題目如下：

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. 1 <= A.length <= 10000
2. -10000 <= A[i] <= 10000
3. 1 <= queries.length <= 10000
4. -10000 <= queries[i][0] <= 10000
5. 0 <= queries[i][1] < A.length

解題思路：題目很簡單，有一點需要註意，就是每次執行query後，不需要遍歷整個數組求出所有偶數的值。只需要記錄上一次偶數的和，執行query前，如果該值是偶數，用上一次的和減去該值；執行query後，如果新值是偶數，用上一次的和加上這個新值，就可以得到這次query執行後的偶數總和。

代碼如下：

class Solution(object):
    def sumEvenAfterQueries(self, A, queries):
        """
        :type A: List[int]
        :type queries: List[List[int]]
        :rtype: List[int]
        """
        res = []
        count = None
        for val,inx in queries:
            before = A[inx]
            A[inx] += val
            after = A[inx]
            if count == None:
                count = sum(filter(lambda x: x % 2 == 0,A))
            else:
                if before % 2 == 0:
                    count -= before
                if after % 2 == 0:
                    count += after
            res.append(count)
        return res

【leetcode】985. Sum of Even Numbers After Queries