题目如下：

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

解题思路：题目很简单，有一点需要注意，就是每次执行query后，不需要遍历整个数组求出所有偶数的值。只需要记录上一次偶数的和，执行query前，如果该值是偶数，用上一次的和减去该值；执行query后，如果新值是偶数，用上一次的和加上这个新值，就可以得到这次query执行后的偶数总和。

代码如下：

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