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

已經說過這種題目多是固定格式，或者去Stack Overflow把題目標題複製上去，就有結果給你

class Solution {
public:
    int subarraysDivByK(vector<int>& A, int K) {
       long long ans = 0;
       long long k=0;
       map
 
<long long,long long>a;
       a[0]=1;
       int len = A.size();
       for(int i=0;i<len;i++){
        ans = ((ans + A[i])%K+K)%K;
        k +=a[ans];
        a[ans]++;
       }
       return k;
    }
};