Smallest Divisor Meeting a Rounded-Sum Limit

### Problem Choose a positive integer divisor d. Divide every nums value by d, round each quotient upward to an integer, then sum them. Return the smallest d whose rounded sum is at most threshold. A feasible divisor exists. ```cpp class Solution { public: int smallestDivisor(vector<int>& nums, int threshold); }; ``` ### Examples **Example 1:** ```text Input: nums = [3,6,7], threshold = 6 Output: 3 Explanation: Divisor3 gives1+2+3=6. Divisor2 gives2+3+4=9, so no smaller divisor works. ``` **Example 2:** ```text Input: nums = [4,8], threshold = 2 Output: 8 Explanation: Each term must round to1; this first happens for the largest value at divisor8. ``` ### Constraints - There are 1 to 50000 values, each an integer in `[1,1000000]`. - `nums.length <= threshold <= 1000000`. - A feasible divisor exists. ### Contract considerations All original values are positive. A feasible divisor is guaranteed under the supplied threshold range. ```hint Read the boundary condition Is a rounded sum equal to threshold permitted, and is the returned value a divisor or the sum? ``` ### Boundary reminders All original values are positive. A feasible divisor is guaranteed under the supplied threshold range.