URI Online Judge Solution 1037 Interval – URI 1037 Solution in C, C++, Java, Python and C#

URI Online Judge Solution URI Solution -Beginner

URI Online Judge Solution 1037 Interval – URI 1037 Solution in C, C++, Java, Python and C#

URI Online Judge Solution 1037 Interval | Beginner
URI Problem Link – https://www.urionlinejudge.com.br/judge/en/problems/view/1037

Problem Name: 1037 Interval
Problem Number : URI – 1037 Interval
Online Judge : URI Online Judge Solution
Category: Beginner
Solution Language : C,C plus plus, java, python, c#(c sharp)

URI Online Judge Solution 1037 Interval - URI 1037 Solution in C, C++, Java, Python and C#
URI Online Judge Solution 1037 Interval – URI 1037 Solution in C, C++, Java, Python and C#

URI 1037 Interval Code in C / URI 1037 solution in C:

#include<stdio.h>

int
main()
{

float
n;
scanf("%f", &n); //take value from the user

if
(n < 0 || n > 100){
printf("Fora de intervalon");
}
else{
if
(n >= 0 && n <= 25){
printf("Intervalo [0,25]n");
}
else if(n > 25 && n <= 50){
printf("Intervalo (25,50]n");
}
else if(n > 50 && n <= 75){
printf("Intervalo (50,75]n");
}
else{
printf("Intervalo (75,100]n");
}
}


return
0;
}

URI 1037 Interval Code in C++ / URI 1037 solution in CPP:

#include<stdio.h>

int
main()
{

float
n;
scanf("%f", &n); //take value from the user

if
(n < 0 || n > 100){
printf("Fora de intervalon");
}
else{
if
(n >= 0 && n <= 25){
printf("Intervalo [0,25]n");
}
else if(n > 25 && n <= 50){
printf("Intervalo (25,50]n");
}
else if(n > 50 && n <= 75){
printf("Intervalo (50,75]n");
}
else{
printf("Intervalo (75,100]n");
}
}


return
0;
}

URI 1037 Interval Code in java/ URI 1037 solution in Java:

import java.util.Scanner;

public class
Main {

public static
void main(String[] args) {
float
N;
Scanner input =new Scanner(System.in);
N = input.nextFloat();
//Set the range
if (N >=0 && N <= 25.0000) {
System.out.printf("Intervalo [0,25]n");
}
else if (N >= 25.00001 && N <= 50.0000000) {
System.out.printf("Intervalo (25,50]n");
}
else if (N >= 50.00001 && N <= 75.0000000) {
System.out.printf("Intervalo (50,75]n");
}
else if (N >= 75.00001 && N<=100.0000000) {
System.out.printf("Intervalo (75,100]n");
}
else {
System.out.print("Fora de intervalon");
}

}

}

URI 1037 Interval Code in Python / URI 1037 solution in Python:

URI 1037 Interval Code / URI 1037 solution in  C# (C Sharp):

Demonstration:

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

Tags: URI Online Judge Solution, URI OJ Solution list, URI Problems Solution, URI solver, URI all problem solution list, URI Interval code in C, URI 1037 code in C++, URI Interval solution in C, URI solution, URI 1037 solution in C,URI 1037 solution in C++-CPP,URI 1037 solution in C# (C sharp),URI 1037 solution in Java,URI 1037 solution in Python, URI Interval code solution, URI Interval code

Leave a Reply

Your email address will not be published.