URI 1037 problem solution in C, C++ and java language || Interval

Main link : URI Web page - Problem No : 1037 Link

URI Online Judge | 1037

Interval

Adapted by Neilor Tonin, URI  Brazil

Timelimit: 1

You must make a program that read a float-point number and print a message saying in which of following intervals the number belongs: [0,25] (25,50], (50,75], (75,100]. If the read number is less than zero or greather than 100, the program must print the message “Fora de intervalo” that means "Out of Interval".

The symbol '(' represents greather than. For example:
[0,25] indicates numbers between 0 and 25.0000, including both.
(25,50] indicates numbers greather than 25 (25.00001) up to 50.0000000.

Input

The input file contains a floating-point number.

Output

The output must be a message like following example.

Input Sample	Output Sample
25.01	Intervalo (25,50]

25.00

Intervalo [0,25]

100.00

Intervalo (75,100]

-25.02

Fora de intervalo

URI SOLUTION 1037 IN C Language

#include<stdio.h>

int main()

{

 float n;

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

 if(n < 0 || n > 100){     //implementing the above logic

  printf("Fora de intervalo\n");

 }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 Solution in java language :

 import java.io.IOException;


import java.util.Scanner;

public class Main {

  

    public static void main(String[] args) throws IOException {

  

        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 intervalo\n");

        }

  

    }

  

}

URI SOLUTION 1037 IN C++ Language


#include <iostream>

using namespace std;

int main()

{

 float n;

 cin >> n;  //take value from the user

 if(n < 0 || n > 100){     //implementing the above logic

  cout << "Fora de intervalo" << endl;

 }else{

  if(n >= 0 && n <= 25){

   cout << "Intervalo [0,25]" << endl;

  }else if(n > 25 && n <= 50){

   cout << "Intervalo (25,50]" << endl;

  }else if(n > 50 && n <= 75){

   cout << "Intervalo (50,75]" << endl;

  }else{

   cout << "Intervalo (75,100]" << endl;

  }

 }

 return 0;

}



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