題目如下：

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

解題思路：本題需要用到一個數學規律，如果a%c = b%c，那麽(a-b)%c=0。我的解法就是從後往前遍歷數組，依次累加每個元素的值並記為sum，同時用字典保存sum%K作為key值出現的次數。同時每累加一個元素，只要去字典中查找歷史sum%K出現的次數，這個次數就是從以這個元素作為起點滿足條件的子數組的個數。特別註意的是，如果sum%K=0，那麽表示這個元素本身就滿足條件，次數要+1。

代碼如下：

class Solution(object):
    
 
def subarraysDivByK(self, A, K):
        """
        :type A: List[int]
        :type K: int
        :rtype: int
        """
        dic = {}
        count = 0
        res = 0
        for i in A[::-1]:
            count += i
            if count%K in dic:
                res += dic[count%K]
            
 
if count % K == 0:
                res += 1
            dic[count%K] = dic.setdefault(count%K,0)+1
        return res
        

【leetcode】974. Subarray Sums Divisible by K