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# 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. `1`

   `<`

   `= A.length`

   `<`

   `= 50000`
2. `0`

   `<`

   `= A[i]`

   `<`

   `= 10^9`

分析

需要存当前的bit or的set，新来的数和所有数bit or,**再加上自己。每次cur都union 入res**

```
class Solution:
    def subarrayBitwiseORs(self, A: List[int]) -> int:        
        cur,res = set(),set()
        for i in A:
            cur = {i|j for j in cur} | {i}
            res |= cur
        return len(res)
```