Algorithm


Problem Name: beecrowd | 1006
 
Algorithm:
media = (a/10 * 2) + (b/10 * 3) + (c/10 * 5);

Average 2

Adapted by Neilor Tonin, URI Brazil

Timelimit: 1

 

Read three values (variables A, B and C), which are the three student's grades. Then, calculate the average, considering that grade A has weight 2, grade B has weight 3 and the grade C has weight 5. Consider that each grade can go from 0 to 10.0, always with one decimal place.

 

Input

 

The input file contains 3 values of floating points (double) with one digit after the decimal point.

 

Output

 

Print the message "MEDIA"(average in Portuguese) and the student's average according to the following example, with a blank space before and after the equal signal.

 

 

 

Input Samples Output Samples

5.0
6.0
7.0

MEDIA = 6.3

 

5.0
10.0
10.0

MEDIA = 9.0

 

10.0
10.0
5.0

MEDIA = 7.5

Code Examples

#1 Code Example with C Programming

Code - C Programming


#include <stdio.h>
int main()
{
   double a, b, c, media;
   scanf("%lf%lf%lf", &a, &b, &c);
   media = (a/10 * 2) + (b/10 * 3) + (c/10 * 5);
   printf("MEDIA = %.1lf\n", media);
   return 0;
}
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Input

x
+
cmd
5.0
6.0
7.0

Output

x
+
cmd
MEDIA = 6.3

#2 Code Example with C++ Programming

Code - C++ Programming


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

int main()
{
    double a, b, c, media;
    cin >> a >> b >> c;
    media = (a/10 * 2) + (b/10 * 3) + (c/10 * 5);
    cout << "MEDIA = "<< fixed << setprecision(1) << media << endl;
    return 0;
}
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Input

x
+
cmd
5.0
6.0
7.0

Output

x
+
cmd
MEDIA = 6.3

#3 Code Example with Java Programming

Code - Java Programming


import java.util.Scanner;

public class Main{
    public static void main(String[] args){

        double a, b, c, med;
        Scanner sc =new Scanner(System.in);

           a =sc.nextDouble();
           b =sc.nextDouble();
           c =sc.nextDouble();

         med = (a/10 * 2) + (b/10 * 3) + (c/10 * 5);
         String media = String.format("MEDIA = %,.1f", med);

        System.out.print(media +"\n");

    }
}
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Input

x
+
cmd
5.0
10.0
10.0

Output

x
+
cmd
MEDIA = 9.0

#4 Code Example with Python Programming

Code - Python Programming


a = float(input())
b = float(input())
c = float(input())

media = (a/10 * 2) + (b/10 * 3) + (c/10 * 5);

print("MEDIA = %0.5f" %media)
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Input

x
+
cmd
10.0
10.0
5.0

Output

x
+
cmd
MEDIA = 7.5

#5 Code Example with C++ Programming

Code - C++ Programming


using System;
class URI {
 
    static void Main(string[] args) {

            double a, b, c;
            a = Convert.ToDouble(Console.ReadLine());
            b = Convert.ToDouble(Console.ReadLine());
            c = Convert.ToDouble(Console.ReadLine());

            Console.WriteLine("MEDIA = " + ((a/10 * 2) + (b/10 * 3) + (c/10 * 5)).ToString("0.00000"));
            Console.ReadKey();
    }
}
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Input

x
+
cmd
10.0
10.0
5.0

Output

x
+
cmd
MEDIA = 7.5

Demonstration


Beecrowd Online Judge Solution 1006 Average 2 - Solution in C, C++, Java, Python and C#

 

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