题目
https://leetcode.com/problems/subarray-sums-divisible-by-k/
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
思路
前缀和而后dp同余即可,任意范围的子数组和(i,j] = sum[j]-sum[i],我们要确保他mod K = 0,根据同余定理可知要求 sum[i] mod K == sum[j] mod K 即 sum[i] 同余 sum[j], 所以使用 hashmap 记录余数的个数,后面直接累加
源码
class Solution {
public:
int sum[30005];
int res = 0;
map<int,int> m;
int subarraysDivByK(vector<int>& A, int K) {
if(A.size()==0)return res;
sum[0] = 0;
m[0] = 1;
for(int i = 1;i <= A.size();++i){
sum[i] = sum[i-1]+A[i-1];
int yu = sum[i]%K;
if(m.count(yu)==0){
m[yu] = 1;
}else{
m[yu]++;
}
res+=(m[yu]-1);
}
return res;
}
};