生成一个长度为 N 的数组，其最大元素最小化且数组元素的总和可被 K 整除

给定两个正整数N和K ，任务是最小化所形成数组的最大元素，使得数组元素的总和为正数且可被K整除。

例子：

Input: N = 4, K = 50
Output: 13
Explanation The generated array is {12, 13, 12, 13}. Sum of the array is 50, which is divisible by K (= 50). Maximum element present in the array is 13, which is minimum possible.

Input: N = 3, K = 3
Output: 1
 

方法：根据以下观察可以解决给定的问题：

• 为了最小化数组的最大元素，每对数组元素的绝对差必须最大为 1 ，数组元素的总和必须为K 。
• 因此，所有N 个元素的值必须至少等于(K/N) 的下限值，并将剩余的(K%N) 个元素加1以得出数组元素K的总和。

根据以上观察，构造数组的最小化最大值是(K/N)的 ceil 值。

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
 
// Function to minimize the maximum
// element present in an N-length array
// having sum of elements divisible by K
int minimumValue(int N, int K)
{
    // Return the ceil value of (K / N)
    return ceil((double)K / (double)N);
}
 
// Driver Code
int main()
{
    int N = 4, K = 50;
    cout << minimumValue(N, K);
 
    return 0;
}

Java

// Java program for the above approach
class GFG{
 
// Function to minimize the maximum
// element present in an N-length array
// having sum of elements divisible by K
static int minimumValue(int N, int K)
{
    
    // Return the ceil value of (K / N)
    return(int)Math.ceil((double)K / (double)N);
}
 
// Driver code
public static void main(String[] args)
{
    int N = 4, K = 50;
      
    System.out.print(minimumValue(N, K));
}
}
 
// This code is contributed by code_hunt.

Python3

# Python3 program for the above approach
import math
 
# Function to minimize the maximum
# element present in an N-length array
# having sum of elements divisible by K
 
 
def minimumValue(N, K):
    # Return the ceil value of (K / N)
    return math.ceil(K / N)
 
 
# Driver Code
N = 4
K = 50
print(minimumValue(N, K))
 
# This code is contributed by abhinavjain194

C#

// C# program for the above approach
using System;
 
class GFG{
 
// Function to minimize the maximum
// element present in an N-length array
// having sum of elements divisible by K
static int minimumValue(int N, int K)
{
   
    // Return the ceil value of (K / N)
    return(int)Math.Ceiling((double)K / (double)N);
}
 
// Driver Code
public static void Main()
{
    int N = 4, K = 50;
     
    Console.WriteLine(minimumValue(N, K));
}
}
 
// This code is contributed by ukasp

Javascript

 
// javascript program for the above approach
 
// Function to minimize the maximum
// element present in an N-length array
// having sum of elements divisible by K
function minimumValue(N, K)
{
     
    // Return the ceil value of (K / N)
    return Math.ceil(K / N);
}
 
// Driver Code
 
    let N = 4, K = 50;
       
    document.write(minimumValue(N, K));
 

输出：

13

时间复杂度： O(1)
辅助空间： O(1)

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