URI - BEECROWD - BEE 1037 - Interval Solution in C,C++,Python

URI - BEECROWD - BEE Online Judge Solution  1037 | Interval - URI - BEECROWD - BEE 1037 Solution in C,C++,Python

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 Online Judge Solution  1036 | Bhaskaras Formula - URI 1036 Solution in C,C++,Python:

  URI Problem 1037 Solution in C :   

URI Online Judge 1037 Solve  in C :

#include<stdio.h>

int main()
{
 float n;
 scanf("%f", &n);

 if(n < 0 || n > 100){
  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 Problem 1037 Solution in C ++:

URI Online Judge 1037 Solve  in C++ :

#include <iostream>

using namespace std;

int main(){
    
    float f;
    
    cin >> f;
    
    if(f < 0.0000)
         cout << "Fora de intervalo\n";
    else if(f >= 0.0000 && f <= 25.0000)
         cout << "Intervalo [0,25]\n";
    else if(f > 25.0000 && f <=50.0000)
         cout << "Intervalo (25,50]\n";
    else if(f > 50.0000 && f <=75.0000)
         cout << "Intervalo (50,75]\n";
    else if(f > 75.0000 && f <=100.0000)
         cout << "Intervalo (75,100]\n";
    else if(f > 100.0000)
         cout << "Fora de intervalo\n";
         
    return 0;
}

  URI Problem 1037 Solution in Python:

URI Online Judge 1037 Solve  in Python 3.8+ :

X = float(input())
if X>75 and X<=100:
    print("Intervalo (75,100]")
elif X>50 and X<=75:
    print("Intervalo (50,75]")
elif X>25 and X<=50:
    print("Intervalo (25,50]")
elif X>=0 and X<=25:
    print("Intervalo [0,25]")
else:
    print("Fora de intervalo")

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