0985. Sum of Even Numbers After Queries

# 985. Sum of Even Numbers After Queries#

## 题目 #

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 #

• 给出一个数组 A 和 query 数组。要求每次 query 操作都改变数组 A 中的元素值，并计算此次操作结束数组 A 中偶数值之和。
• 简单题，先计算 A 中所有偶数之和。再每次 query 操作的时候，动态维护这个偶数之和即可。

## 代码 #

``````
package leetcode

func sumEvenAfterQueries(A []int, queries [][]int) []int {
cur, res := 0, []int{}
for _, v := range A {
if v%2 == 0 {
cur += v
}
}
for _, q := range queries {
if A[q[1]]%2 == 0 {
cur -= A[q[1]]
}
A[q[1]] += q[0]
if A[q[1]]%2 == 0 {
cur += A[q[1]]
}
res = append(res, cur)
}
return res
}

``````

Apr 8, 2023