URI Online Judge Solution | 1072 Interval 2 - Solution in C, C++, Java, Python and C#
URI Online Judge Solution 1072 Interval 2 | BeginnerURI 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 in\n", in);
printf("%d out\n", 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 << " in\n";
cout << out << " out\n";
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+" in\n"+out +" out\n");
}
}
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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