最多可以购买的糖果数量

给定一个大小为n的数组arr[] ，其中arr[i]是i类型的糖果数量。你有无限量的钱。任务是尽可能多地购买满足以下条件的糖果：
如果您购买x(i)种i型糖果（显然，0 ≤ x(i) ≤ arr[i]），那么对于所有j (1 ≤ j ≤ i)，至少必须满足以下一项：

1. x(j) < x(i) （你买的 j 类糖果比 i 类少）
2. x(j) = 0 （你买了 0 个 j 类型的糖果）

例子：

Input:arr[] = {1, 2, 1, 3, 6} 
Output: 10 
x[] = {0, 0, 1, 3, 6} where x[i] is the number of candies bought of type i
Input: arr[] = {3, 2, 5, 4, 10} 
Output: 20
Input: arr[] = {1, 1, 1, 1} 
Output: 1 
 

方法：我们可以使用贪心方法，从数组的末尾开始。如果我们吃了x 个i + 1类型的糖果，那么我们只能吃min(arr[i], x – 1) 个i类型的糖果。如果此值为负，我们将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.
?>

Javascript

 
// Javascript implementation of the approach
 
// Function to return the maximum candies
// that can be bought
function maxCandies(arr, n)
{
   
    // Buy all the candies of the last type
    let prevBought = arr[n - 1];
    let candies = prevBought;
   
    // Starting from second last
    for (let i = n - 2; i >= 0; i--)
    {
   
        // Amount of candies of the current
        // type that can be bought
        let 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
 
      let arr = [ 1, 2, 1, 3, 6 ];
    let n = arr.length;
    document.write(maxCandies(arr, n));
             

输出：

10