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1. #基础写法,效率慢
import datetime
start = datetime.datetime.now()
prime_number = []
n = 100000
for i in range(2, n):
for j in range(2,i):
if i%j == 0:
break
else:
prime_number.append(i)
t_time = (datetime.datetime.now()-start).total_seconds()
print('用时：{}s,{}内的素数一共有{}个'.format(t_time, n, len(prime_number)))
print(prime_number)
用时：49.140267s，100000内的素数一共有9592个
改进1.
除了2 的偶数都不是素数 ，所以将2直接放在列表中，第7行改为for i in range(3,n,2)，将步长设为2，跳过所有偶数；同时因为奇数不能被偶数整队，所以将第8行改成for j in range(3, i, 2)
import datetime
start = datetime.datetime.now()
prime_number = [2]
n = 100000
for i in range(3, n, 2):
for j in range(3, i, 2):
if i%j == 0:
break
else:
prime_number.append(i)
t_time = (datetime.datetime.now()-start).total_seconds(
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