CF 230B. T-primes

题目

time limit per test：2 seconds；memory limit per test：256 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".

 思路

代码

import math
def euler_sieve(n):
    prime = [True] * (n+1)
    prime[0] = prime[1] = False

    p = 2
    while p*p <= n:
        if prime[p]:
            for i in range(p*p, n+1, p):
                prime[i] = False
        p += 1

    return prime

# 判断一个数是否为素数
def is_prime(n):
    if n < 2:
        return False
    return prime[n]

t=int(input())
nums=list(map(int,input().split()))
n=max(nums)
prime = euler_sieve(int(math.sqrt(n))+1)
# print(prime)
for x in nums:
    r=int(x**0.5)
    if r**2!=x:
        print("NO")
    else:
        if is_prime(r):
            print("YES")
        else:
            print("NO")
 

import math
def judgePri(N):
    isPrime = [True] * (N+1)    #用于标注一个整数是否为质数
    primes = [] #用于存放素数，一开始是空的
    cnt = 0
    isPrime[0]=isPrime[1] = False
    # M=int(math.sqrt(N))
    for i in range(2,N):
        if isPrime[i]:  #如果i为质数，就添加到primes中
            primes.append(i)
            cnt += 1    #标记primes中元素的个数
        for j in range(cnt):
            temp = i * primes[j]
            if temp >= N:   #若合数i * primes[j]越界，则不作处理，结束本轮遍历
                break
            isPrime[temp] = False
            if i % primes[j] == 0:
                break
    return isPrime

t=int(input())
nums=list(map(int,input().split()))
n=max(nums)
isPrime= judgePri(int(math.sqrt(n))+1)
# print(prime)
for x in nums:
    r=int(math.sqrt(x))
    if r**2!=x:
        print("NO")
    else:
        if isPrime[r]:
            print("YES")
        else:
            print("NO")