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Beecrowd 1014 solution ||BEECROWD 1014 – Consumption Solution in C ,Python || URI 1014 || Free Code Center

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Beecrowd 1014 solution || BEECROWD 1014 – Consumption Solution

Question

BEECROWD 1014 – Consumption Solution in C

Solution in C

#include<stdio.h>

int main() {

int X;
float Y;
scanf("%d%f",&X,&Y);
double a=X/Y;
printf("%.3lf km/l\n",a);//Beecrowd 1014 solution

return 0;

}

Explanation:

The code provided is a C program that calculates the fuel efficiency of the vehicle and prints kilometers per liter (km/litre). Here is a step-by-step explanation of the code.

The #include directive is used in conjunction with standard input/output libraries, which provide functions such as scanf and printf.

The main() function is the entry point of the function. Where program execution begins and ends. It has a return type of int, which means that an integer value should be returned when the program finishes execution.

The following changes are declared in the main() function.

X: Integer variable used to store the value of the distance traveled (in kilometers).

Y: A floating variable used to store fuel consumption value (in liters).

The scarf (“%d%f”, &X, &Y) statement is used to read the user input. Requires two values ​​separated by a space or other character. The first value is stored in X and the second value is stored in Y.

The next rule calculates fuel efficiency by dividing the distance traveled (X) by fuel consumption (Y). The result is stored in a double-precision variable called a.

Finally, the print (“%.3lf km/l\n”, a) statement is used to print the calculated fuel consumption. The “%.3lf” format specifier is used to print a value with three decimal places. The string “km/l” is included to indicate the unit of measurement.

sand after 0; The statement indicates the end of the main() function and indicates that the operation was successful. A value of 0 is returned to the operating system, indicating that the operation terminated without error.

In summary, this code lets the user record the distance traveled (in kilometers) and the amount of fuel used (in liters). then calculates the effect of fuel efficiency in km/l and prints the result with three decimal places.

Beecrowd 1014 solution in Python

The solution in Python:

X = int(input())
Y = float(input())
a = X / Y
print("{:.3f} km/l".format(a))

Explanation:

  1. In Python, there is no need to include any header file like in C. We can directly start writing the code.
  2. The input() function is used to read user input from the console. In Python, it automatically converts the input into a string.
  3. We read the values for X and Y using input() and convert them to the appropriate data types using int() and float() functions, respectively.
  4. The line a = X / Y performs the division operation and stores the result in the variable a.
  5. The print() function is used to display the calculated fuel efficiency. We use string formatting to specify the number of decimal places using the :.3f format specifier, which rounds the value to three decimal places.
  6. The output will be displayed on the console, showing the fuel efficiency in km/l.
  7. The return 0 statement is not necessary in Python, so it can be omitted.

Please note that Python is an interpreted language, while C is a compiled language. The code structures and syntax differ between the two languages, so it’s important to understand the specific syntax and conventions when translating code from one language to another.

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Beecrowd 1013 solution- The Greatest

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Beecrowd 1013- The Greatest solution Question

Make a program that reads 3 integer values and present the greatest one followed by the message “eh o maior”. Use the following formula:

Input

The input file contains 3 integer values.

Output

Print the greatest of these three values followed by a space and the message “eh o maior”.

Input SamplesOutput Samples
7 14 106106 eh o maior

Beecrowd 1013 solution- The Greatest solution in C

#include<stdio.h>

int main() {

int a,b,c,d;
scanf("%d%d%d",&a,&b,&c);
d=(a+b+abs(a-b))/2;
d=(d+c+abs(d-c))/2;
printf("%d eh o maior\n",d);

return 0;

}

Beecrowd 1013 solution- The Greatest solution in C++ 

#include<bits/stdc++.h>
using namespace std;

int main() {

int a,b,c,d;
cin>>a>>b>>c;

d=(a+b+abs(a-b))/2;
d=(d+c+abs(d-c))/2;
cout<<d<<" eh o maior"<<endl;


return 0;

}

The Greatest Solution in Python

a, b, c = map(int, input().split())

d = (a + b + abs(a - b)) // 2
d = (d + c + abs(d - c)) // 2

print(f"{d} eh o maior")

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URI 1011 solution in C || Beecrowd 1011 Sphere solution in C || Beecrowd 1011 || Free Code Center

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URI 1011 Solution in C || Beecrowd 1011 Sphere Solution

Beecrowd 1011 Sphere Solution in C: A Comprehensive Guide

At its core, programming is about solving issues. In this text, we’ll take a deep dive into the answer for the Beecrowd 1011 Sphere hassle in C. We’ll start with a quick evaluation of the problem, then walk through the solution step-with the aid of-step, and ultimately offer some guidelines for improving the efficiency of the code.

Overview of the URI/Beecrowd online decide solution1011 Sphere solution in C
The Beecrowd 1011/URI 1011 Sphere problem is a traditional problem in programming. It involves calculating the extent of a sphere with a given radius. Here’s the hassle assertion:

You must calculate and print the volume of a sphere given its radius R. The system to calculate the volume is (four/3) * pi * R^three. Consider (assign) for this trouble that π = three.14159.

URI 1011 Sphere solution in C

#include<stdio.h>

int main() {

double R,pi,v;
pi=3.14159;
scanf("%lf",&R);
v=(4.0piRRR)/3;
printf("VOLUME = %.3lf\n",v);//URI 1011

return 0;

}

URI 1011 Sphere solution in C

Tips for Improving the Efficiency of the Code

While the above solution works perfectly fine for the URI 1011 Sphere solution in C, there are a few ways we can optimize the code for efficiency. Here are some tips:

  1. Use the pow() function to calculate the cube of the radius instead of multiplying it by itself three times. This can improve the readability of the code and make it easier to modify in the future.
  2. Use a constant instead of initializing pi every time the program runs. This can improve the efficiency of the program by reducing the amount of memory and CPU time required.
  3. Use meaningful variable names to make the code easier to read and understand. For example, we could use “radius” instead of “R” to make the purpose of the variable more clear.
  4. Use comments to explain the purpose and functionality of the code. This can make it easier for others to understand and modify the code in the future.

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Beecrowd 1002 solution || Beecrowd 1002 – Area of a Circle solution in C, C++, Python

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BEECROWD 1002 – Area of a Circle Question

BEECROWD 1002 Solution – Area of a Circle solution  | Area of a Circle- URI – BEECROWD – BEE Problem 1002 Solution in C,C++,Python 

Introduction

In this article, we’ll give a thorough C, C++, Java, C#, and Python solution to URI Online Judge problem 1002. We must determine the area of a circle given its radius in order to solve this issue. Our solution is streamlined for effectiveness and precision and is easily adaptable to other programming languages and platforms.Beecrowd 1002 solution

Overview of the Solution

One function that receives the radius as an argument and outputs the circle’s area constitutes our solution. The area is calculated using the mathematical formula A = . R2, where has a constant value of 3.14159. Large-scale calculations and real-time applications can benefit from the function’s implementation, which increases performance and reduces errors.

Application of the Solution

This is how it’s done Solution in C
#include<stdio.h> 

int main() {

double R,A;

scanf("%lf",&R);
A=3.14159*R*R;
printf("A=%.4lf\n",A);

return 0;

}

Beecrowd 1002 solution

This is how it’s done Solution in C++
#include <iostream>
#include <iomanip>
#include <cmath>

using namespace std;

double calculate_area(double radius) {
    const double pi = 3.14159;
    return pi * pow(radius, 2);
}

int main() {
    double radius;
    cin >> radius;
    double area = calculate_area(radius);
    cout << fixed << setprecision(4) << "A=" << area << endl;
    return 0;
}

Beecrowd 1002 solution

This is how it’s done Solution in Java 
import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner input = new Scanner(System.in);
        double radius = input.nextDouble();
        double area = calculateArea(radius);
        System.out.printf("A=%.4f%n", area);
    }

    public static double calculateArea(double radius) {
        final double pi = 3.14159;
        return pi * Math.pow(radius, 2);
    }
}

Beecrowd 1002 solution

This is how it’s done Solution in Python

import math

def calculate_area(radius):
    pi = 3.14159
    return pi * radius ** 2

radius = float(input())
area = calculate_area(radius)
print(f"A={area:.4f}")

Conclusion

Our solution provides a fast and accurate way to calculate the area of a circle given its radius in C, C++, Java, and Python. By following the implementation examples we have provided, you can easily adapt our solution to your specific programming language and platform. With our optimized approach, you can confidently calculate circle areas with high precision, making it suitable for various real-life applications.

Thank you for considering our solution. We are confident that it will help you outrank the article you provided and establish your website as a reliable source of information for URI Online Judge problem 1002 and similar programming challenges.

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Uri 1009 Solution in C || Beecrowd1009 – Salary with Bonus solution in C,Python || Beecrowd1009 || Free Code Center

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Beecrowd1009 – Salary with Bonus solution

problem question

Uri 1009 solution in C- Salary with Bonus solution 

#include<stdio.h>

int main() {

double a,b,c;
char name;

scanf("%s %lf %lf",&name,&a,&b);
 c=b*.15+a;

printf("TOTAL = R$ %.2lf\n",c);

return 0;

}

Uri 1009 solution in Python || Beecrowd1009 – Salary with Bonus solution with Python

name = input()
a, b = map(float, input().split())
c = b * 0.15 + a

print(f"TOTAL = R$ {c:.2f}")

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