URI Online Judge Solution | 1072 Interval 2 – Solution in C, C++, Java, Python and C#

URI Online Judge Solution | 1072 Interval 2 – Solution in C, C++, Java, Python and C#

URI Online Judge Solution 1072 Interval 2 | Beginner
URI Problem Link – https://www.urionlinejudge.com.br/judge/en/problems/view/1072

Problem Name: 1072 Interval 2
Problem Number : URI – 1072 Interval 2
Online Judge : URI Online Judge Solution
Category: Beginner
Solution Language : C,C plus plus, java, python, c#(c sharp)

URI Solution 1072 Interval 2 Code in C / URI 1072 solution in C:

#include <stdio.h>
int main()
{
    int x, a, i;

    int in = 0;
    int out = 0;

    scanf("%d", &x);
    for(i = 0; i < x; i++)
    {
        scanf("%d", &a);
        if(a >= 10 && a <= 20){
            in++;
        }else{
            out++;
        }

    }
    printf("%d inn", in);
    printf("%d outn", out);

    return 0;
}

URI Solution 1072 Interval 2 Code / URI 1072 Interval 2 solution in CPP:

#include <iostream>

using namespace std;

int main(){
    int x, a;
    
    int in = 0;
    int out = 0;
    
    cin >> x;
    for(int i = 0; i < x; i++){
            cin >> a;
            if(a >= 10 && a <= 20) in++;
            else out++;
    }
    cout << in << " inn";
    cout << out << " outn";
    
    return 0;
}

/**
4
    14
    123
    10
    -25
        2 in
        2 out
*/
import java.util.Scanner;

public class Main {

 public static void main(String[] args) {
  int N , X, in = 0, out = 0;
  int interval_start =10,interval_end =20 ;

  Scanner input =new Scanner(System.in);

  N =input.nextInt();
  for (int i = 1; i <= N; i++) {
   X =input.nextInt();
   if (X >= interval_start && X <= interval_end) {
    in += 1;
   }else {
    out += 1;
   }
  }
  System.out.print(in+" inn"+out +" outn");


 }

}

URI Solution 1072 Interval 2  Code / URI 1072 Interval 2 solution in  C# (C Sharp):

Demonstration:

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

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