Continuous Subarray Sum

题目描述：
Given a list of non-negative numbers and a target integer k, write a function to check if the array has a continuous subarray of size at least 2 that sums up to the multiple of k, that is, sums up to n*k where n is also an integer.

Example 1:
Input: [23, 2, 4, 6, 7], k=6
Output: True
Explanation: Because [2, 4] is a continuous subarray of size 2 and sums up to 6.
Example 2:
Input: [23, 2, 6, 4, 7], k=6
Output: True
Explanation: Because [23, 2, 6, 4, 7] is an continuous subarray of size 5 and sums up to 42.
Note:
The length of the array won’t exceed 10,000.
You may assume the sum of all the numbers is in the range of a signed 32-bit integer.

分析：
显然，当前i个的和对k的取余，等于前i+x,x>=2对k的取余时，有解，以此动态规划，其中特判0和第一位。

代码如下：

    bool checkSubarraySum(vector<int>& nums, int k) {       // if(k<0) k=-k;            if(k==0){            for(int i=1;i<nums.size();i++){                if(nums[i]==0&&nums[i-1]==0)                    return 1;            }         return 0;            }        int len=nums.size();        int now=0;        int temp;        set<int>ans;        ans.insert(0);        for(int i=0;i<len;i++){            now+=nums[i];            if(ans.count(now%k)&&i!=0) return 1;            ans.insert(temp);             temp=now%k;        }        return 0;    }