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

URI Online Judge Solution URI Solution -Beginner

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;
}

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


/**
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  Python:

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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