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], index = queries[i]`, 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]` 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, the array is [2,2,3,4], and the sum of even values is 2 + 2 + 4 = 8.
After adding -3 to A, the array is [2,-1,3,4], and the sum of even values is 2 + 4 = 6.
After adding -4 to A, the array is [-2,-1,3,4], and the sum of even values is -2 + 4 = 2.
After adding 2 to A, 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] <= 10000`
5. `0 <= queries[i] < A.length`

985. Sum of Even Numbers After Queries
``````struct Solution;

impl Solution {
fn sum_even_after_queries(mut a: Vec<i32>, queries: Vec<Vec<i32>>) -> Vec<i32> {
let mut sum = a.iter().filter(|&x| x % 2 == 0).sum();
let mut res: Vec<i32> = vec![];
for query in queries {
let v = query;
let i = query as usize;
let x = a[i];
let y = a[i] + v;
if x % 2 == 0 {
sum -= x;
}
if y % 2 == 0 {
sum += y;
}
a[i] = y;
res.push(sum);
}
res
}
}

#[test]
fn test() {
let a = vec![1, 2, 3, 4];
let queries: Vec<Vec<i32>> = vec_vec_i32![[1, 0], [-3, 1], [-4, 0], [2, 3]];
let res = vec![8, 6, 2, 4];
assert_eq!(Solution::sum_even_after_queries(a, queries), res);
}
``````