数组所有子序列之和的按位或

给定长度为N的数组arr [] ，任务是从给定数组中找到所有可能的子序列之和的按位或。

例子：

Input: arr[] = {4, 2, 5}
Output: 15
Explanation: All subsequences from the given array and their corresponding sums:
{4} – 4
{2} – 2
{5} – 5
{4, 2} – 6
{4, 5} – 9
{2, 5} – 7
{4, 2, 5} -11
Therefore, the Bitwise OR of all sums = 4 | 2 | 5 | 6 | 9 | 7 | 11 = 15.

Input: arr[] = {1, 9, 8}
Output: 27
Explanation: All subsequences from the given array and their corresponding sums:
{1} – 1
{9} – 9
{8} – 8
{1, 9} – 10
{9, 8} – 17
{1, 8} – 9
{1, 9, 8} – 18
Therefore, Bitwise OR of all sums = 1 | 9 | 8 | 10 | 17 | 9 | 18 = 27.

天真的方法：最简单的方法是从给定数组生成所有可能的子序列，并找到它们各自的和。现在，在计算了它们的总和之后，打印所有获得的总和的按位或。

时间复杂度： O(2 ^N )
辅助空间： O(1)

高效的方法：为了优化上述方法，该想法是基于以下观察结果：

• 数组元素中的所有设置位也都设置在最终结果中。
• 给定数组的前缀和数组中设置的所有位也都设置在最终结果中。

请按照以下步骤解决以上问题：

• 用0初始化一个变量结果，该结果存储给定数组arr []的每个子序列之和的按位或。
• 用0初始化一个变量prefixSum ，该变量在任何时刻存储arr []的前缀和。
• 使用变量i遍历[0，N]范围内的数组元素。
  • 更新prefixSum作为prefixSum + = ARR [i]中。
  • 结果更新结果| = arr [i] | prefixSum。
• 完成上述步骤后，将结果值打印为答案。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to calculate Bitwise OR of
// sums of all subsequences
int findOR(int nums[], int N)
{
    // Stores the prefix sum of nums[]
    int prefix_sum = 0;
 
    // Stores the bitwise OR of
    // sum of each subsequence
    int result = 0;
 
    // Iterate through array nums[]
    for (int i = 0; i < N; i++) {
 
        // Bits set in nums[i] are
        // also set in result
        result |= nums[i];
 
        // Calculate prefix_sum
        prefix_sum += nums[i];
 
        // Bits set in prefix_sum
        // are also set in result
        result |= prefix_sum;
    }
 
    // Return the result
    return result;
}
 
// Driver Code
int main()
{
    // Given array arr[]
    int arr[] = { 4, 2, 5 };
 
    int N = sizeof(arr) / sizeof(arr[0]);
 
    // Function Call
    cout << findOR(arr, N);
 
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
 
class GFG{
   
// Function to calculate Bitwise OR of
// sums of all subsequences
static int findOR(int nums[], int N)
{
    // Stores the prefix sum of nums[]
    int prefix_sum = 0;
 
    // Stores the bitwise OR of
    // sum of each subsequence
    int result = 0;
 
    // Iterate through array nums[]
    for (int i = 0; i < N; i++) {
 
        // Bits set in nums[i] are
        // also set in result
        result |= nums[i];
 
        // Calculate prefix_sum
        prefix_sum += nums[i];
 
        // Bits set in prefix_sum
        // are also set in result
        result |= prefix_sum;
    }
 
    // Return the result
    return result;
}
 
// Driver Code
public static void main(String[] args)
{
    // Given array arr[]
    int arr[] = { 4, 2, 5 };
    int N = arr.length;
    System.out.print(findOR(arr, N));
}
}

Python3

# Python3 program for the
# above approach
 
# Function to calculate
# Bitwise OR of sums of
# all subsequences
def findOR(nums,  N):
 
    # Stores the prefix
    # sum of nums[]
    prefix_sum = 0
 
    # Stores the bitwise OR of
    # sum of each subsequence
    result = 0
 
    # Iterate through array nums[]
    for i in range(N):
 
        # Bits set in nums[i] are
        # also set in result
        result |= nums[i]
 
        # Calculate prefix_sum
        prefix_sum += nums[i]
 
        # Bits set in prefix_sum
        # are also set in result
        result |= prefix_sum
 
    # Return the result
    return result
 
# Driver Code
if __name__ == "__main__":
 
    # Given array arr[]
    arr = [4, 2, 5]
 
    N = len(arr)
 
    # Function Call
    print(findOR(arr, N))
 
# This code is contributed by Chitranayal

C#

// C# program for the above approach
using System;
 
class GFG{
 
// Function to calculate Bitwise OR of
// sums of all subsequences
static int findOR(int[] nums, int N)
{
     
    // Stores the prefix sum of nums[]
    int prefix_sum = 0;
 
    // Stores the bitwise OR of
    // sum of each subsequence
    int result = 0;
 
    // Iterate through array nums[]
    for(int i = 0; i < N; i++)
    {
         
        // Bits set in nums[i] are
        // also set in result
        result |= nums[i];
 
        // Calculate prefix_sum
        prefix_sum += nums[i];
 
        // Bits set in prefix_sum
        // are also set in result
        result |= prefix_sum;
    }
 
    // Return the result
    return result;
}
 
// Driver Code
public static void Main()
{
     
    // Given array arr[]
    int[] arr = { 4, 2, 5 };
     
    // Size of array
    int N = arr.Length;
     
    // Function call
    Console.Write(findOR(arr, N));
}
}
 
// This code is contributed by code_hunt

输出：

15

时间复杂度： O(N)
辅助空间： O(1)