CF 230B. T-primes

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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

根据题目，一个数如果是完全平方数，并且这个数的平方根是质数，那就符合 Т -prime 的定义，这是可以

证明的。可以用欧拉筛法判断质数，或者埃氏筛法。不知为何，在 CF 上提交之后，用埃氏筛法甚至比欧

拉筛法要快，可能是没有想到优化的方法 , 或者某些操作耗费了时间。

代码

python">import math
def euler_sieve(n):prime = [True] * (n+1)prime[0] = prime[1] = Falsep = 2while p*p <= n:if prime[p]:for i in range(p*p, n+1, p):prime[i] = Falsep += 1return prime# 判断一个数是否为素数
def is_prime(n):if n < 2:return Falsereturn 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")

python">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 = 0isPrime[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breakisPrime[temp] = Falseif i % primes[j] == 0:breakreturn isPrimet=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")