在给定数组中的所有对 (i, j) 上最大化表达式 [ij – K.(Ai | Aj)] 的值

给定一个长度为N的数组A [] 和一个整数K，任务是在给定数组中的所有对(i, j)上最大化表达式[ij – K.(A _i | A _j )]的值，其中 ( 1 ≤ i < j ≤ N)和|表示按位或运算符。

例子：

Input: A[] = {5, 20, 1, 0, 8, 11}, K = 10
Output: 2
Explanation: The maximum value of the expression f(i, j) = i.j – K.(A[i] | A[j]) can be found for below pair:
f(3, 4) = 3.4 – 10.(1 | 0) = 2

Input: A[] = {1, 5, 6, 7, 8, 19}, K = 3
Output: -9

方法：必须进行以下观察：

Let f(i, j) = i.j – K.(A_i | A_j). 

=> Notice that for f(i, j) to be maximum, i.j should be maximum and K.(A_i | A_j) should be minimum.
=> Thus, in the best case, f(i, j) will be maximum if 
      => K.(A_i | A_j) is equal to 0
      => i = (N-1) and j = N.
=> So, the maximum value of the expression can be f(i, j) = (N-1)*N.

=> Also, the minimum value will be obtained by subtracting the maximum value of K.(A_i | A_j) from (N-1)*N. 

=> It is known that 
a | b < 2*max(a, b) where | is the bitwise OR operator

从上面的属性和给定的约束，可以推断：

• (A _i | A _j )的最大值可以是2*N 。因为(0 ≤ A _i ≤ N)
• K的最大值可以是 100，因为(1 ≤ K ≤ min(N, 100)) 。
• 因此， f(i, j)的最小值将是：

f(i, j) = (N-1)*N – K*(A_i | Aj)
       = (N-1)*N – 100*2*N
       = N*(N – 201)

• 可以很容易地观察到，得到的答案总是介于：

N*(N-201) <= ans <= (N-1)*N

• 另外，请注意对于i = N – 201和j = N ，
  • f(i, j)的最大值将为N*(N – 201) ，
  • 这又是i = N – 1和j = N的f(i, j)的最小值。
  • 因此，必须从i = N – 201到i = N和j = i+1到j = N检查表达式的最大值。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to find the maximum
// value of the given expression
long long int maxValue(int N, int K,
                       long long int A[])
{
    // Stores the maximum value of
    // the given expression
    long long int ans = LLONG_MIN;
 
    // Nested loops to find the maximum
    // value of the given expression
    for (long long int i = max(0, N - 201);
         i < N; ++i) {
        for (long long int j = i + 1;
             j < N; ++j) {
            ans = max(ans, (i + 1) * (j + 1)
                               - K * (A[i] | A[j]));
        }
    }
 
    // Return the answer
    return ans;
}
 
// Driver Code
int main()
{
    // Given input
    int N = 6, K = 10;
    long long int A[N]
        = { 5, 20, 1, 0, 8, 11 };
 
    // Function Call
    cout << maxValue(N, K, A);
    return 0;
}

Java

// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
 
class GFG {
 
  // Function to find the maximum
  // value of the given expression
  static int maxValue(int N, int K,
                      int A[])
  {
    // Stores the maximum value of
    // the given expression
    int ans = Integer.MIN_VALUE;
 
    // Nested loops to find the maximum
    // value of the given expression
    for (int i = Math.max(0, N - 201);
         i < N; ++i) {
      for (int j = i + 1;
           j < N; ++j) {
        ans = Math.max(ans, (i + 1) * (j + 1)
                       - K * (A[i] | A[j]));
      }
    }
 
    // Return the answer
    return ans;
  }
 
  // Driver Code
  public static void main (String[] args)
  {
 
    // Given input
    int N = 6, K = 10;
    int  A[]  = { 5, 20, 1, 0, 8, 11 };
 
    // Function Call
    System.out.println( maxValue(N, K, A));
  }
}
 
// This code is contributed by hrithikgarg03188.

Python3

# Python3 program for the above approach
LLONG_MIN = -9223372036854775808
 
# Function to find the maximum
# value of the given expression
def maxValue(N, K, A):
 
    # Stores the maximum value of
    # the given expression
    ans = LLONG_MIN
 
    # Nested loops to find the maximum
    # value of the given expression
    for i in range(max(0, N - 201), N):
        for j in range(i + 1, N):
            ans = max(ans, (i + 1) * (j + 1)
                      - K * (A[i] | A[j]))
 
    # Return the answer
    return ans
 
# Driver Code
if __name__ == "__main__":
 
    # Given input
    N, K = 6, 10
    A = [5, 20, 1, 0, 8, 11]
 
    # Function Call
    print(maxValue(N, K, A))
 
# This code is contributed by rakeshsahni

C#

// C# program for the above approach
using System;
 
class GFG {
 
  // Function to find the maximum
  // value of the given expression
  static int maxValue(int N, int K,
                      int []A)
  {
     
    // Stores the maximum value of
    // the given expression
    int ans = Int32.MinValue;
 
    // Nested loops to find the maximum
    // value of the given expression
    for (int i = Math.Max(0, N - 201);
         i < N; ++i) {
      for (int j = i + 1;
           j < N; ++j) {
        ans = Math.Max(ans, (i + 1) * (j + 1)
                       - K * (A[i] | A[j]));
      }
    }
 
    // Return the answer
    return ans;
  }
 
  // Driver Code
  public static void Main ()
  {
 
    // Given input
    int N = 6, K = 10;
    int  []A  = { 5, 20, 1, 0, 8, 11 };
 
    // Function Call
    Console.WriteLine(maxValue(N, K, A));
  }
}
 
// This code is contributed by Samim Hossain Mondal.

Javascript

        // JavaScript code for the above approach
 
        // Function to find the maximum
        // value of the given expression
        function maxValue(N, K, A)
        {
         
            // Stores the maximum value of
            // the given expression
            let ans = Number.MIN_VALUE;
 
            // Nested loops to find the maximum
            // value of the given expression
            for (let i = Math.max(0, N - 201);
                i < N; ++i) {
                for (let j = i + 1;
                    j < N; ++j) {
                    ans = Math.max(ans, (i + 1) * (j + 1)
                        - K * (A[i] | A[j]));
                }
            }
 
            // Return the answer
            return ans;
        }
 
        // Driver Code
 
        // Given input
        let N = 6, K = 10;
        let A
            = [5, 20, 1, 0, 8, 11];
 
        // Function Call
        document.write(maxValue(N, K, A));
 
     // This code is contributed by Potta Lokesh
    

输出

2

时间复杂度： O(N ² )
辅助空间： O(1)