Given an array of integers ​​nums​​​ and an integer ​​threshold​​, we will choose a positive integer divisor and divide all the array by it and sum the result of the division. Find the smallest divisor such that the result mentioned above is less than or equal to ​​threshold​​.

Each result of division is rounded to the nearest integer greater than or equal to that element. (For example: 7/3 = 3 and 10/2 = 5).

It is guaranteed that there will be an answer.

 

Example 1:

Input: nums = [1,2,5,9], threshold = 6
Output: 5
Explanation: We can get a sum to 17 (1+2+5+9) if the divisor is 1. 
If the divisor is 4 we can get a sum to 7 (1+1+2+3) and if the divisor is 5 the sum will be 5 (1+1+1+2).

Example 2:

Input: nums = [2,3,5,7,11], threshold = 11
Output: 3

Example 3:

Input: nums = [19], threshold = 5
Output: 4

 

Constraints:

  • ​​1 <= nums.length <= 5 * 10^4​​
  • ​​1 <= nums[i] <= 10^6​​
  • ​​nums.length <= threshold <= 10^6​​

​​题解：​​

​​标准二分搜索。​​

class Solution {
public:
    int calDiv(vector<int>& nums, int div) {
        int sum = 0;
        for (int i = 0; i < nums.size(); i++) {
            int num = 0;
            if (nums[i] % div == 0) {
                num = nums[i] / div;
            }
            else {
                num = nums[i] / div + 1;
            }
            sum += num;
        }
        return sum;
    }
    int smallestDivisor(vector<int>& nums, int threshold) {
        int n = nums.size();
        if (n == 0) {
            return 0;
        }
        int left = 1, right = threshold;
        for (int i = 0; i < n; i++) {
            right = max(right, nums[i]);
        }
        while (left < right) {
            int mid = (left + right) / 2;
            int sum = calDiv(nums, mid);
            if (sum <= threshold) {
                right = mid;
            }
            else {
                left = mid + 1;
            }
        }
        return left;
    }
};