Python质数计算器

Question

prime = [2]
while len(prime) <= 1000:
    i=3
    a = 0
    for number in prime:
        testlist= []
        testlist.append(i%number)
    if 0 in testlist:
        i=i+1
    else:
        prime.append(i)
        i=i+1
print(prime[999])

[尝试制作一个计算在线课程素数的程序。该程序永远不会结束，但是我看不到代码中的无限循环。

素数是只能被其自身除以一个的数。

我的逻辑是，如果数字可以除以前面的质数，那么它就不是质数。

Answer 1

正如您对问题的评论所指出的，您的代码中存在多个错误。

这是您的代码正常工作的版本。

prime = [2]
i = 3
while len(prime) <= 1000:
    testlist = []
    for number in prime:
        testlist.append(i % number)
    if 0 not in testlist:
        prime.append(i)
    i = i + 1
print prime

Answer 2

我尚未测试，但是您可以创建如下所示的方法：

def get_prime_no_upto(number):
  start = 2
  primes = list(range(start,number)).to_a
  for no in range(start,number):
    for num in range(start,no):
      if ( no % num  == 0) and (num != no):
        primes.delete(no)
        break
  primes

并且可以像使用它

print primeno(100)

欢呼！

Answer 3

我编写了可以帮助您的代码。

import math
import time
start=time.perf_counter()#starting timer
try:
    x=int(input('Please enter a number:'))#asking for input
    print('')#spacing 
    b=x % 2#checking divisibility
    l=x % 3
    f=x % 5
    c=x % 7
    if x==2 or x==3 or x==5 or x==7:
        message='The number %s is prime!'#telling if prime
        print(message % x)
    elif f==0 or c==0:
        message='The number %s is not prime'
        print(message % x)
    elif x==0:
      print('0 is not prime by definition!')
    elif x==1:
        print('1 is not prime by definition!')
    elif x<2:
        print('please enter a positive number greater than one')#error statement
    elif b==0 or l==0:
        message='The number %s is not prime'
        print(message % x)
    else:
        message='The number %s is prime!'
        print(message % x)
except Exception:
        print('please enter a valid number')#error statement 
finally:
    print(' ')
    print('Prime numbers are positive non-zero numbers that have exactly two factors')
    end=time.perf_counter()#ending timer
    final_time=end-start
    final_time=math.floor(final_time)#rounding
    if final_time==1:
        message='The time taken is about %s second'#if statement deciding what it will show
        print('')
        print(message % final_time)
    else:
        message='The time taken is about %s seconds'
        print('')
        print(message % final_time) 

谢谢，Anikait Sota（中环）