# 985. Sum of Even Numbers After Queries

**<font color=red>难度: Easy</font>**

## 刷题内容

> 原题连接

* https://leetcode.com/problems/sum-of-even-numbers-after-queries/

> 内容描述

```
We have an array A of integers, and an array queries of queries.

For the i-th query val = queries[i][0], index = queries[i][1], we add val to A[index].  Then, the answer to the i-th query is the sum of the even values of A.

(Here, the given index = queries[i][1] is a 0-based index, and each query permanently modifies the array A.)

Return the answer to all queries.  Your answer array should have answer[i] as the answer to the i-th query.

 

Example 1:

Input: A = [1,2,3,4], queries = [[1,0],[-3,1],[-4,0],[2,3]]
Output: [8,6,2,4]
Explanation: 
At the beginning, the array is [1,2,3,4].
After adding 1 to A[0], the array is [2,2,3,4], and the sum of even values is 2 + 2 + 4 = 8.
After adding -3 to A[1], the array is [2,-1,3,4], and the sum of even values is 2 + 4 = 6.
After adding -4 to A[0], the array is [-2,-1,3,4], and the sum of even values is -2 + 4 = 2.
After adding 2 to A[3], the array is [-2,-1,3,6], and the sum of even values is -2 + 6 = 4.
 

Note:

1 <= A.length <= 10000
-10000 <= A[i] <= 10000
1 <= queries.length <= 10000
-10000 <= queries[i][0] <= 10000
0 <= queries[i][1] < A.length
```

## 解题方案

> 思路 1
******- 时间复杂度: O(N)******- 空间复杂度: O(1)******

sb题没什么好说的


```python
class Solution:
    def sumEvenAfterQueries(self, A: 'List[int]', queries: 'List[List[int]]') -> 'List[int]':
        res = []
        sums = sum(i for i in A if i % 2 == 0)
        for val, idx in queries:
            if A[idx] % 2 == 0:
                sums -= A[idx]
            A[idx] += val
            if A[idx] % 2 == 0:
                sums += A[idx]
            res.append(sums)
        return res
```































