Codeforces Round #142 (Div. 2)——B

B. T-primes

time limit per test
2 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

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.

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".

分析：只有当这个数的平方根是素数时，满足题意。

#include <iostream>
#include <cmath>
#include <cstdio>
#define sqr(x) ((x)*(x))
using namespace std;

bool composite[1000010];
int n;
long long x;

int main()
{
composite[1]=true;
//素数打表
for(int i=2; i<=1000000; i++)
if(!composite[i])
{
for(int j=2; i*j<=1000000; j++)
composite[i*j]=true;
}
cin>>n;
while(n--)
{
long long x;
scanf("%I64d",&x);
long long y=(long long)sqrt(x);
if(y*y==x&&composite[y]==false) printf("YES\n");
else printf("NO\n");
}
return 0;
}