Given an arrayAof integers, return the number of (contiguous, non-empty) subarrays that have a sum divisible byK.

Example 1:

Input: A = [4,5,0,-2,-3,1], K = 5
Output: 7
Explanation: There are 7 subarrays with a sum divisible by K = 5:
[4, 5, 0, -2, -3, 1], [5], [5, 0], [5, 0, -2, -3], [0], [0, -2, -3], [-2, -3]
 

Note:

1. 1 <= A.length <= 30000
2. -10000 <= A[i] <= 10000
3. 2 <= K <= 10000

class Solution {
    public int subarraysDivByK(int[] A, int K) {
        Map<Integer, Integer> map = new HashMap<>();
        map.put(0, 1);
        int count = 0, sum = 0;
        for(int a : A) {
            sum 
 
= (sum + a) % K;
            if(sum < 0) sum += K;  // Because -1 % 5 = -1, but we need the positive mod 4
            count += map.getOrDefault(sum, 0);
            map.put(sum, map.getOrDefault(sum, 0) + 1);
        }
        return count;
    }
}

誒？？？？把presum（premod）存起來，如果當前mod出現過，說明中間有能到cur的形成sum % k == 0的點，多少個呢？用map存了頻率。