查找所有可能的子数组的按位与(＆)

给定大小为N的数组A ，其中， $1\leq N \leq 10^{5}$ 。任务是找到所有可能的子阵列，然后将这些结果和所有的和。

例子：

Input : 1 2 3
Output : 0
All possible subarrays are 
{1}, {2}, {3}, {1, 2}, {2, 3} and {1, 2, 3}
ANDs of these subarrays are 1, 2, 3, 0, 2, 0.
AND of these ANDs is 0.

Input : 100 500 1000
Output : 96

方法：
朴素的解决办法是找到所有的子阵列的AND，然后打印其结果AND。这将导致O(N ² )解。

最佳解决方案：使用$X\&X\&...\&X=X$ i：e元素出现多少次无关紧要，并且AND ing仅算为一个。因此，我们的问题简化为仅找到数组所有元素的AND 。
下面是上述方法的实现。

C++

// C++ program to find of all the sub-arrays
#include 
using namespace std;
  
// function to return AND of sub-arrays
int AND(int a[], int n)
{
    int ans = a[0];
    for (int i = 0; i < n; ++i) 
        ans &= a[i];  
    return ans;
}
  
// Driver program
int main()
{
    int a[] = { 1, 2, 3 };
  
    // size of the array
    int n = sizeof(a) / sizeof(a[0]);
  
    // print and of all subarrays
    cout << AND(a, n);
  
    return 0;
}

Java

//Java program to find of all the sub-arrays
public class GFG {
  
    //function to return AND of sub-arrays
    static int AND(int a[], int n)
    {
     int ans = a[0];
     for (int i = 0; i < n; ++i) 
         ans &= a[i];  
     return ans;
    }
  
    // Driver code
    public static void main(String[] args) {
      
        int a[] = { 1, 2, 3 };
  
         // size of the array
         int n = a.length;
  
         // print and of all subarrays
         System.out.println(AND(a, n));
    }
}

Python 3

# Python 3 Program  to find of all the sub-arrays 
  
# function to return AND of sub-arrays
def AND(a, n) :
  
    ans = a[0]
    for i in range(n) :
        ans &= a[i]
          
    return ans
  
  
  
# Driver Code
if __name__ == "__main__" :
  
    a = [ 1, 2, 3]
  
    # size of the array
    n = len(a)
  
    # print and of all subarrays
    print(AND(a, n))
  
# This code is contributed by ANKITRAI1

C#

//C# program to find of all the sub-arrays  
  
using System;
  
public class GFG { 
    
    //function to return AND of sub-arrays 
    static int AND(int []a, int n) 
    { 
     int ans = a[0]; 
     for (int i = 0; i < n; ++i)  
         ans &= a[i];   
     return ans; 
    } 
    
    // Driver code 
    public static void Main() { 
        
        int []a = { 1, 2, 3 }; 
    
         // size of the array 
         int n = a.Length; 
    
         // print and of all subarrays 
         Console.WriteLine(AND(a, n)); 
    } 
}

输出：

0

时间复杂度： O(N)