装满K杯所需的最少瓶子数

给定N杯有水，并且每个杯子的容量A。任务是找到可准确装满K杯所需的最少瓶子数。每个瓶子的容量为100个单位。

例子：

Input: N = 4, K = 3, arr[] = {200, 150, 140, 300}
Output: 5
We have to fill out exactly 3 glasses.
So we fill out 140, 150, 200 whose sum is 490 so we need 5 bottles for this.

Input: N = 5, K = 4, arr[] = {1, 2, 3, 2, 1}
Output: 1

方法：要准确地填写K杯，请拿最小容量的K杯。因此，对于这种排序，列出了给定的容量。最终答案将是(第一个k杯容量总和)/(一瓶容量)的上限值。

下面是上述方法的实现：

C/C++

// C++ implementation of the above approach
#include 
using namespace std;
  
// function to calculate minimum glasses
int Min_glass(int n, int k, int a[])
{
    // sort the array based on
    // their capacity
    sort(a, a + n);
  
    int sum = 0;
  
    // taking sum of capacity
    // of first k glasses
    for (int i = 0; i < k; i++)
        sum += a[i];
  
    // calculate the answer
    int ans = ceil((double)sum / (double)100);
  
    return ans;
}
  
// Driver code
int main()
{
  
    int n = 4;
    int k = 3;
    int a[] = { 200, 150, 140, 300 };
  
    cout << Min_glass(n, k, a);
  
    return 0;
}

Java

// Java implementation of the 
// above approach
import java.util.*;
  
class GFG
{
// function to calculate minimum glasses
public static double Min_glass(int n, int k, 
                            int[] a)
{
    // sort the array based on
    // their capacity
  
    int sum = 0;
  
    // taking sum of capacity
    // of first k glasses
    for (int i = 0; i < k; i++)
        sum += a[i];
  
    // calculate the answer
    double ans = Math.ceil((double)sum /
                            (double)100);
  
    return ans;
}
  
// Driver code 
public static void main(String[] args)
{
    int n = 4;
    int k = 3;
    int[] a = { 200, 150, 140, 300 };
    Arrays.sort(a);
    System.out.println(Min_glass(n, k, a));
}
}
  
// This code is contributed by mits

Python3

# Python3 implementation of the above approach 
from math import ceil
  
# Function to calculate minimum glasses 
def Min_glass(n, k, a): 
   
    # sort the array based on their capacity 
    a.sort() 
  
    # calculate the answer 
    return ceil(sum(a[:k]) / 100) 
   
# Driver code 
if __name__ == "__main__": 
   
    n, k = 4, 3 
    a = [200, 150, 140, 300]  
  
    print(Min_glass(n, k, a)) 
  
# This code is contributed by Rituraj Jain

C#

// C# implementation of the 
// above approach
using System;
  
class GFG
{
// function to calculate minimum glasses
public static double Min_glass(int n, int k, 
                               int []a)
{
    // sort the array based on
    // their capacity
  
    int sum = 0;
  
    // taking sum of capacity
    // of first k glasses
    for (int i = 0; i < k; i++)
        sum += a[i];
  
    // calculate the answer
    double ans = Math.Ceiling((double)sum /
                              (double)100);
  
    return ans;
}
  
// Driver code 
public static void Main()
{
    int n = 4;
    int k = 3;
    int[] a = { 200, 150, 140, 300 };
    Array.Sort(a);
    Console.WriteLine(Min_glass(n, k, a));
}
}
  
// This code is contributed
// by Soumik Mondal

输出：

5