在糖果店里,有 N 种不同类型的糖果可供选择,并且提供了所有 N 种不同类型糖果的价格。糖果店也提供有吸引力的优惠。我们可以从商店购买一个糖果,最多可以免费获得 K 个其他糖果(都是不同类型的)。
- 找出购买所有 N 种不同糖果所必须花费的最低金额。
- 找出购买所有 N 种不同糖果必须花费的最大金额。
在这两种情况下,我们都必须利用报价并尽可能多地取回糖果。如果有 k 个或更多糖果可用,我们必须为每次购买糖果取 k 个糖果。如果可用的糖果少于 k 个,我们必须将所有糖果都用于购买糖果。
例子:
Input :
price[] = {3, 2, 1, 4}
k = 2
Output :
Min = 3, Max = 7
Explanation :
Since k is 2, if we buy one candy we can take
atmost two more for free.
So in the first case we buy the candy which
costs 1 and take candies worth 3 and 4 for
free, also you buy candy worth 2 as well.
So min cost = 1 + 2 = 3.
In the second case we buy the candy which
costs 4 and take candies worth 1 and 2 for
free, also We buy candy worth 3 as well.
So max cost = 3 + 4 = 7.
需要注意的一件重要事情是,我们必须使用该优惠,并为每次购买糖果获得最多的糖果。所以如果我们想把钱最小化,我们必须以最低成本购买糖果,免费获得最高成本的糖果。为了使金钱最大化,我们必须反过来做。下面是一个基于此的算法。
First Sort the price array.
For finding minimum amount :
Start purchasing candies from starting
and reduce k free candies from last with
every single purchase.
For finding maximum amount :
Start purchasing candies from the end
and reduce k free candies from starting
in every single purchase.
下图是上述方法的说明:
最低金额 :
最大金额:
下面是上述方法的实现:
C++
// C++ implementation to find the minimum
// and maximum amount
#include
using namespace std;
// Function to find the minimum amount
// to buy all candies
int findMinimum(int arr[], int n, int k)
{
int res = 0;
for (int i = 0; i < n; i++) {
// Buy current candy
res += arr[i];
// And take k candies for free
// from the last
n = n - k;
}
return res;
}
// Function to find the maximum amount
// to buy all candies
int findMaximum(int arr[], int n, int k)
{
int res = 0, index = 0;
for (int i = n - 1; i >= index; i--)
{
// Buy candy with maximum amount
res += arr[i];
// And get k candies for free from
// the starting
index += k;
}
return res;
}
// Driver code
int main()
{
int arr[] = { 3, 2, 1, 4 };
int n = sizeof(arr) / sizeof(arr[0]);
int k = 2;
sort(arr, arr + n);
// Function call
cout << findMinimum(arr, n, k) << " "
<< findMaximum(arr, n, k) << endl;
return 0;
}
Java
// Java implementation to find the
// minimum and maximum amount
import java.util.*;
class GFG {
// Function to find the minimum
// amount to buy all candies
static int findMinimum(int arr[], int n, int k)
{
int res = 0;
for (int i = 0; i < n; i++) {
// Buy current candy
res += arr[i];
// And take k candies for free
// from the last
n = n - k;
}
return res;
}
// Function to find the maximum
// amount to buy all candies
static int findMaximum(int arr[], int n, int k)
{
int res = 0, index = 0;
for (int i = n - 1; i >= index; i--)
{
// Buy candy with maximum amount
res += arr[i];
// And get k candies for free from
// the starting
index += k;
}
return res;
}
// Driver code
public static void main(String[] args)
{
int arr[] = { 3, 2, 1, 4 };
int n = arr.length;
int k = 2;
Arrays.sort(arr);
// Function call
System.out.println(findMinimum(arr, n, k) + " "
+ findMaximum(arr, n, k));
}
}
// This code is contributed by prerna saini
Python3
# Python implementation
# to find the minimum
# and maximum amount
# Function to find
# the minimum amount
# to buy all candies
def findMinimum(arr, n, k):
res = 0
i = 0
while(n):
# Buy current candy
res += arr[i]
# And take k
# candies for free
# from the last
n = n-k
i += 1
return res
# Function to find
# the maximum amount
# to buy all candies
def findMaximum(arr, n, k):
res = 0
index = 0
i = n-1
while(i >= index):
# Buy candy with
# maximum amount
res += arr[i]
# And get k candies
# for free from
# the starting
index += k
i -= 1
return res
# Driver code
arr = [3, 2, 1, 4]
n = len(arr)
k = 2
arr.sort()
# Function call
print(findMinimum(arr, n, k), " ",
findMaximum(arr, n, k))
# This code is contributed
# by Anant Agarwal.
C#
// C# implementation to find the
// minimum and maximum amount
using System;
public class GFG {
// Function to find the minimum
// amount to buy all candies
static int findMinimum(int[] arr, int n, int k)
{
int res = 0;
for (int i = 0; i < n; i++)
{
// Buy current candy
res += arr[i];
// And take k candies for
// free from the last
n = n - k;
}
return res;
}
// Function to find the maximum
// amount to buy all candies
static int findMaximum(int[] arr, int n, int k)
{
int res = 0, index = 0;
for (int i = n - 1; i >= index; i--)
{
// Buy candy with maximum
// amount
res += arr[i];
// And get k candies for free
// from the starting
index += k;
}
return res;
}
// Driver code
public static void Main()
{
int[] arr = { 3, 2, 1, 4 };
int n = arr.Length;
int k = 2;
Array.Sort(arr);
// Function call
Console.WriteLine(findMinimum(arr, n, k) + " "
+ findMaximum(arr, n, k));
}
}
// This code is contributed by Sam007.
PHP
= $index; $i--)
{
// Buy candy with maximum amount
$res += $arr[$i];
// And get k candies
// for free from
// the starting
$index += $k;
}
return $res;
}
// Driver Code
$arr = array(3, 2, 1, 4);
$n = sizeof($arr);
$k = 2;
sort($arr); sort($arr,$n);
// Function call
echo findMinimum($arr, $n, $k)," "
,findMaximum($arr, $n, $k);
return 0;
// This code is contributed by nitin mittal.
?>
Javascript
C++
// C++ implementation
// to find the minimum
// and maximum amount
#include
using namespace std;
// function to find the maximum
// and the minimum cost required
void find(vector arr, int n, int k)
{
// Sort the array
sort(arr.begin(), arr.end());
int b = ceil(n / k * 1.0);
int min_sum = 0, max_sum = 0;
for(int i = 0; i < b; i++)
min_sum += arr[i];
for(int i = 2; i < arr.size(); i++)
max_sum += arr[i];
// print the minimum cost
cout << "minimum " << min_sum << endl;
// print the maximum cost
cout << "maximum " << max_sum << endl;
}
// Driver code
int main()
{
vector arr = {3, 2, 1, 4};
int n = arr.size();
int k = 2;
// Function call
find(arr,n,k);
}
// This code is contributed by mohit kumar 29.
Java
// Java implementation to find the minimum
// and maximum amount
import java.io.*;
import java.util.Arrays;
import java.lang.Math;
class GFG{
// Function to find the maximum
// and the minimum cost required
static void find(int[] arr, int n, int k)
{
// Sort the array
Arrays.sort(arr);
int b = (int)Math.ceil(n / k * 1.0);
int min_sum = 0, max_sum = 0;
for(int i = 0; i < b; i++)
min_sum += arr[i];
for(int i = 2; i < arr.length; i++)
max_sum += arr[i];
// Print the minimum cost
System.out.println("minimum " + min_sum);
// Print the maximum cost
System.out.println("maximum " + max_sum);
}
// Driver code
public static void main (String[] args)
{
int[] arr = { 3, 2, 1, 4 };
int n = arr.length;
int k = 2;
// Function call
find(arr, n, k);
}
}
// This code is contributed by shivanisinghss2110
Python3
# Python implementation
# to find the minimum
# and maximum amount
#import ceil function
from math import ceil
# function to find the maximum
# and the minimum cost required
def find(arr,n,k):
# Sort the array
arr.sort()
b = int(ceil(n/k))
# print the minimum cost
print("minimum ",sum(arr[:b]))
# print the maximum cost
print("maximum ", sum(arr[-b:]))
# Driver Code
arr = [3, 2, 1, 4]
n = len(arr)
k = 2
# Function call
find(arr,n,k)
C#
// C# implementation to find the minimum
// and maximum amount
using System;
class GFG{
// Function to find the maximum
// and the minimum cost required
static void find(int[] arr, int n, int k)
{
// Sort the array
Array.Sort(arr);
int b = (int)Math.Ceiling(n / k * 1.0);
int min_sum = 0, max_sum = 0;
for(int i = 0; i < b; i++)
min_sum += arr[i];
for(int i = 2; i < arr.Length; i++)
max_sum += arr[i];
// Print the minimum cost
Console.WriteLine("minimum " + min_sum);
// Print the maximum cost
Console.WriteLine("maximum " + max_sum);
}
// Driver code
public static void Main()
{
int[] arr = { 3, 2, 1, 4 };
int n = arr.Length;
int k = 2;
// Function call
find(arr, n, k);
}
}
// This code is contributed by ukasp
Javascript
输出
3 7
时间复杂度: O(n log n)
另一个实现:
我们可以借助内置的 ceil()函数,借助最小整数函数(Ceiling函数)来实现:
下面是在Python的实现:
C++
// C++ implementation
// to find the minimum
// and maximum amount
#include
using namespace std;
// function to find the maximum
// and the minimum cost required
void find(vector arr, int n, int k)
{
// Sort the array
sort(arr.begin(), arr.end());
int b = ceil(n / k * 1.0);
int min_sum = 0, max_sum = 0;
for(int i = 0; i < b; i++)
min_sum += arr[i];
for(int i = 2; i < arr.size(); i++)
max_sum += arr[i];
// print the minimum cost
cout << "minimum " << min_sum << endl;
// print the maximum cost
cout << "maximum " << max_sum << endl;
}
// Driver code
int main()
{
vector arr = {3, 2, 1, 4};
int n = arr.size();
int k = 2;
// Function call
find(arr,n,k);
}
// This code is contributed by mohit kumar 29.
Java
// Java implementation to find the minimum
// and maximum amount
import java.io.*;
import java.util.Arrays;
import java.lang.Math;
class GFG{
// Function to find the maximum
// and the minimum cost required
static void find(int[] arr, int n, int k)
{
// Sort the array
Arrays.sort(arr);
int b = (int)Math.ceil(n / k * 1.0);
int min_sum = 0, max_sum = 0;
for(int i = 0; i < b; i++)
min_sum += arr[i];
for(int i = 2; i < arr.length; i++)
max_sum += arr[i];
// Print the minimum cost
System.out.println("minimum " + min_sum);
// Print the maximum cost
System.out.println("maximum " + max_sum);
}
// Driver code
public static void main (String[] args)
{
int[] arr = { 3, 2, 1, 4 };
int n = arr.length;
int k = 2;
// Function call
find(arr, n, k);
}
}
// This code is contributed by shivanisinghss2110
蟒蛇3
# Python implementation
# to find the minimum
# and maximum amount
#import ceil function
from math import ceil
# function to find the maximum
# and the minimum cost required
def find(arr,n,k):
# Sort the array
arr.sort()
b = int(ceil(n/k))
# print the minimum cost
print("minimum ",sum(arr[:b]))
# print the maximum cost
print("maximum ", sum(arr[-b:]))
# Driver Code
arr = [3, 2, 1, 4]
n = len(arr)
k = 2
# Function call
find(arr,n,k)
C#
// C# implementation to find the minimum
// and maximum amount
using System;
class GFG{
// Function to find the maximum
// and the minimum cost required
static void find(int[] arr, int n, int k)
{
// Sort the array
Array.Sort(arr);
int b = (int)Math.Ceiling(n / k * 1.0);
int min_sum = 0, max_sum = 0;
for(int i = 0; i < b; i++)
min_sum += arr[i];
for(int i = 2; i < arr.Length; i++)
max_sum += arr[i];
// Print the minimum cost
Console.WriteLine("minimum " + min_sum);
// Print the maximum cost
Console.WriteLine("maximum " + max_sum);
}
// Driver code
public static void Main()
{
int[] arr = { 3, 2, 1, 4 };
int n = arr.Length;
int k = 2;
// Function call
find(arr, n, k);
}
}
// This code is contributed by ukasp
Javascript
输出
('minimum ', 3)
('maximum ', 7)
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