Day 21 B. T-primes

Problem
We know that prime numbers are positive integers that have exactly two distinct positive divisors. Similarly, we’ll call a positive integer t Т-prime, if t has exactly three distinct positive divisors.

You are given an array of n positive integers. For each of them determine whether it is Т-prime or not.

Input
The first line contains a single positive integer, n (1 ≤ n ≤ 105), showing how many numbers are in the array. The next line contains n space-separated integers xi (1 ≤ xi ≤ 1012).

Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is advised to use the cin, cout streams or the %I64d specifier.

Output
Print n lines: the i-th line should contain “YES” (without the quotes), if number xi is Т-prime, and “NO” (without the quotes), if it isn’t.

Examples
input
3
4 5 6
output
YES
NO
NO

Note
The given test has three numbers. The first number 4 has exactly three divisors — 1, 2 and 4, thus the answer for this number is “YES”. The second number 5 has two divisors (1 and 5), and the third number 6 has four divisors (1, 2, 3, 6), hence the answer for them is “NO”.

题目大致意思为：给你一个数 判断这个数的因子个数是否为3个

#include<iostream>
#include<algorithm>
#include<string.h>
#include<vector>
#include<cmath>
#include<set>
#include<map>
#define ll long long
using namespace std;
//如果要判断一个数的因子个数是否为三个 
//那么这个数除了自身和1以外 
//必须是一个完全平方数并且开根号后的这个数是质数
bool Isprime(int a)
{
	if (a < 2)
	{
		return false;
	}
	for (int i = 2; i <= sqrt(a); i++)
	{
		if (a % i == 0)
		{
			return false;
		}
	}
	return true;
}
int main()
{
	int x;
	cin >> x;
	for (int i = 0; i < x; i++)
	{
		ll a;
		ll b;
		cin >> a;
		b = sqrt(a);
		if (b * b == a)
		{
			if (Isprime(sqrt(a)))
			{
				cout << "YES" << endl;
			}
			else
			{
				cout << "NO" << endl;
			}
		}
		else
		{
			cout << "NO" << endl;
		}
	}
	return 0;
}