985. 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 <= A.length <= 10000
-10000 <= A[i] <= 10000
1 <= queries.length <= 10000
-10000 <= queries[i][0] <= 10000
0 <= queries[i][1] < A.length

这题也没什么，不断的维护总和即可，不能每次操作都重新计算综合，肯定会超时。

class SumOfEvenNumbersAfterQueries : public Solution {
public:
    void Execute() {

    }

    vector<int> sumEvenAfterQueries(vector<int>& A, vector<vector<int>>& queries) {
        int sum = 0;
        for (auto v : A) {
            if (v % 2 == 0) {
                sum += v;
            }
        }
        vector<int> sums;
        for (auto &query : queries) {
            int prev = A[query[1]];
            int now = prev + query[0];
            if (prev % 2 == 0 && now % 2 == 0) {
                sum += now - prev;
            } else if (prev % 2 == 0 && now % 2 != 0) {
                sum -= prev;
            } else if (prev % 2 != 0 && now % 2 == 0) {
                sum += now;
            }
            sums.push_back(sum);
            A[query[1]] = now;
        }
        return sums;
    }
};