Here is my c++ solution, you can watch video explanation here : https://youtu.be/Qs9rxqfzTs8

bool is_smart_number(int num) {
    int val = (int) sqrt(num);
    if(val * val == num)
        return true;
    return false;
}

For Java coders: change line 10 to:

        if(val * val == num)

A smart number is only smart if its a perfect square.

bool is_smart_number(int num) { if(num<=0) return false;

int val = (int) sqrt(num);
if(val * val == num)
    return true;
return false;

}

Its correct but compiling false

Why this problem doesnt support JavaScript language

function isSmartNumber(num) {
    var val = Math.floor(Math.sqrt(num));
    if (val * val === num) {
        return true;
    }
    return false;
}

import java.io.; import java.util.; import java.text.; import java.math.; import java.util.regex.*;

public class Solution { public static boolean isSmartNumber(int num) { int val = (int) Math.sqrt(num);
if(num / val == 1) return true; return false; }

public static void main(String[] args) {
    int test_cases;
    Scanner in = new Scanner(System.in);
    test_cases = in.nextInt();
    int num;
    for(int i = 0; i < test_cases; i++){
        num = in.nextInt();
        boolean ans = isSmartNumber(num);
        if(ans){
            System.out.println("YES");
        }
        else System.out.println("NO");
    }
}

}