Determine if Number is a Perfect Square

The program must accept an integer N as the input. The program should determine if the given number is a perfect square. If the number is a perfect square, the program must print YES. Otherwise, the program must print NO.

Boundary Condition(s):
1 <= N <= 10^12

Input Format:
The first line contains the integer N.

Output Format:
The first line contains either YES or NO.

Example Input/Output 1:
Input:
25

Output:
YES

Explanation:
The number 25 is a perfect square of 5. Hence the output is YES.

Example Input/Output 2:
Input:
20

Output:
NO

Explanation:
The number 20 is not a perfect square. Hence the output is NO.

n = int(input().strip())
sqrt_n = int(n**0.5)
print("YES" if sqrt_n*sqrt_n == n else "NO")

#include<stdio.h>
#include<math.h>

int main() {
    long long int n;
    scanf("%lld", &n);
    long long int root = (long long int)sqrt(n);
    printf("%s", (root * root == n) ? "YES" : "NO");
    return 0;
}

#include <iostream>
#include <cmath>
using namespace std;

int main() {
    long long int n;
    cin >> n;
    long long int root = (long long int)sqrt(n);
    cout << ((root * root == n) ? "YES" : "NO");
    return 0;
}

import java.util.Scanner;

public class PerfectSquare {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        long n = sc.nextLong();
        long sqrtN = (long) Math.sqrt(n);
        System.out.println((sqrtN * sqrtN == n) ? "YES" : "NO");
    }
}