leetcode 523. Continuous Subarray Sum

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.

板梯题意很简单，就在一个暴力遍历即可

代码如下：

#include <iostream>#include <vector>#include <map>#include <set>#include <queue>#include <stack>#include <string>#include <climits>#include <algorithm>#include <sstream>#include <functional>#include <bitset>#include <numeric>#include <cmath>using namespace std;class Solution {public:    bool checkSubarraySum(vector<int>& nums, int k)     {        vector<int> dp(nums.size()+1, 0);        for (int i = 1; i <= nums.size(); i++)        {            dp[i] = dp[i-1] + nums[i-1];        }        for (int i = 0; i < nums.size(); i++)        {            for (int j = i + 2; j <= nums.size(); j++)            {                if (k == 0)                {                    if (dp[j] - dp[i] == 0)                        return true;                }                else if ((dp[j] - dp[i]) % k == 0)                    return true;            }        }        return false;    }};