985. Sum of Even Numbers After Queries

问题：

给定原数组A，和操作数组queries

假设q=queries[i]

那么q的操作为：对A[q[1]]+=q[0]

evensum为A的所有偶数元素之和，求每次操作后evensum的值，所组成的列表res

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

解法：

先求的原数组的偶数和evensum

接着遍历操作数组queries

判断操作前A[q[1]]

若为偶数，evensum-=A[q[1]]

进行操作：A[q[1]]+=q[0]

再判断操作后的A[q[1]]

若为偶数，evensum+=A[q[1]]

把evensum放入结果数组res。

代码参考：

 1 class Solution {
 2 public:
 3     vector<int> sumEvenAfterQueries(vector<int>& A, vector<vector<int>>& queries) {
 4         vector<int> res;
 5         int evensum=0;
 6         for(int a:A){
 7             if(!(a%2))evensum+=a;
 8         }
 9         for(vector<int> q:queries){
10             if(!(A[q[1]]%2)) evensum-=A[q[1]];
11             A[q[1]]+=q[0];
12             if(!(A[q[1]]%2)) evensum+=A[q[1]];
13             res.push_back(evensum);
14         }
15         return res;
16     }
17 };

这里，也可以判断操作前A[q[1]]和要追加值q[0]的奇偶性

若为偶+偶，则evensum+=q[0]即可

偶+奇，则evensum-=A[q[1]]即可

奇+奇，则evensum+=(A[q[1]]+q[0])

奇+偶，不变化。

代码参考：

 1 class Solution {
 2 public:
 3     vector<int> sumEvenAfterQueries(vector<int>& A, vector<vector<int>>& queries) {
 4         vector<int> res;
 5         int evensum=0;
 6         for(int a:A){
 7             if(!(a%2))evensum+=a;
 8         }
 9         for(vector<int> q:queries){
10             if(!(A[q[1]]%2)){//原数字为偶数
11                 if(!(q[0]%2)){
12                     evensum+=q[0];
13                 }else{
14                     evensum-=A[q[1]];
15                 }
16             }else{//原数字为奇数
17                 if(q[0]%2){
18                     evensum+=(A[q[1]]+q[0]);
19                 }
20             }
21             res.push_back(evensum);
22             A[q[1]]+=q[0];
23         }
24         return res;
25     }
26 };