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📜  可以购买的最大糖果数量

📅  最后修改于: 2021-06-27 00:25:00             🧑  作者: Mango

给定大小为n的数组arr [] ,其中arr [i]i型糖果的数量。你有无限量的钱。任务是购买满足以下条件的尽可能多的糖果:
如果您购买类型的X(I)糖果(显然,0≤X(I)≤改编[1]),那么对于所有j(1≤Ĵ≤I)中的必须持有至少一项:

  1. x(j) (购买的j型糖果比i型糖果少)
  2. x(j)= 0 (您购买了0个j型糖果)

例子:

方法:我们可以使用贪婪方法,并从数组末尾开始。如果我们采取了类型i + 1X糖果那么我们只能采取分钟(ARR [I]中,x – 1)I型的糖果。如果该值为负,则我们无法购买当前类型的糖果。

下面是上述方法的实现:

C++
// C++ implementation of the approach
#include 
using namespace std;
  
// Function to return the maximum candies
// that can be bought
int maxCandies(int arr[], int n)
{
  
    // Buy all the candies of the last type
    int prevBought = arr[n - 1];
    int candies = prevBought;
  
    // Starting from second last
    for (int i = n - 2; i >= 0; i--) {
  
        // Amount of candies of the current
        // type that can be bought
        int x = min(prevBought - 1, arr[i]);
  
        if (x >= 0) {
  
            // Add candies of current type
            // that can be bought
            candies += x;
  
            // Update the previous bought amount
            prevBought = x;
        }
    }
  
    return candies;
}
  
// Driver code
int main()
{
    int arr[] = { 1, 2, 1, 3, 6 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << maxCandies(arr, n);
  
    return 0;
}


Java
// Java implementation of the approach
class GFG
{
      
// Function to return the maximum candies
// that can be bought
static int maxCandies(int arr[], int n)
{
  
    // Buy all the candies of the last type
    int prevBought = arr[n - 1];
    int candies = prevBought;
  
    // Starting from second last
    for (int i = n - 2; i >= 0; i--) 
    {
  
        // Amount of candies of the current
        // type that can be bought
        int x = Math.min(prevBought - 1, arr[i]);
  
        if (x >= 0) 
        {
  
            // Add candies of current type
            // that can be bought
            candies += x;
  
            // Update the previous bought amount
            prevBought = x;
        }
    }
  
    return candies;
}
  
// Driver code
public static void main(String[] args)
{
    int arr[] = { 1, 2, 1, 3, 6 };
    int n = arr.length;
    System.out.println(maxCandies(arr, n));
}
}
  
// This code is contributed by Code_Mech.


Python3
# Python3 implementation of the approach 
  
# Function to return the maximum candies 
# that can be bought 
def maxCandies(arr, n) :
      
    # Buy all the candies of the last type 
    prevBought = arr[n - 1];
    candies = prevBought; 
      
    # Starting from second last 
    for i in range(n - 2, -1, -1) :
          
        # Amount of candies of the current
        # type that can be bought 
        x = min(prevBought - 1, arr[i]); 
        if (x >= 0) :
              
            # Add candies of current type 
            # that can be bought
            candies += x; 
              
            # Update the previous bought amount 
            prevBought = x; 
              
    return candies; 
  
# Driver code 
if __name__ == "__main__" : 
      
    arr = [ 1, 2, 1, 3, 6 ];
    n = len(arr)
    print(maxCandies(arr, n)); 
  
# This code is contributed by Ryuga


C#
// C# implementation of the approach
using System;
  
class GFG
{
      
// Function to return the maximum candies
// that can be bought
static int maxCandies(int[] arr, int n)
{
  
    // Buy all the candies of the last type
    int prevBought = arr[n - 1];
    int candies = prevBought;
  
    // Starting from second last
    for (int i = n - 2; i >= 0; i--) 
    {
  
        // Amount of candies of the current
        // type that can be bought
        int x = Math.Min(prevBought - 1, arr[i]);
  
        if (x >= 0) 
        {
  
            // Add candies of current type
            // that can be bought
            candies += x;
  
            // Update the previous bought amount
            prevBought = x;
        }
    }
  
    return candies;
}
  
// Driver code
public static void Main()
{
    int[] arr= { 1, 2, 1, 3, 6 };
    int n = arr.Length;
    Console.WriteLine(maxCandies(arr, n));
}
}
  
// This code is contributed by Code_Mech.


PHP
= 0; $i--) 
    {
  
        // Amount of candies of the current
        // type that can be bought
        $x = min($prevBought - 1, $arr[$i]);
  
        if ($x >= 0) 
        {
  
            // Add candies of current type
            // that can be bought
            $candies += $x;
  
            // Update the previous bought amount
            $prevBought = $x;
        }
    }
  
    return $candies;
}
  
// Driver code
$arr = array(1, 2, 1, 3, 6 );
$n = sizeof($arr);
echo(maxCandies($arr, $n));
  
// This code is contributed by Code_Mech.
?>


输出:
10

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