URI-1048 Solve

Salary Increase

By Neilor Tonin, URI  Brazil

Timelimit: 1

The company ABC decided to give a salary increase to its employees, according to the following table:

Salary	Readjustment Rate
0 - 400.00 400.01 - 800.00 800.01 - 1200.00 1200.01 - 2000.00 Above 2000.00	15% 12% 10% 7% 4%

Read the employee's salary, calculate and print the new employee's salary, as well the money earned and the increase percentual obtained by the employee, with corresponding messages in Portuguese, as the below example.

Input

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

Output

Print 3 messages followed by the corresponding numbers (see example) informing the new salary, the among of money earned and the percentual obtained by the employee. Note:
Novo salario:  means "New Salary"
Reajuste ganho: means "Money earned"
Em percentual: means "In percentage"

Input Sample	Output Sample
400.00	Novo salario: 460.00 Reajuste ganho: 60.00 Em percentual: 15 %

800.01

Novo salario: 880.01 Reajuste ganho: 80.00 Em percentual: 10 %

2000.00

Novo salario: 2140.00 Reajuste ganho: 140.00 Em percentual: 7 %

Problem link [URI 1048##link##]

Solution:

• [tab]
• C

• #include<stdio.h>
  int main()
  {
   float n;
   scanf("%f", &n);
   if (n <= 400.0)
     printf("Novo salario: %.2f\nReajuste ganho: %.2f\nEm percentual: 15 %%\n", n * 1.15, n * 0.15);
   else if (n <= 800.0)
     printf("Novo salario: %.2f\nReajuste ganho: %.2f\nEm percentual: 12 %%\n", n * 1.12, n * 0.12);
   else if (n <= 1200.0)
     printf("Novo salario: %.2f\nReajuste ganho: %.2f\nEm percentual: 10 %%\n", n * 1.10, n * 0.10);
   else if (n <= 2000.0)
     printf("Novo salario: %.2f\nReajuste ganho: %.2f\nEm percentual: 7 %%\n", n * 1.07, n * 0.07);
   else
     printf("Novo salario: %.2f\nReajuste ganho: %.2f\nEm percentual: 4 %%\n", n * 1.04, n * 0.04);
   return 0;
  }

• C++

• #include <iostream>
  #include <iomanip>
  
  using namespace std;
  
  int main()
  {
   float n, tmp, per;
  
   cin >> n;
  
   if(n <= 400){
    per = 15;
    tmp = n + ((n * per) / 100);
   }else if(n > 400 && n <= 800){
    per = 12;
    tmp = n + ((n * per) / 100);
   }else if(n > 800 && n <= 1200){
    per = 10;
    tmp = n + ((n * per) / 100); 
   }else if(n > 1200 && n <= 2000){
    per = 7;
    tmp = n + ((n * per) / 100);
   }else{
    per = 4;
    tmp = n + ((n * per) / 100);
   }
  
   cout << "Novo salario: " << fixed << setprecision(2) << tmp << endl;
   cout << "Reajuste ganho: " << fixed << setprecision(2) << (tmp - n) << endl;
   cout << "Em percentual: " << fixed << setprecision(0) << per << " %" << endl;
  
   return 0;
  }