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

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Friday, May 26, 2017

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

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


URI Online Judge Solution 1006 Average 2 | Beginner
URI Problem Link -1006 Average 2 - https://www.urionlinejudge.com.br/judge/en/problems/view/1006

Problem Name: 1006 Average 2 solution
Problem Number : URI - 1006 Average 2 code
Online Judge : URI Online Judge Solution
Category: Beginner
Solution Language : C,C plus plus, java, python, c#(c sharp)
URI 1006 problem screenshot:

URI Online Judge Solution 1006 Average 2 - Solution in C, C++, Java, Python and C#
URI Online Judge Solution 1006 Average 2 - Solution in C, C++, Java, Python and C#
See a video if you have some time - 
URI 1006 solution Video - 




URI Solution 1006 Average 2 Code in C:

#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;
}

URI Solution 1006 Average 2 Code C++/ URI 1006 Average 2 solution in CPP:

#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;
}


URI Solution 1006 Average 2 Code in java / URI 1006 Average 2 solution in Java:

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");

    }
}


URI Solution 1006 Average 2 Code / URI 1006 Average 2 solution in  Python:

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

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

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


URI Solution 1006 Average 2 Code / URI 1006 Average 2 solution in  C# (C Sharp):




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();

    }
}


Demonstration:

Since in the question, it has told that each grade has grown from 0 to 10, so devide it with 10 and multiplication into with their weight. Just main line is this,

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

Just implement this in coding. Since having any problem just put a comment below. Thanks



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