x中[l，r]范围内所有元素的OR和。
例子 ：

Input : A[] = {1, 2, 4, 3, 2}
        B[] = {2, 3, 3, 12, 1}
Output : 22
Explanation: Here, one way to get maximum
sum is to select sub-array [l = 2, r = 4]
f(A, 2, 4) = 2|4|3 = 7
f(B, 2, 4) = 3|3|12 = 15
So, f(A, 2, 4) + f(B, 2, 4) = 7 + 15 = 22.
This sum can be achieved in many other ways.

Input : A[] = {1, 2, 2}
        B[] = {2, 1, 3}
Output : 6

观察按位OR运算符。如果我们取两个整数X和Y，则(X | Y> = X)。可以通过一些例子来证明。让我们使用以上方程式得出一个公式。
$f(a, 1, i-1) | f(a, i, j) | f(a, j+1, n) >= f(a, i, j)$
并且$f(a, 1, i-1) | f(a, i, j) | f(a, j+1, n) = f(a, 1, n)$
根据以上两个方程， $f(a, 1, n) >= f(a, i, j).$
因此，当我们对整个数组进行“或”运算时，我们将得到最大的总和。 $f(a, 1, n) + f(b, 1, n)$
下面是上述方法的实现：

C++

// CPP program to find maximum OR sum
#include 
using namespace std;
 
// function to find maximum OR sum
void MaximumSum(int a[], int b[], int n)
{
    int sum1 = 0, sum2 = 0;
     
    // OR sum of all the elements
    // in both arrays
    for (int i = 0; i < n; i++) {
        sum1 |= a[i];
        sum2 |= b[i];
    }
    cout << sum1 + sum2 << endl;
}
 
// Driver Code
int main()
{
    int A[] = { 1, 2, 4, 3, 2 };
    int B[] = { 2, 3, 3, 12, 1 };
    int n = sizeof(A) / sizeof(A[0]);
    MaximumSum(A, B, n);
    return 0;
}

Java

// Java program to find maximum OR sum
 
class GFG {
     
// function to find maximum OR sum
static void MaximumSum(int a[], int b[], int n)
{
    int sum1 = 0, sum2 = 0;
 
    // OR sum of all the elements
    // in both arrays
    for (int i = 0; i < n; i++) {
    sum1 |= a[i];
    sum2 |= b[i];
    }
    System.out.println(sum1 + sum2);
}
 
// Driver code
public static void main(String arg[])
{
    int A[] = {1, 2, 4, 3, 2};
    int B[] = {2, 3, 3, 12, 1};
    int n = A.length;
    MaximumSum(A, B, n);
}
}
 
// This code is contributed by Anant Agarwal.

Python3

# Python 3 program to
# find maximum OR sum
 
# function to find
# maximum OR sum
def MaximumSum(a, b, n):
 
    sum1 = 0
    sum2 = 0
     
    # OR sum of all the
    # elements in both arrays
    for i in range(0, n):
        sum1 |= a[i]
        sum2 |= b[i]
     
    print(sum1 + sum2)
 
# Driver Code
A = [ 1, 2, 4, 3, 2 ]
B = [ 2, 3, 3, 12, 1 ]
n = len(A)
 
MaximumSum(A, B, n)
 
# This code is contributed by Smitha Dinesh Semwal

C#

// C# program to find maximum OR sum
using System;
 
class GFG {
     
    // function to find maximum OR sum
    static void MaximumSum(int []a, int []b, int n)
    {
        int sum1 = 0, sum2 = 0;
     
        // OR sum of all the elements
        // in both arrays
        for (int i = 0; i < n; i++)
        {
            sum1 |= a[i];
            sum2 |= b[i];
        }
        Console.WriteLine(sum1 + sum2);
    }
     
    // Driver code
    public static void Main()
    {
        int []A = {1, 2, 4, 3, 2};
        int []B = {2, 3, 3, 12, 1};
        int n = A.Length;
        MaximumSum(A, B, n);
    }
}
 
// This code is contributed by Vt_m.

Javascript

 
// JavaScript program to find maximum OR sum
 
// function to find maximum OR sum
function MaximumSum(a, b, n)
{
    let sum1 = 0, sum2 = 0;
   
    // OR sum of all the elements
    // in both arrays
    for (let i = 0; i < n; i++) {
    sum1 |= a[i];
    sum2 |= b[i];
    }
    document.write(sum1 + sum2);
}
 
// Driver code
         
    let A = [1, 2, 4, 3, 2];
    let B = [2, 3, 3, 12, 1];
    let n = A.length;
    MaximumSum(A, B, n);
                   

输出 ：

22