题目要求
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.
题目分析及思路
题目给出一个整数数组A和一个query数组，query数组的每一个元素是一个列表，该列表由两个整数组成，分别对应一个值和一个A索引。要求把该值加到该索引对应的A数组的值，然后将A数组中的偶数求和。最后返回所有query对应的结果。首先对A中所有偶数求和，之后遍历query数组，判断某索引对应A中的值在加值前后的奇偶性。
python代码
class Solution:
def sumEvenAfterQueries(self, A: ‘List[int]‘, queries: ‘List[List[int]]‘) -> ‘List[int]‘:
res = []
s = sum([i for i in A if i % 2 == 0])
for v, i in queries:
if A[i] % 2 == 0:
s -= A[i]
A[i] += v
if A[i] % 2 == 0:
s += A[i]
res.append(s)
return res
原文地址：https://www.cnblogs.com/yao1996/p/10356224.html