URI Online Judge Solution 1015 Distance Between Two Points – URI 1015 Solution in C, C++, Java, Python and C#

URI Online Judge Solution URI Solution -Beginner

URI Online Judge Solution 1015 Distance Between Two Points – URI 1015 Solution in C, C++, Java, Python and C#

URI Online Judge Solution 1015 Distance Between Two Points | Beginner
URI Problem Link – https://www.urionlinejudge.com.br/judge/en/problems/view/1015

Problem Name: 1015 Distance Between Two Points solution
Problem Number : URI – 1015 Distance Between Two Points code
Online Judge : URI Online Judge Solution
Category: Beginner
Solution Language : C,C plus plus, java, python, c#(c sharp)

URI Solution 1015 Distance Between Two Points Code in C / URI 1015 code in C:

#include <stdio.h>
#include <math.h>

int
main()
{

double
x1, x2, y1, y2, dist;

scanf("%lf %lf %lf %lf", &x1, &y1, &x2, &y2);

dist = sqrt(pow(x2-x1,2)+pow(y2-y1,2));

printf("%.4lfn", dist);
return
0;
}

URI Solution 1015 Distance Between Two Points Code / URI 1015 solution in CPP:

#include <stdio.h>
#include <math.h>

int
main()
{

double
x1, x2, y1, y2, dist;

scanf("%lf %lf %lf %lf", &x1, &y1, &x2, &y2);

dist = sqrt(pow(x2-x1,2)+pow(y2-y1,2));

printf("%.4lfn", dist);
return
0;
}

URI Solution 1015 Distance Between Two Points Code / URI 1015 solution in Java:

import java.util.Scanner;


public class
Main {


public static
void main(String[] args) {
double
x1, x2, y1, y2, dist;

Scanner sc = new Scanner(System.in);
x1 = sc.nextDouble();
y1 = sc.nextDouble();
x2 = sc.nextDouble();
y2 = sc.nextDouble();

dist = Math.sqrt(Math.pow(x2-x1,2)+Math.pow(y2-y1,2));

System.out.printf("%.4fn", dist);

}

}

URI Solution 1015 Distance Between Two Points Code / URI 1015 solution in  Python:

import math

linha1 = input().split(" ")
linha2 = input().split(" ")

x1,y1 = linha1
x2,y2 = linha2

distancia = math.sqrt(((float(x2) - float(x1))*(float(x2) - float(x1))) + ((float(y2)-float(y1)) *(float(y2)-float(y1))))

print("%0.4f" %distancia)

URI Solution 1015 Code / URI 1015 solution in  C# (C Sharp):

Demonstration:

Look for this problems ,

URI Online Judge Solution 1015 Distance Between Two Points - URI 1015 solution

So, our code is:

        scanf("%lf %lf %lf %lf", &x1, &y1, &x2, &y2);

dist = sqrt(pow(x2-x1,2)+pow(y2-y1,2));

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

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