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 <= A.length <= 30000
-10000 <= A[i] <= 10000
2 <= K <= 10000
已经说过这种题目多是固定格式,或者去Stack Overflow把题目标题复制上去,就有结果给你
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; } };