2026-05-31：减小数组使其满足条件的最小 K 值。用go语言，给定一个正整数数组 nums。对任意正整数 k，定义函数 nonPositive(nums, k)：

package main

import (
    "fmt"
    "math"
    "sort"
)

func minimumK(nums []int)int {
    n := len(nums)
    nonPositive := func(k int)int {
        sum := n
        for _, x := range nums {
            sum += (x - 1) / k
        }
        return sum
    }

    sum := 0
    for _, x := range nums {
        sum += x
    }

    left := max(int(math.Ceil(math.Sqrt(float64(n)))), int(math.Ceil(math.Cbrt(float64(sum))))) // 答案的下界
    right := int(math.Ceil(math.Sqrt(float64(nonPositive(left)))))                              // 答案的上界
    ans := left + sort.Search(right-left, func(k int)bool {
        k += left
        return nonPositive(k) <= k*k
    })
    return ans
}

func main() {
    nums := []int{3, 7, 5}
    result := minimumK(nums)
    fmt.Println(result)
}

# -*-coding:utf-8-*-

import math

def minimumK(nums):
    n = len(nums)
    
    def nonPositive(k):
        total = n
        for x in nums:
            total += (x - 1) // k
        return total
    
    total_sum = sum(nums)
    
    left = max(
        math.ceil(math.sqrt(n)),
        math.ceil(total_sum ** (1/3))
    )
    
    # Python 需要确保 left 是整数
    left = int(left)
    
    right = int(math.ceil(math.sqrt(nonPositive(left))))
    
    # 二分查找
    low, high = 0, right - left
    while low < high:
        mid = (low + high) // 2
        k = left + mid
        if nonPositive(k) <= k * k:
            high = mid
        else:
            low = mid + 1
    
    ans = left + low
    return ans

if __name__ == "__main__":
    nums = [3, 7, 5]
    result = minimumK(nums)
    print(result)

#include <iostream>
#include <vector>
#include <cmath>
#include <algorithm>

using namespace std;

int minimumK(vector<int>& nums) {
    int n = nums.size();

    auto nonPositive = [&](int k) -> int {
        int sum = n;
        for (int x : nums) {
            sum += (x - 1) / k;
        }
        return sum;
    };

    int sum = 0;
    for (int x : nums) {
        sum += x;
    }

    int left = max(
        (int)ceil(sqrt((double)n)),
        (int)ceil(cbrt((double)sum))
        );

    int right = (int)ceil(sqrt((double)nonPositive(left)));

    // 二分查找
    int ans = left;
    int low = 0, high = right - left;
    while (low < high) {
        int mid = (low + high) / 2;
        int k = left + mid;
        if (nonPositive(k) <= k * k) {
            high = mid;
        } else {
            low = mid + 1;
        }
    }

    ans = left + low;
    return ans;
}

int main() {
    vector<int> nums = {3, 7, 5};
    int result = minimumK(nums);
    cout << result << endl;
    return0;
}