Calculation of salary needs to limit houses

Question

An employee of a company receives salary increase annually: It is known that:

This employee was hired in 1995, with initial salary of R $ 1.000,00;

In 1996 he received a 1.5% increase over his initial salary;

From 1997 (inclusive), wage increases always correspond to twice the percentage of the previous year.

Make a program that determines the employee’s current salary.

After completing this, change the program allowing the user to enter the employee’s initial salary.

My code:

salario = int(input('Digite o salário do funcionário: '))
ano = 1996
porcentagem = 1.5/100
salario_1 = salario
while ano <= 2020:
    salario = salario + (porcentagem * salario_1)
    ano += 1
    porcentagem += porcentagem
print(salario)

My real doubt is how to limit the print of an integer (Ex.1234567, and only appear --> 123), because the output of my code comes out a little strange sometimes. With the salary being 1000, the output is: 503317465.0 (and I doubt that’s even the right result).

Answer 1

There are some problems in this code. Besides being too complex, it does not do what the statement asks and produces the correct result. The issue of formatting is just a point that won’t make it correct.

There are several ways to format, I will put one, it is not the most modern, but the whole code is not the most idiomatic even. Just like you can use a more elaborate formula, but I believe that’s not the goal.

Just remembering that for real monetary values can not do this way, since Python has by default numeric type with binary floating type that has no accuracy and they do not lend themselves to if this type of task, would have to use another type or juggle with integer (which wouldn’t even work when asking for simple data entry), but for exercise is ok.

salario = 1000
ano = 1996
while ano <= 2020:
    salario *= 1.015
    ano += 1
print("{0:.2f}".format(salario, 2))

Behold working in the ideone. And in the repl it.. Also put on the Github for future reference.

When the exercise is given so it makes little sense because you don’t need a code to do that, you have all the data and it can be calculated without code, but if you’re going to do the code I would make it as simple as possible, only artificially keeping the values already known to calculate, then would use the compound interest formula.

print(f"{1000 * (1.015 ** (2020 - 1996 + 1)):.2f}")

And continuing what you ask later I could do just this:

print(f"{int(input('Digite o salário do funcionário: ')) * 1.45095:.2f}")

I did not treat possible typing error that would invalidate the entry and generate exception.