T-primes(预处理判断素数)

B. T-primes

(CF-230B)

time limit per test : 2 seconds
memory limit per test : 256 megabytes
input : standard input
output : standard output

Description

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 ≤ 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.

Examples

Input

3
4 5 6

Output

YES
NO
NO

其实就是个判断素数的水题，时间为两秒，但用了常规的方法(直接用for遍历)之后发现一直tle，可能有个样例是数量很多并且数字都很大，于是就对数字先进行预处理，把非素数给标记，然后对于每一次的输入用O(1)的时间来判断是否为素数。

#include 
using namespace std;
typedef long long ll;
const int maxn=1e6+5;
int a[maxn+1];

void isprime()
{
     
    a[0]=a[1]=1;
    memset(a,0,sizeof(a));
    for(int i=2;i<=maxn;i++){
     
        if(!a[i]){
     
            for(int j=i+i;j<=maxn;j+= i){
     
                a[j]=1;
            }
        }
    }
}

int main(){
     
    int t;
    scanf("%d",&t);
    isprime();
    while(t--){
     
        ll n;
        scanf("%I64d",&n);
        ll x=sqrt(n);
        if(x>1&&x*x==n&&a[x]==0)
            printf("YES\n");
        else printf("NO\n");
    }
    return 0;
}

小结：对于素数的判断不只有for遍历，还可以先预处理再直接判断