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] <= 100002 <= K <= 10000class Solution { public int subarraysDivByK(int[] A, int K) { Map<Integer, Integer> map = new HashMap<>(); map.put(0, 1); int count = 0, sum = 0; for(int a : A) { sum = (sum + a) % K; if(sum < 0) sum += K; // Because -1 % 5 = -1, but we need the positive mod 4 count += map.getOrDefault(sum, 0); map.put(sum, map.getOrDefault(sum, 0) + 1); } return count; } }
诶？？？？把presum（premod）存起来，如果当前mod出现过，说明中间有能到cur的形成sum % k == 0的点，多少个呢？用map存了频率。
974. Subarray Sums Divisible by K
原文：https://www.cnblogs.com/wentiliangkaihua/p/13462443.html