题目

思路：滑动窗口

• 右指针right，每轮循环往前走一步。for(int right = 0; right < len_size; ++right){}
• 如果出现不同时符合两个条件（1.2.）的数组，那么左边的指针需要不断往右走，直到满足条件才可以
• 为什么滑动窗口是有效的？因为滑动窗口遍历了所有满足条件的子数组

  代码：

  ```cpp

  include

  include

  include

using namespace std;

int main() { int n, m, k; cin >> n >> m >> k;

// input array
vector<int> nums;
for (int i = 0; i < n; ++i) {
    int input = 0;
    cin >> input;
    nums.push_back(input);
}
int cnt_nega = 0, max_comb = 0;
for (int right = 0, left = 0; right < n; ++right) {
    int right_num = nums[right], left_num = nums[left];
    if (right_num < 0) {cnt_nega++;}
    while (left < right && right_num + left_num > k && cnt_nega > m) {
        if (left_num < 0) {cnt_nega--;}
        left++;
        left_num = nums[left];
    }
    if (left != right) {max_comb = max(max_comb, left_num + right_num);}
}
cout << "max_comb is: " << max_comb << endl; 
return 0;

}

```