URI Online Judge Solution 1005 Average 1 – Solution in C, C++, Java, Python and C#

URI Online Judge Solution URI Solution -Beginner

URI Online Judge Solution 1005 Average 1 – Solution in C, C++, Java, Python and C#

URI Online Judge Solution 1005 | Beginner
URI Problem Link – https://www.urionlinejudge.com.br/judge/en/problems/view/1005
Problem Name: URI 1005 Average 1 code
Problem Number : URI – 1005 Average 1 Solution
Online Judge : URI Online Judge Solution
Category: Beginner
Solution Language : C,C plus plus, java, python, c#(c sharp)
Problem’s Screenshot:

URI 1005 solution in C,C++

See a 4 minutes video – URI online judge solution 1005 in C programming :

If you have a little more time, then see the video, – URI 1005 solution video

URI Solution 1005 Code in C / URI 1005 solution in C:

#include <stdio.h>
int main(){
float
x,y;
scanf("%f %f", &x, &y);
printf("MEDIA = %.5fn", (x*3.5+y*7.5)/(3.5+7.5));
return
0;
}

URI Solution 1005 Code / URI 1005 solution in CPP:

#include <cstdio>

int
main()
{

double
a,b,m;

scanf("%lf %lf", &a, &b);

m = (a/11 * 3.5) + (b/11 * 7.5);

printf("MEDIA = %.5lfn", m);

return
0;
}

URI Solution 1005 Code / URI 1005 solution in Java:

import java.util.Scanner;
public class
URI_1005 {
public static
void main(String[] args){

float
a, b, m;
Scanner sc = new Scanner(System.in);
a = sc.nextFloat();
b = sc.nextFloat();

m = (float) ((a/11 * 3.5) + (b/11 * 7.5));

System.out.printf("MEDIA = %.5fn", m);


}
}

URI Solution 1005 Code / URI 1005 solution in  Python:

nota1 = float(input())
nota2 = float(input())

media = (((nota1 * 3.5) + (nota2 * 7.5)) / 11)

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

URI Solution 1005 Code / URI 1005 solution in  C# (C Sharp):


using System;
class
URI {

static
void Main(string[] args) {

double
a, b;
a = Convert.ToDouble(Console.ReadLine());
b=Convert.ToDouble(Console.ReadLine());
Console.WriteLine("MEDIA = " + ((a*3.5 + b*7.5)/(3.5+7.5)).ToString("0.00000"));
Console.ReadKey();

}
}

Demonstration:

This is a simple problem.
Steps to solve URI 1005 Average solution in C:

First take the value of a and b –

 scanf("%f %f", &x, &y);

Second Get the median and print –

printf("MEDIA = %.5fn", (x*3.5 + y*7.5) / (3.5 + 7.5));

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

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