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# 985.Sum of Even Numbers After Queries

难度:Easy

**985.Sum of Even Numbers After Queries**

> 给出一个整数数组 A 和一个查询数组 queries。 对于第 i 次查询，有 val = queries\[i]\[0], index = queries\[i]\[1]，我们会把 val 加到 A\[index] 上。然后，第 i 次查询的答案是 A 中偶数值的和。 （此处给定的 index = queries\[i]\[1] 是从 0 开始的索引，每次查询都会永久修改数组 A。） 返回所有查询的答案。你的答案应当以数组 answer 给出，answer\[i] 为第 i 次查询的答案。

示例：

```
输入：A = [1,2,3,4], queries = [[1,0],[-3,1],[-4,0],[2,3]]
输出：[8,6,2,4]
解释：
开始时，数组为 [1,2,3,4]。
将 1 加到 A[0] 上之后，数组为 [2,2,3,4]，偶数值之和为 2 + 2 + 4 = 8。
将 -3 加到 A[1] 上之后，数组为 [2,-1,3,4]，偶数值之和为 2 + 4 = 6。
将 -4 加到 A[0] 上之后，数组为 [-2,-1,3,4]，偶数值之和为 -2 + 4 = 2。
将 2 加到 A[3] 上之后，数组为 [-2,-1,3,6]，偶数值之和为 -2 + 6 = 4。
 
提示：
1 <= A.length <= 10000
-10000 <= A[i] <= 10000
1 <= queries.length <= 10000
-10000 <= queries[i][0] <= 10000
0 <= queries[i][1] < A.length
```

代码如下：

```
class Solution {
public:
    vector<int> sumEvenAfterQueries(vector<int>& A, vector<vector<int>>& queries) {
        int sum=0;
        for(auto i:A)
            sum += i%2 ?0:i;
       // cout<<sum<<endl;
        vector<int>res;
        for(auto q:queries)
        {
            if(A[q[1]]%2)
                q[0]%2 ? res.push_back(sum+=A[q[1]]+q[0]) : res.push_back(sum);
            else
                q[0]%2 ?  res.push_back(sum-=A[q[1]]) :res.push_back(sum+=q[0]) ;
            A[q[1]]+=q[0];
            //cout<<q[0]<<A[q[1]]<<endl;
           //cout<<sum<<endl;
        }
        return res;
    }
};
```

> 执行用时 : 360 ms, 在Sum of Even Numbers After Queries的C++提交中击败了38.36% 的用户 内存消耗 : 33.1 MB, 在Sum of Even Numbers After Queries的C++提交中击败了100.00% 的用户


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