Codeforces--230B--T-primes

题目描述：
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.
输入描述:
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.
输出描述:
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.
输入：
3
4 5 6
输出：
YES
NO
NO
题意：
判断一个数是不是只有三个因子
题解：
只有三个因子的数，肯定是素数，且是个完全平方数，欧拉筛筛素数，判断就完事了。
代码：

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<cmath>
using namespace std;

const int maxn = 10000005;
typedef long long ll;
bool vis[maxn];
int prime[maxn * 2];

void init(){
    int x = 0;
    memset(vis,0,sizeof(vis));
    vis[1] = 1;
    int k = 0;
        for(int i = 2;i < maxn;i ++)
        {
            if(!vis[i])prime[++ k] = i;
            for(int j = 1;j <= k && prime[j] * i< maxn;j ++)
            {
                vis[prime[j] * i] = 1;
                if(i % prime[j] == 0)break;
            }
        }
}

int main(){
    int t;
    ll n;
    init();
    scanf("%d",&t);
    while(t--){
        scanf("%lld",&n);
        ll cnt = sqrt(double(n));
        if(cnt * cnt == n && vis[cnt] == 0) printf("YES\n");
        else printf("NO\n");
    }
}