最小化构造数组中的最大元素，总和可被 K 整除

给定两个整数N和K ，任务是找到一个大小为N的数组的最大元素的最小值，该数组由元素之和可被K整除的正整数组成。

例子：

Input: N = 4, K = 3
Output: 2
Explanation: 
Let the array be [2, 2, 1, 1]. Here, sum of elements of this array is divisible by K=3, and maximum element is 2.

Input: N = 3, K = 5
Output: 2

方法：要找到大小为 N 且总和可被 K 整除的数组的最小最大值，请尝试创建一个总和可能最小的数组。

• 能被 K 整除的 N 个元素（每个元素的值都大于 0 ）的最小总和是：
  

sum = K * ceil(N/K)

• 现在，如果总和可被 N 整除，则最大元素将为sum/N ，否则为(sum/N + 1)。

下面是上述方法的实现。

C++

// C++ program for the above approach.
 
#include 
using namespace std;
 
// Function to find smallest maximum number
// in an array whose sum is divisible by K.
int smallestMaximum(int N, int K)
{
    // Minimum possible sum possible
    // for an array of size N such that its
    // sum is divisible by K
    int sum = ((N + K - 1) / K) * K;
 
    // If sum is not divisible by N
    if (sum % N != 0)
        return (sum / N) + 1;
 
    // If sum is divisible by N
    else
        return sum / N;
}
 
// Driver code.
int main()
{
    int N = 4;
    int K = 3;
 
    cout << smallestMaximum(N, K) << endl;
    return 0;
}

Java

// Java program for the above approach
import java.io.*;
import java.util.*;
 
class GFG{
     
 
// Function to find smallest maximum number
// in an array whose sum is divisible by K.
static int smallestMaximum(int N, int K)
{
    // Minimum possible sum possible
    // for an array of size N such that its
    // sum is divisible by K
    int sum = ((N + K - 1) / K) * K;
 
    // If sum is not divisible by N
    if (sum % N != 0)
        return (sum / N) + 1;
 
    // If sum is divisible by N
    else
        return sum / N;
}
 
// Driver code
public static void main(String args[])
{
    int N = 4;
    int K = 3;
 
    System.out.println(smallestMaximum(N, K));
}
}
 
// This code is contributed by code_hunt.

Python3

# python program for the above approach.
# Function to find smallest maximum number
# in an array whose sum is divisible by K.
def smallestMaximum(N,K):
   
    # Minimum possible sum possible
    # for an array of size N such that its
    # sum is divisible by K
    sum = ((N + K - 1) // K) * K
 
    # If sum is not divisible by N
    if (sum % N != 0):
        return (sum // N) + 1
 
    # If sum is divisible by N
    else:
        return sum // N
 
# Driver code.
if __name__ == "__main__":
    N = 4
    K = 3
 
    print(smallestMaximum(N, K))
     
# This code is contributed by anudeep23042002.

C#

// C# program for the above approach.
using System;
using System.Collections.Generic;
 
class GFG{
 
// Function to find smallest maximum number
// in an array whose sum is divisible by K.
static int smallestMaximum(int N, int K)
{
    // Minimum possible sum possible
    // for an array of size N such that its
    // sum is divisible by K
    int sum = ((N + K - 1) / K) * K;
 
    // If sum is not divisible by N
    if (sum % N != 0)
        return (sum / N) + 1;
 
    // If sum is divisible by N
    else
        return sum / N;
}
 
// Driver code.
public static void Main()
{
    int N = 4;
    int K = 3;
    Console.Write(smallestMaximum(N, K));
}
}
 
// This code is contributed by SURENDRA_GANGWAR.

Javascript

        // JavaScript Program to implement
        // the above approach
 
        // Function to find smallest maximum number
        // in an array whose sum is divisible by K.
        function smallestMaximum(N, K)
        {
         
            // Minimum possible sum possible
            // for an array of size N such that its
            // sum is divisible by K
            let sum = Math.floor((N + K - 1) / K) * K;
 
            // If sum is not divisible by N
            if (sum % N != 0)
                return Math.floor(sum / N) + 1;
 
            // If sum is divisible by N
            else
                return Math.floor(sum / N);
        }
 
        // Driver code.
 
        let N = 4;
        let K = 3;
 
        document.write(smallestMaximum(N, K));
 
// This code is contributed by Potta Lokesh
    

输出：

2

时间复杂度： O(1)
辅助空间： O(1)