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

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

代码：

class Solution {

public:

    vector<int> sumEvenAfterQueries(vector<int>& A, vector<vector<int>>& queries) {

        int na = A.size(), nq = queries.size();

        int sum = 0;

        vector<int> ans;

        for(int i = 0; i < na; i ++) {

            if(A[i] % 2 == 0) sum += A[i];

        }

        for(int i = 0; i < nq; i ++) {

            int pos = queries[i][1], val = queries[i][0];

            if(A[pos] % 2 == 0) {

                if((A[pos] + val) % 2 == 0)

                    sum += val;

                else sum -= A[pos];

            } else {

                if((A[pos] + val) % 2 == 0)

                    sum += val + A[pos];

            }

            A[pos] += val;

            ans.push_back(sum);

        }

        return ans;

    }

};