Codey and Textbooks

https://www.hackerrank.com/contests/codenection-2023-preliminary-round-closed-category/challenges/cn-c4

Question

Codey, who is enthusiastic when it comes to studies, is eager to purchase n textbooks from the university bookstore. These textbooks are priced in an interesting way:

• The first textbook costs $m$ dollars.

• The second textbook costs $2 * m$ dollars.

• The $i$-th textbook costs $i * m$ dollars.

Codey has k dollars in his wallet, but it's wondering how much it needs to borrow from the E-bee to cover the costs of purchasing n textbooks.

Your task is to calculate and output the amount of money that Codey has to borrow from the E-bee. If it doesn't need to borrow any money, the answer should be $0$.

Can you assist Codey in figuring out how much it needs to borrow from E-Bee, if at all?

Input Format

The first line contains three integers n, m, k, where n represents the number of textbooks Codey wants, m represents the cost of the first book, and k represents the initial number of ringgit Codey has.

Constraints

$$0 \le n, m, k \le 10^5$$

Output Format

Outputs only one integer, the amount of money Codey has to borrow from E-bee. If it doesn't have to borrow money from E-bee, output 0.

Sample Inputs:

Input 0

Input

5 2 3

Output

27

Explanation

Codey wants 5 textbooks, and the first book costs 2 ringgit. The cost of the books is:

$$(1 * 2) + (2 * 2) + (3 * 2) + (4 * 2) + (5 * 2) = 30$$

Codey has 3 ringgit, therefore Codey needs to borrow 30 - 3 = 27 ringgit.

Input 1

Input

3 2 15

Output

0

Explanation

Codey doesn't have to borrow money from E-bee, because Codey has sufficient money.

---

Solution - Sn AP

This is a typical "find the sum of AP" question. The formula of Sum of AP series is:$\frac{n}{2} * (2a + (n-1) * d)$

we know that d is always 1, so we can omit it. Substitute a to m, and we have the total price. Lastly, subtract k to see if it is more than 0. If it isn't, print 0; otherwise, print that value. The value is the amount of money that Codey has to borrow.

Here's the solution:

n, m, k = map(int, input().strip().split())
ans = ((n * (n + 1) // 2) * m) - k
if ans <= 0:
    print("0")
else:
    print(int(ans))