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