https://leetcode.com/problems/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][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:

  1. Input: A = [1,2,3,4], queries = [[1,0],[-3,1],[-4,0],[2,3]]
  2. Output: [8,6,2,4]
  3. Explanation:
  4. At the beginning, the array is [1,2,3,4].
  5. After adding 1 to A[0], the array is [2,2,3,4], and the sum of even values is 2 + 2 + 4 = 8.
  6. After adding -3 to A[1], the array is [2,-1,3,4], and the sum of even values is 2 + 4 = 6.
  7. After adding -4 to A[0], the array is [-2,-1,3,4], and the sum of even values is -2 + 4 = 2.
  8. 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

代码:

  1. class Solution {
  2. public:
  3. vector<int> sumEvenAfterQueries(vector<int>& A, vector<vector<int>>& queries) {
  4. int na = A.size(), nq = queries.size();
  5. int sum = 0;
  6. vector<int> ans;
  7. for(int i = 0; i < na; i ++) {
  8. if(A[i] % 2 == 0) sum += A[i];
  9. }
  10.  
  11. for(int i = 0; i < nq; i ++) {
  12. int pos = queries[i][1], val = queries[i][0];
  13. if(A[pos] % 2 == 0) {
  14. if((A[pos] + val) % 2 == 0)
  15. sum += val;
  16. else sum -= A[pos];
  17. } else {
  18. if((A[pos] + val) % 2 == 0)
  19. sum += val + A[pos];
  20. }
  21. A[pos] += val;
  22. ans.push_back(sum);
  23. }
  24. return ans;
  25. }
  26. };

  

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