## Description

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

## Solutions

题意比较简单，每次 query，都把对应的数加到对应位置上，然后计算数列中所有偶数的和。

### 1. Array

# Time: O(n)
# Space: O(1)
class Solution:
def sumEvenAfterQueries(self, A: List[int], queries: List[List[int]]) -> List[int]:
res = []
pre_sum = self.get_even_sum(A)
for val, idx in queries:
if A[idx] % 2 == 0 and val % 2 == 0:
pre_sum += val
elif A[idx] % 2 != 0 and val % 2 != 0:
pre_sum += val + A[idx]
elif A[idx] % 2 == 0 and val % 2 != 0:
pre_sum -= A[idx]
else: # A[idx] % 2 != 0 and val % 2 == 0
pass
A[idx] += val
res.append(pre_sum)
return res

def get_even_sum(self, A):
if not A:
return 0
sum_v = 0
for a in A:
if a % 2 == 0:
sum_v += a
return sum_v

# 58/58 cases passed (548 ms)
# Your runtime beats 82.58 % of python3 submissions
# Your memory usage beats 5.88 % of python3 submissions (18.8 MB)