B. T-primes_b - primes gym - 102267b csdn

作者：Gausst松鼠会 | 2024-02-10 19:13:12

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b - primes gym - 102267b csdn

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 x_i (1 ≤ x_i ≤ 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 x_i is Т-prime, and "NO" (without the quotes), if it isn't.

   Sample test(s) 
 

     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".

解题说明：此题就是判断一个数是否仅含有3个因子，其实也就是除了1和它自身外，还存在另一个因子。如果找到这个另外的因子就输出yes，否则为no。

首先来分析这个数的情况，由于只有一个不同的因子，那这个数肯定是某个数的平方数，否则不可能只有一个因子。这样问题缩小为在2到n^(1/2)-1之间求因子的个数，如果因子个数不为0，那证明还存在其他因子，不符合题意，我们就证明其中不存在因子即可，问题进一步转换为证明n^(1/2)是个素数。

证明素数的问题很常见，令m=n^(1/2)，然后判断m^(1/2)之间有没有因子即可。


#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cmath>
using namespace std;
 
int main()
{
	int n,i,j;
	long long a,b;
	cin>>n;
	for(i=1;i<=n;i++)
	{
		cin>>a;
		b=sqrtl(a); //long double sqrtl( long double x );
		for(j=2;j*j<=b;j++) 
		{	
			if(b%j==0)
			{
				break;
			}
		}
		if(j*j>b && b*b==a && a>1)
		{
			cout<<"YES\n";
		}
		else
		{
			cout<<"NO\n";
		}
	}
	return 0;
}

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