Algorithm


Problem Name: beecrowd | 1051
Problem Link: https://www.beecrowd.com.br/judge/en/problems/view/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

 

4520.00

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 & Try With Live Editor

Input

x
+
cmd
3002.00

Output

x
+
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 & Try With Live Editor

Input

x
+
cmd
1701.12

Output

x
+
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 & Try With Live Editor

Input

x
+
cmd
4520.00

Output

x
+
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
resultado = resto * 0.08
    print "R$ %.2f" %resultado
elif(salario > 3000 and salario < 4500):
    resto = salario - 3000
resultado = (resto * 0.18) + (1000 * 0.08)
    print "R$ %.2f" %resultado
else:
    resto = salario - 4500
resultado = (resto * 0.28) + (1500 * 0.18) + (1000 * 0.08>

print "R$ %.2f" %resultado
Copy The Code & Try With Live Editor

Input

x
+
cmd
3002.00

Output

x
+
cmd
R$ 80.36
Advertisements

Demonstration


Previous
#1050 Beecrowd Online Judge Solution 1050 DDD- Solution in C, C++, Java, Python and C#
Next
#1052 Beecrowd Online Judge Solution 1052 Month- Solution in C, C++, Java, Python and C#