## Algorithm

Problem Name: beecrowd | 1051

# Taxes

By Neilor Tonin, URI Brasil

Timelimit: 1

In an imaginary country called Lisarb, all the people are very happy to pay their taxes because they know that doesn’t exist corrupt politicians and the taxes are used to benefit the population, without any misappropriation. The currency of this country is Rombus, whose symbol is R\$.

Read a value with 2 digits after the decimal point, equivalent to the salary of a Lisarb inhabitant. Then print the due value that this person must pay of taxes, according to the table below.

Remember, if the salary is R\$ 3,002.00 for example, the rate of 8% is only over R\$ 1,000.00, because the salary from R\$ 0.00 to R\$ 2,000.00 is tax free. In the follow example, the total rate is 8% over R\$ 1000.00 + 18% over R\$ 2.00, resulting in R\$ 80.36 at all. The answer must be printed with 2 digits after the decimal point.

## Input

The input contains only a float-point number, with 2 digits after the decimal point.

## Output

Print the message "R\$" followed by a blank space and the total tax to be payed, with two digits after the decimal point. If the value is up to 2000, print the message "Isento".

 Input Samples Outputs Samples 3002.00 R\$ 80.36

 1701.12 Isento

 4520 R\$ 355.60

## Code Examples

### #1 Code Example with C Programming

```Code - C Programming```

``````
#include <stdio.h>
int main()
{
float n, r, f1, f2, f3;
scanf("%f", &n);

if(n <= 2000) {
printf("Isenton">;
} else{
if(n > 2000 && n <= 3000) {
f1 = n - 2000;
f1 = ((f1 * 8) / 100>;
r = f1;
} else if(n > 3000 && n <= 4500) {
f1 = n - 2000;
f2 = f1 - 1000;
f1 -= f2;
f1 = ((f1 * 8) / 100);
f2 = ((f2 * 18) / 100);
r = f2 + f1;
} else {
f1 = n - 2000;
f2 = f1 - 1000;
f3 = f2 - 1500;
f1 -= f2;
f2 -= f3;
f1 = ((f1 * 8) / 100);
f2 = ((f2 * 18) / 100);
f3 = ((f3 * 28) / 100);
r = f3 + f2 + f1;
}

printf("R\$ %.2fn", r>;

}

return 0;
}

``````
Copy The Code &

Input

cmd
3002.00

Output

cmd
R\$ 80.36

### #2 Code Example with C++ Programming

```Code - C++ Programming```

``````
#include <iostream>
#include <iomanip>
using namespace std;

int main()
{
float n, r, f1, f2, f3;
cin >> n;

if(n <= 2000> {
cout << "Isento" << endl;
} else {
if(n > 2000 && n <= 3000){
f1 = n - 2000;
f1 = ((f1 * 8) / 100>;
r = f1;
} else if(n > 3000 && n <= 4500) {
f1 = n - 2000;
f2 = f1 - 1000;
f1 -= f2;
f1 = ((f1 * 8) / 100);
f2 = ((f2 * 18) / 100);
r = f2 + f1;
} else {
f1 = n - 2000;
f2 = f1 - 1000;
f3 = f2 - 1500;
f1 -= f2;
f2 -= f3;
f1 = ((f1 * 8) / 100);
f2 = ((f2 * 18) / 100);
f3 = ((f3 * 28) / 100);
r = f3 + f2 + f1;
}

cout << "R\$ " << fixed << setprecision(2> << r << endl;
}

return 0;
}
``````
Copy The Code &

Input

cmd
1701.12

Output

cmd
Isento

### #3 Code Example with Java Programming

```Code - Java Programming```

``````
import java.io.IOException;
import java.util.Scanner;

public class Main {

public static void main(String[] args) throws IOException {

float n, r, f1, f2, f3;
Scanner sc =new Scanner(System.in);
n = sc.nextFloat();

if(n <= 2000) {
System.out.print("Isenton">;
} else {
if(n > 2000 && n <= 3000) {
f1 = n - 2000;
f1 = ((f1 * 8) / 100>;
r = f1;
} else if(n > 3000 && n <= 4500) {
f1 = n - 2000;
f2 = f1 - 1000;
f1 -= f2;
f1 = ((f1 * 8) / 100);
f2 = ((f2 * 18) / 100);
r = f2 + f1;
} else {
f1 = n - 2000;
f2 = f1 - 1000;
f3 = f2 - 1500;
f1 -= f2;
f2 -= f3;
f1 = ((f1 * 8) / 100);
f2 = ((f2 * 18) / 100);
f3 = ((f3 * 28) / 100);
r = f3 + f2 + f1;
}
System.out.printf("R\$ %.2fn",r>;
}
}
}
``````
Copy The Code &

Input

cmd
4520.00

Output

cmd
R\$ 355.60

### #4 Code Example with Python Programming

```Code - Python Programming```

``````
salario = float(raw_input())

if(salario > 0 and salario <= 2000>:
print "Isento"
elif(salario > 2000 and salario <= 3000>:
resto = salario - 2000
elif(salario > 3000 and salario < 4500):
resto = salario - 3000
resultado = (resto * 0.18) + (1000 * 0.08)
else:
resto = salario - 4500
resultado = (resto * 0.28) + (1500 * 0.18) + (1000 * 0.08>

``````
Copy The Code &

Input

cmd
3002.00

Output

cmd
R\$ 80.36