# PROBLEM LINK:

Practice

Contest: Division 1

Contest: Division 2

Contest: Division 3

Contest: Division 4

* Author:* notsoloud

*mexomerf*

**Tester:***iceknight1093*

**Editorialist:**# DIFFICULTY:

TBD

# PREREQUISITES:

None

# PROBLEM:

Chef has Y coins, a gym membership costs X.

Chef can spend one coin to attend a trial session, which gets him a D\% discount on the membership.

Find the minimum number of trials Chef must attend to be able to afford a membership.

# EXPLANATION:

Suppose Chef attends exactly k trials, and then buys a membership.

He then receives a total discount of (k\cdot D)\% on the membership cost, while spending k coins beforehand.

So, the total cost to Chef is

If this quantity is \leq Y, Chef can afford it; otherwise he cannot.

Chef cannot attend more than Y trials (since he canâ€™t spend more than Y coins), so simply try every k = 0, 1, 2, \ldots, Y and find the first one among them such that

If they all fail the check, the answer is -1.

# TIME COMPLEXITY:

\mathcal{O}(Y) per testcase.

# CODE:

## Editorialist's code (Python)

```
for _ in range(int(input())):
d, x, y = map(int, input().split())
ans = -1
for i in range(y+1):
coins = y - i
discount = i*d*x/100
if coins >= x - discount:
ans = i
break
print(ans)
```