Python3 找素数

# -*- coding:utf-8 -*-
import math
import time

ss = []  #  放可能是素数的列表
fss = []  # 放可能是非素数的列表
result = []  # 最终结果
x = 300

print("start!" )
start = time.clock()
# 遍历所有小于X，大于2的数
for xx in range(2, x+1):
    # 只要xx的数，不能被2至xx的平方根的所有数整除，就是素数
    for i in range(2, int(math.sqrt(xx)+1)):
        if (xx % i) != 0:
            i = i + 1
            # print("素数: ",xx)
            ss.append(xx)
        else:
            # print("非素数", xx)
            fss.append(xx)

# 只要x中的数没有出现在非素数列表中，则它就是素数
for j in range(2, x+1):
    if j not in fss:
        result.append(j)

print("result: ", result)

end = time.clock()
times = ''.join("%f s" % (end - start))  # 看用时多久
print("end!\n耗时：", times)

start!
result:  [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283, 293]
end! 
耗时：0.002066 s

# 判断是否为素数
def is_prime(n):
    if n == 1:
        return False
    for i in range(2, int(math.sqrt(n)+1)):
        if n % i == 0:
            return False
    return True

start = time.clock()
print("start:" )
for n in range(2, x+1):
    if is_prime(n) is True:
        result.append(n)
print(result)
end = time.clock()
times = ''.join("%f s" % (end - start))
print("end!\n耗时：", times)

start:
[2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283, 293]
end!
耗时： 0.000434 s