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Algorithm for calculating the total cost in groups N

I have a limited supply of objects, and as the objects are purchased, the price increases accordingly in groups of N (each time N objects are bought, the price increases). When you're trying to purchase multiple properties, what's the easiest way to calculate the total cost?

Example:
I have 24 foo. For every N (use case 3) that was bought, the price increases by 1.
So if I buy 1 at price 1, there are 23 left and 2 are left at price 1.
After 1 has been bought, someone wants to buy 6. Well, the total cost would be = (2 * 1) + (3 * 2) + (1 * 3)

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2 answers


Ok, I think this is right now ...

Given:

X  = Total Number of Items Bought
N  = Number of Items per Price Increment
B  = Base Item Price, before any Increments
I  = Price Increment per [N]


Set:

J  = FLOOR((X-1)/N)+1

Then:

TotalCost = X*(B-I) + I*(X*J - N*J*(J-1)/2)
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Borrowing RBarryYoung notation, the first N items are worth B each, the second N items are worth B + i each, the third N items are worth B + 2 * i each, etc.

To buy items X: Q: = X div N (gender division), whole groups are bought, plus R: = X mod N additional items. The first cost Q * N * (B + (B + (Q - 1) * I)) / 2, since with linearly increasing item costs, the average item cost is equal to the average cost of the first item, B and the last item cost, B + ( Q - 1) * I. The last items cost R * (B + Q * I), so the resulting function f (X) is



f(X) := (Q * N * (B + (B + (Q - 1) * I))) div 2 + R * (B + Q*I).

To compute the cost of elements (zero) indexed from X inclusive to X 'exclusive, use f(X') - f(X)

.

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