We have an array nums
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 nums[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 nums
.)
Return the answer to all queries. Your answer
array should have answer[i]
as the answer to the i
-th query.
Example 1:
Input: nums = [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 nums[0], the array is [2,2,3,4], and the sum of even values is 2 + 2 + 4 = 8. After adding -3 to nums[1], the array is [2,-1,3,4], and the sum of even values is 2 + 4 = 6. After adding -4 to nums[0], the array is [-2,-1,3,4], and the sum of even values is -2 + 4 = 2. After adding 2 to nums[3], the array is [-2,-1,3,6], and the sum of even values is -2 + 6 = 4.
Note:
1 <= nums.length <= 10000
-10000 <= nums[i] <= 10000
1 <= queries.length <= 10000
-10000 <= queries[i][0] <= 10000
0 <= queries[i][1] < nums.length