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

思路：two pointer

class Solution:
    def subarraysDivByK(self, A, K):
        """
        :type A: List[int]
        :type K: int
        :rtype: int
        """
        su = 0
        a = []
        for i in A:
            su+=i
            su%=K
            a.append(su)
        
        d={0:[-1]}
        for i,v in enumerate(a):
            if v not in d: d[v]=[]
            d[v].append(i)
            
        res=0
        for i in d:
            l = len(d[i])
            res+=l*(l-1)//2
        return res