注意这里的子数组是连续的非空的数组，我们要学会与处理就是求交集

class Solution {
public:
    long long countSubarrays(vector<int>& nums, int k) {
        long long ans = 0;
        for (int i = 0; i < nums.size(); i++) {
            int x = nums[i];
            for (int j = i - 1; j >= 0 && (nums[j] & x) != nums[j]; j--) {
                nums[j] &= x;
            }
            ans += upper_bound(nums.begin(), nums.begin() + i + 1, k) -
                   lower_bound(nums.begin(), nums.begin() + i + 1, k);
        }
        return ans;
    }
};

class Solution {
public:
    long long countSubarrays(vector<int>& nums, int k) {
        long long ans = 0;
        int left = 0, right = 0;
        for (int i = 0; i < nums.size(); i++) {
            int x = nums[i];
            for (int j = i - 1; j >= 0 && (nums[j] & x) != nums[j]; j--) {
                nums[j] &= x;
            }
            while (left <= i && nums[left] < k) {
                left++;
            }
            while (right <= i && nums[right] <= k) {
                right++;
            }
            ans += right - left;
        }
        return ans;
    }
};

这种不会减少的边界就可以用指针来做