python算法——求10万以内的质数
1、使用质数定义计算
# -*- coding:utf-8 -*- # version:python3.7 ''' @ file :prime_number @ author:zhangyangyang @ create:2020/3/22 @ remark: ''' #version1
import datetime #导入模块计算效率
start = datetime.datetime.now() count = 0 for x in range(2,100000): #求指定范围内的质数 for i in range(2,x): #除以1和本身之外的数 if x % i == 0: break else: #print(x) count += 1 delta = (datetime.datetime.now() - start).total_seconds() #total_seconds()总秒数 print('count=',count,'delta=',delta) #墙上的时间 执行结果: count= 9592 delta= 148.146291 #效率极差
2、优化1:经计算,临界值为开方值
#version2:优化
import datetime #导入模块计算效率 start = datetime.datetime.now() count = 0 for x in range(2,100000): for i in range(2,int(x ** 0.5 + 1)): #优化1,经测试:临界值为开方值 if x % i == 0: break else: #print(x) count += 1 delta = (datetime.datetime.now() - start).total_seconds() #total_seconds()总秒数 print('count=',count,'delta=',delta) 执行结果: count= 9592 delta= 1.084154 #效率极大提高
3、优化2:大于2的偶数全是合数
#version3:优化+ import datetime #导入模块计算效率 start = datetime.datetime.now() count = 1 #print(2) #从3开始,自己打印2 for x in range(3,100000,2): #优化2:从3开始的奇数 #for i in range(3,int(x ** 0.5 + 1)): #优化3:奇数不用和2取模 for i in range(3, int(x ** 0.5) + 1,2): #优化4:即也不用和偶数取模 if x % i == 0: break else: #print(x) count += 1 delta = (datetime.datetime.now() - start).total_seconds() #total_seconds()总秒数 print('count=',count,'delta=',delta) #墙上的时间 执行结果: count= 9592 delta= 0.553471 #性能进一步提高
4、优化3:5的倍数全是合数,剔除5的倍数
#version4:优化++ import datetime #导入模块计算效率 start = datetime.datetime.now() count = 1 #print(2) #从3开始,自己打印2 for x in range(3,100000,2): #优化2:从3开始的奇数 if x > 10 and x % 5 == 0: continue #优化5:剔除5的倍数 #for i in range(3,int(x ** 0.5 + 1)): #优化3:奇数不用和2取模 for i in range(3, int(x ** 0.5) + 1,2): #优化4:即也不用和偶数取模 if x % i == 0: break else: #print(x) count += 1 delta = (datetime.datetime.now() - start).total_seconds() #total_seconds()总秒数 print('count=',count,'delta=',delta) #墙上的时间 执行结果: count= 9592 delta= 0.493866
5、思考,总结,再优化:
质数:所有的质数除过2,都是奇数;
质数:临界值(开方值);
质数:质数*质数肯定不是质数,给定列表存放已知质数,使用该列表值进行判断,在该值的基础上锁定临界值;
孪生质数:大于3的质数只有6N-1和6N+1两种形式,如果6N-1和6N+1都是素数,成为孪生素数(效率也挺高)
# -*- coding:utf-8 -*- # version:python3.7 import datetime n = 100000 count = 2 primenumber = [3] start = datetime.datetime.now() for i in range(5,n + 1,2): flag = False x = int(i ** 0.5) for j in primenumber: if j > x: flag = True break if i % j == 0: flag = False break if flag: count += 1 #print(i) primenumber.append(i) end = (datetime.datetime.now() - start).total_seconds() print("count=",count,' ',"time=",end) 执行结果: count= 9592 time= 0.449377
# -*- coding:utf-8 -*- # version:python3.7 import datetime n = 100000 count = 3 primenumbers = [3,5] start = datetime.datetime.now() x = 7 step = 4 while x < n: flag = False j = int(x ** 0.5) if x % 5 != 0: for i in primenumbers: if i > j: flag = True break if x % i == 0: flag = False break if flag: count += 1 primenumbers.append(x) x += step step = 4 if step == 2 else 2 end = (datetime.datetime.now() - start).total_seconds() print("count=",count,' ',"time=",end) 执行结果: count= 9592 time= 0.380034
6、质数的应用:
应用在密码学领域,都要使用大素数