Leetcode523. 连续的子数组和--前缀和、动态规划

#include<iostream>
#include<vector>

using namespace std;

class Solution {
public:
    bool checkSubarraySum(vector<int>& nums, int k) {
        vector<int> presum(nums.size());
        presum[0]=nums[0];
        for(int i=1; i<nums.size(); i++){
            presum[i]=presum[i-1]+nums[i];
        }
        for (int i=0; i<nums.size()-1; i++){
            for(int j=i+1; j<nums.size(); j++){
                int sum=presum[j]-presum[i]+nums[i];
                if (sum==k || (k!=0 && sum%k==0)){
                    return true;
                }
            }
        }
        return false;
    }
};

int main(){
    vector<int> nums={23,2,6,4,7};
    int k=6;
    Solution S;
    bool ans=S.checkSubarraySum(nums, k);
    cout<<ans<<endl;
    return 0;
}
输出结果：
1