human
I found an interesting problem on https://leetcode.com/problems/bitwise-ors-of-subarrays/:
We have an array A of non-negative integers.
For every (contiguous) subarray B = [A[i], A[i+1], ..., A[j]] (with i <= j), we take the bitwise OR of all the elements in B, obtaining a result A[i] | A[i+1] | ... | A[j].
Return the number of possible results.  (Results that occur more than once are only counted once in the final answer.)
 

Example 1:
Input: [0]
Output: 1
Explanation: 
There is only one possible result: 0.

Example 2:
Input: [1,1,2]
Output: 3
Explanation: 
The possible subarrays are [1], [1], [2], [1, 1], [1, 2], [1, 1, 2].
These yield the results 1, 1, 2, 1, 3, 3.
There are 3 unique values, so the answer is 3.

Example 3:
Input: [1,2,4]
Output: 6
Explanation: 
The possible results are 1, 2, 3, 4, 6, and 7.



 
Note:

1 <= A.length <= 50000
0 <= A[i] <= 10^9

I tried it in Python, but could not do it. Can you solve it?

 ... Write me a the Python code you requested?
assistant
#Here is the Python code you requested.

class Solution:
    def subarrayBitwiseORs(self, A: List[int]) -> int:
        s, ans = set(), set()
        for a in A:
            s = {a} | {a | b for b in s}
            ans |= s
        return len(ans)