#include<stdio.h>
int main(){
double x1,x2,y1,y2,d;
scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);
d=sqrt(pow(x2-x1,2)+pow(y2-y1,2));
printf("%.4lf\n",d);
return 0;
}
import math
x1, y1, x2, y2 = map(float, input().split())
d = math.sqrt((x2 - x1) ** 2 + (y2 - y1) ** 2)
print(f"{d:.4f}")
#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;
}
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.
X = int(input())
Y = float(input())
a = X / Y
print("{:.3f} km/l".format(a))
input() function is used to read user input from the console. In Python, it automatically converts the input into a string.X and Y using input() and convert them to the appropriate data types using int() and float() functions, respectively.a = X / Y performs the division operation and stores the result in the variable a.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.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.
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:
The input file contains 3 integer values.
Print the greatest of these three values followed by a space and the message “eh o maior”.
| Input Samples | Output Samples |
| 7 14 106 | 106 eh o maior |
#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;
}
#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;
}
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")
beecrowd | 1012
Make a program that reads three floating point values: A, B and C. Then, calculate and show:
a) the area of the rectangled triangle that has base A and height C.
b) the area of the radius’s circle C. (pi = 3.14159)
c) the area of the trapezium which has A and B by base, and C by height.
d) the area of the square that has side B.
e) the area of the rectangle that has sides A and B.
The input file contains three double values with one digit after the decimal point.
The output file must contain 5 lines of data. Each line corresponds to one of the areas described above, always with a corresponding message (in Portuguese) and one space between the two points and the value. The value calculated must be presented with 3 digits after the decimal point.
| Input Samples | Output Samples |
| 3.0 4.0 5.2 | TRIANGULO: 7.800 CIRCULO: 84.949 TRAPEZIO: 18.200 QUADRADO: 16.000 RETANGULO: 12.000 |
| 12.7 10.4 15.2 | TRIANGULO: 96.520 CIRCULO: 725.833 TRAPEZIO: 175.560 QUADRADO: 108.160 RETANGULO: 132.080 |
#include<stdio.h>
int main() {
double A,B,C,t,pi,c,tra,s,rec;
scanf("%lf%lf%lf",&A,&B,&C);
pi=3.14159;
t=.5AC;
c=piCC;
tra=.5(A+B)C;
s=BB; rec=AB;
printf("TRIANGULO: %.3lf\nCIRCULO: %.3lf\nTRAPEZIO: %.3lf\nQUADRADO: %.3lf\nRETANGULO: %.3lf\n",t,c,tra,s,rec);
return 0;
}
import math
A, B, C = map(float, input().split())
pi = 3.14159
t = 0.5 * A * C
c = pi * (C ** 2)
tra = 0.5 * (A + B) * C
s = B ** 2
rec = A * B
print(f"TRIANGULO: {t:.3f}")
print(f"CIRCULO: {c:.3f}")
print(f"TRAPEZIO: {tra:.3f}")
print(f"QUADRADO: {s:.3f}")
print(f"RETANGULO: {rec:.3f}")
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.
#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
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:
For the first time, I have developed a C++ application that compiles and runs flawlessly. I’ve made the decision to learn C++, and in order to become comfortable with the language, I’ve developed my first C++ program, which I wrote all by myself after only a few hours of instruction. (Also, no “Hello, World.” program has been written by me.
#include<iostream> using namespace std; void primeSieve(int n) { int prime[100]={0}; for(int i=2;i<=n;i++) { if(prime[i]==0){ for(int j=i*i;j<=n;j+=i){ prime[i]==1; } } } for(int i=2;i<=n;i++) { if(prime[i]==0) { cout<<i<<" ";
}cout<<endl;}} int main() { int a; cin>>a; primeSieve(a); }
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.
#include<stdio.h>
int main() {
double R,A;
scanf("%lf",&R);
A=3.14159*R*R;
printf("A=%.4lf\n",A);
return 0;
}
#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;
}
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);
}
}
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}")
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.
#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;
}
name = input()
a, b = map(float, input().split())
c = b * 0.15 + a
print(f"TOTAL = R$ {c:.2f}")