> For the complete documentation index, see [llms.txt](https://dfine.gitbook.io/leetcode/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://dfine.gitbook.io/leetcode/974.subarray_sums_divisible_by_k.md).

# 974.Subarray Sums Divisible By K

**974.Subarray Sums Divisible By K**

难度:Medium

> 给定一个整数数组 A，返回其中元素之和可被 K 整除的（连续、非空）子数组的数目。

```
示例：

输入：A = [4,5,0,-2,-3,1], K = 5
输出：7
解释：
有 7 个子数组满足其元素之和可被 K = 5 整除：
[4, 5, 0, -2, -3, 1], [5], [5, 0], [5, 0, -2, -3], [0], [0, -2, -3], [-2, -3]
 

提示：

1 <= A.length <= 30000
-10000 <= A[i] <= 10000
2 <= K <= 10000
```

如果某个子列能整除，则从零开始到头尾的子列和模K相同，最后找出相同的模的组合。 模为0的时候的统计数应该+1，默认为索引0之前有一个模为0的起点。

```
class Solution {
public:
    int subarraysDivByK(vector<int>& A, int K) {
        if(A.empty()) return 0;
        vector<int> allMod(K,0);
        int sum=0;
        allMod[0]=1;
        for(int i=0;i<A.size();i++)
        {
            sum+=A[i];
            sum %=K;
            if(sum<0) 
            {
                sum+=K;
            }
            allMod[sum]++;
         //cout<<sum<<endl;
        }
        int res=0;
        for(auto m:allMod)
        {
            if(m>1)
            res+=m*(m-1)/2;
        }

        return res;
    }
};
```

> 执行用时 :92 ms, 在所有 C++ 提交中击败了82.90%的用户\
> 内存消耗 :28.7 MB, 在所有 C++ 提交中击败了50.00%的用户


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