codeforces230 B. T-primes 【思维 + 打表】_input 3 4 5 6 output yes no no note the given test

B. T-primes

time limit per test2 seconds
memory limit per test256 megabytes

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”.

题意： 给你n个数范围$[1,10^{12}]$,让你判断该数是否只包含三个因子

分析： 我们首先知道除了1之外的任意数，最少包含2个因子（1和本身），当一个数含有一个非平方根之外的因子时，必然存在与之对应的另一个因子，比如当 15 的一个因子为3，存在另一因子 5,与之对应，所以我们只需判断该数的平方根是否为素数即可，比如 36，该数的平方根为6，则必存在其他因子，我们只需打 十万之内的素数表判断即可

参考函数

#include <bits/stdc++.h>

using namespace std;
typedef long long ll;
const int maxn = 1e6 + 7;

bool isp[maxn];

void init() {
    isp[0] = isp[1] = true;
    for(int i = 2;i <= 1000;i++) {
        for(int j = i*2;j < maxn;j += i) {
            isp[j] = true;
        }
    }
}

int main() {
    int n;cin>>n;
    init();
    while(n--) {
        ll x;scanf("%lld",&x);
        ll m = (ll)sqrt(x+0.5);
        if(m*m == x && !isp[m]) {
            puts("YES");
        } else {
            puts("NO");
        }
    }
    return 0;
}
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