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")