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# Subarray Sums Divisible by K

Given an array `A` of integers, return the number of (contiguous, non-empty) subarrays that have a sum divisible by `K`.

**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`

分析

sum(0,i)%k = sum(0,j)%k, i+1->j之间可以被K整除。 这里注意先把当前cnt\[cur] 加入res， 再increase cnt\[cur] 因为每次当前Index只需要之前的数字，不是更新后数字

```
class Solution:
    def subarraysDivByK(self, A: List[int], K: int) -> int:
        n = len(A)
        mm = {}
        mm[0] = 1
        res = prefix = 0
        for a in A:
            prefix += a
            rem = prefix%K
            if rem in mm:
                res += mm[rem]
                mm[rem] +=1
            else:
                mm[rem] = 1
        return res
```