Python 中的簡單素數生成器

Question

有人可以告訴我這段代碼我做錯了什么嗎？ 無論如何，它只是打印“計數”。 我只想要一個非常簡單的素數生成器（沒什么花哨的）。

import math

def main():
    count = 3
    one = 1
    while one == 1:
        for x in range(2, int(math.sqrt(count) + 1)):
            if count % x == 0: 
                continue
            if count % x != 0:
                print count

        count += 1

Answer 1

有一些問題：

• 為什么在沒有除以 x 時打印出計數？ 這並不意味着它是素數，它只是意味着這個特定的 x 不會整除它
• continue移動到下一個循環迭代 - 但你真的想使用break停止它

這是你的代碼，有一些修復，它只打印出素數：

import math

def main():
    count = 3
    
    while True:
        isprime = True
        
        for x in range(2, int(math.sqrt(count) + 1)):
            if count % x == 0: 
                isprime = False
                break
        
        if isprime:
            print count
        
        count += 1

對於更有效的素數生成，請參閱埃拉托色尼篩，正如其他人所建議的那樣。 這是一個很好的優化實現，有很多評論：

# Sieve of Eratosthenes
# Code by David Eppstein, UC Irvine, 28 Feb 2002
# http://code.activestate.com/recipes/117119/

def gen_primes():
    """ Generate an infinite sequence of prime numbers.
    """
    # Maps composites to primes witnessing their compositeness.
    # This is memory efficient, as the sieve is not "run forward"
    # indefinitely, but only as long as required by the current
    # number being tested.
    #
    D = {}
    
    # The running integer that's checked for primeness
    q = 2
    
    while True:
        if q not in D:
            # q is a new prime.
            # Yield it and mark its first multiple that isn't
            # already marked in previous iterations
            # 
            yield q
            D[q * q] = [q]
        else:
            # q is composite. D[q] is the list of primes that
            # divide it. Since we've reached q, we no longer
            # need it in the map, but we'll mark the next 
            # multiples of its witnesses to prepare for larger
            # numbers
            # 
            for p in D[q]:
                D.setdefault(p + q, []).append(p)
            del D[q]
        
        q += 1

請注意，它返回一個生成器。

Answer 2

def is_prime(num):
    """Returns True if the number is prime
    else False."""
    if num == 0 or num == 1:
        return False
    for x in range(2, num):
        if num % x == 0:
            return False
    else:
        return True

>> filter(is_prime, range(1, 20))
  [2, 3, 5, 7, 11, 13, 17, 19]

我們將在一個列表中得到所有不超過 20 的質數。 我本可以使用埃拉托色尼篩法，但你說你想要一些非常簡單的東西。 ;)

Answer 3

re 很強大：

import re


def isprime(n):
    return re.compile(r'^1?$|^(11+)\1+$').match('1' * n) is None

print [x for x in range(100) if isprime(x)]

###########Output#############
[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]

Answer 4

def primes(n): # simple sieve of multiples 
   odds = range(3, n+1, 2)
   sieve = set(sum([list(range(q*q, n+1, q+q)) for q in odds], []))
   return [2] + [p for p in odds if p not in sieve]

>>> primes(50)
[2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47]

測試一個數是否為素數：

>>> 541 in primes(541)
True
>>> 543 in primes(543)
False

Answer 5

print [x for x in range(2,100) if not [t for t in range(2,x) if not x%t]]

Answer 6

這是一個簡單的（Python 2.6.2）解決方案......它符合 OP 的原始請求（現在六個月大）； 並且應該是任何“編程 101”課程中完全可以接受的解決方案......因此這篇文章。

import math

def isPrime(n):
    for i in range(2, int(math.sqrt(n)+1)):
        if n % i == 0: 
            return False;
    return n>1;

print 2
for n in range(3, 50):
    if isPrime(n):
        print n

這種簡單的“蠻力”方法對於現代 PC 上大約 16,000 的數字“足夠快”（在我的 2GHz 機器上大約需要 8 秒）。

顯然，這可以更有效地完成，通過不重新計算每個偶數的素數，或每個數字的 3、5、7 等的每個倍數......參見Eratosthenes 篩（參見上面 eliben 的實現），如果您感覺特別勇敢和/或瘋狂，甚至可以使用阿特金篩。

警告 Emptor：我是一個蟒蛇菜鳥。 請不要把我說的任何話當成福音。

Answer 7

SymPy是一個用於符號數學的 Python 庫。 它提供了幾個函數來生成素數。

isprime(n)              # Test if n is a prime number (True) or not (False).

primerange(a, b)        # Generate a list of all prime numbers in the range [a, b).
randprime(a, b)         # Return a random prime number in the range [a, b).
primepi(n)              # Return the number of prime numbers less than or equal to n.

prime(nth)              # Return the nth prime, with the primes indexed as prime(1) = 2. The nth prime is approximately n*log(n) and can never be larger than 2**n.
prevprime(n, ith=1)     # Return the largest prime smaller than n
nextprime(n)            # Return the ith prime greater than n

sieve.primerange(a, b)  # Generate all prime numbers in the range [a, b), implemented as a dynamically growing sieve of Eratosthenes. 

這里有些例子。

>>> import sympy
>>> 
>>> sympy.isprime(5)
True
>>> list(sympy.primerange(0, 100))
[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]
>>> sympy.randprime(0, 100)
83
>>> sympy.randprime(0, 100)
41
>>> sympy.prime(3)
5
>>> sympy.prevprime(50)
47
>>> sympy.nextprime(50)
53
>>> list(sympy.sieve.primerange(0, 100))
[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]

Answer 8

這是一個 numpy 版本的 Eratosthenes 篩法，具有良好的復雜性（低於對長度為 n 的數組進行排序）和矢量化。

import numpy as np 
def generate_primes(n):
    is_prime = np.ones(n+1,dtype=bool)
    is_prime[0:2] = False
    for i in range(int(n**0.5)+1):
        if is_prime[i]:
            is_prime[i*2::i]=False
    return np.where(is_prime)[0]

時間：

import time    
for i in range(2,10):
    timer =time.time()
    generate_primes(10**i)
    print('n = 10^',i,' time =', round(time.time()-timer,6))

>> n = 10^ 2  time = 5.6e-05
>> n = 10^ 3  time = 6.4e-05
>> n = 10^ 4  time = 0.000114
>> n = 10^ 5  time = 0.000593
>> n = 10^ 6  time = 0.00467
>> n = 10^ 7  time = 0.177758
>> n = 10^ 8  time = 1.701312
>> n = 10^ 9  time = 19.322478

Answer 9

python 3（生成素數）

import math

i = 2
while True:
    for x in range(2, int(math.sqrt(i) + 1)):
        if i%x==0:
            break
    else:
        print(i)
    i += 1

Answer 10

在我看來，最好采用函數式方法，

所以我首先創建一個函數來確定數字是否為素數，然后根據需要在循環或其他地方使用它。

def isprime(n):
      for x in range(2,n):
        if n%x == 0:
            return False
    return True

然后運行一個簡單的列表理解或生成器表達式來獲取素數列表，

[x for x in range(1,100) if isprime(x)]

Answer 11

另一個簡單的例子，只考慮奇數的簡單優化。 一切都用惰性流（python 生成器）完成。

用法：primes = list(create_prime_iterator(1, 30))

import math
import itertools

def create_prime_iterator(rfrom, rto):
    """Create iterator of prime numbers in range [rfrom, rto]"""
    prefix = [2] if rfrom < 3 and rto > 1 else [] # include 2 if it is in range separately as it is a "weird" case of even prime
    odd_rfrom = 3 if rfrom < 3 else make_odd(rfrom) # make rfrom an odd number so that  we can skip all even nubers when searching for primes, also skip 1 as a non prime odd number.
    odd_numbers = (num for num in xrange(odd_rfrom, rto + 1, 2))
    prime_generator = (num for num in odd_numbers if not has_odd_divisor(num))
    return itertools.chain(prefix, prime_generator)

def has_odd_divisor(num):
    """Test whether number is evenly divisable by odd divisor."""
    maxDivisor = int(math.sqrt(num))
    for divisor in xrange(3, maxDivisor + 1, 2):
        if num % divisor == 0:
            return True
    return False

def make_odd(number):
    """Make number odd by adding one to it if it was even, otherwise return it unchanged"""
    return number | 1

Answer 12

這是我的實現。 我確定有一種更有效的方法，但似乎有效。 基本標志使用。

def genPrime():
    num = 1
    prime = False
    while True:
        # Loop through all numbers up to num
        for i in range(2, num+1):
            # Check if num has remainder after the modulo of any previous numbers
            if num % i == 0:
                prime = False
                # Num is only prime if no remainder and i is num
                if i == num:
                    prime = True
                break

        if prime:
            yield num
            num += 1
        else:
            num += 1

prime = genPrime()
for _ in range(100):
    print(next(prime))

Answer 13

剛剛研究了這個主題，在線程中查找示例並嘗試制作我的版本：

from collections import defaultdict
# from pprint import pprint

import re


def gen_primes(limit=None):
    """Sieve of Eratosthenes"""
    not_prime = defaultdict(list)
    num = 2
    while limit is None or num <= limit:
        if num in not_prime:
            for prime in not_prime[num]:
                not_prime[prime + num].append(prime)
            del not_prime[num]
        else:  # Prime number
            yield num
            not_prime[num * num] = [num]
        # It's amazing to debug it this way:
        # pprint([num, dict(not_prime)], width=1)
        # input()
        num += 1


def is_prime(num):
    """Check if number is prime based on Sieve of Eratosthenes"""
    return num > 1 and list(gen_primes(limit=num)).pop() == num


def oneliner_is_prime(num):
    """Simple check if number is prime"""
    return num > 1 and not any([num % x == 0 for x in range(2, num)])


def regex_is_prime(num):
    return re.compile(r'^1?$|^(11+)\1+$').match('1' * num) is None


def simple_is_prime(num):
    """Simple check if number is prime
    More efficient than oneliner_is_prime as it breaks the loop
    """
    for x in range(2, num):
        if num % x == 0:
            return False
    return num > 1


def simple_gen_primes(limit=None):
    """Prime number generator based on simple gen"""
    num = 2
    while limit is None or num <= limit:
        if simple_is_prime(num):
            yield num
        num += 1


if __name__ == "__main__":
    less1000primes = list(gen_primes(limit=1000))
    assert less1000primes == list(simple_gen_primes(limit=1000))
    for num in range(1000):
        assert (
            (num in less1000primes)
            == is_prime(num)
            == oneliner_is_prime(num)
            == regex_is_prime(num)
            == simple_is_prime(num)
        )
    print("Primes less than 1000:")
    print(less1000primes)

    from timeit import timeit

    print("\nTimeit:")
    print(
        "gen_primes:",
        timeit(
            "list(gen_primes(limit=1000))",
            setup="from __main__ import gen_primes",
            number=1000,
        ),
    )
    print(
        "simple_gen_primes:",
        timeit(
            "list(simple_gen_primes(limit=1000))",
            setup="from __main__ import simple_gen_primes",
            number=1000,
        ),
    )
    print(
        "is_prime:",
        timeit(
            "[is_prime(num) for num in range(2, 1000)]",
            setup="from __main__ import is_prime",
            number=100,
        ),
    )
    print(
        "oneliner_is_prime:",
        timeit(
            "[oneliner_is_prime(num) for num in range(2, 1000)]",
            setup="from __main__ import oneliner_is_prime",
            number=100,
        ),
    )
    print(
        "regex_is_prime:",
        timeit(
            "[regex_is_prime(num) for num in range(2, 1000)]",
            setup="from __main__ import regex_is_prime",
            number=100,
        ),
    )
    print(
        "simple_is_prime:",
        timeit(
            "[simple_is_prime(num) for num in range(2, 1000)]",
            setup="from __main__ import simple_is_prime",
            number=100,
        ),
    )

運行此代碼的結果顯示了有趣的結果：

$ python prime_time.py
Primes less than 1000:
[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, 307, 311, 313, 317, 331, 337, 347, 349, 353, 359, 367, 373, 379, 383, 389, 397, 401, 409, 419, 421, 431, 433, 439, 443, 449, 457, 461, 463, 467, 479, 487, 491, 499, 503, 509, 521, 523, 541, 547, 557, 563, 569, 571, 577, 587, 593, 599, 601, 607, 613, 617, 619, 631, 641, 643, 647, 653, 659, 661, 673, 677, 683, 691, 701, 709, 719, 727, 733, 739, 743, 751, 757, 761, 769, 773, 787, 797, 809, 811, 821, 823, 827, 829, 839, 853, 857, 859, 863, 877, 881, 883, 887, 907, 911, 919, 929, 937, 941, 947, 953, 967, 971, 977, 983, 991, 997]

Timeit:
gen_primes: 0.6738066330144648
simple_gen_primes: 4.738092333020177
is_prime: 31.83770858097705
oneliner_is_prime: 3.3708438930043485
regex_is_prime: 8.692703998007346
simple_is_prime: 0.4686249239894096

所以我可以看到我們在這里對不同的問題有正確的答案； 對於素數生成器gen_primes看起來是正確的答案； 但是對於質數檢查， simple_is_prime函數更適合。

這有效，但我總是is_prime接受更好的方法來制作is_prime函數。

Answer 14

這是我所擁有的：

def is_prime(num):
    if num < 2:         return False
    elif num < 4:       return True
    elif not num % 2:   return False
    elif num < 9:       return True
    elif not num % 3:   return False
    else:
        for n in range(5, int(math.sqrt(num) + 1), 6):
            if not num % n:
                return False
            elif not num % (n + 2):
                return False

    return True

對於大數，它非常快，因為它只檢查已經是素數的數的除數。

現在，如果要生成素數列表，可以執行以下操作：

# primes up to 'max'
def primes_max(max):
    yield 2
    for n in range(3, max, 2):
        if is_prime(n):
            yield n

# the first 'count' primes
def primes_count(count):
    counter = 0
    num = 3

    yield 2

    while counter < count:
        if is_prime(num):
            yield num
            counter += 1
        num += 2

為了提高效率，可能需要在這里使用生成器。

僅供參考，而不是說：

one = 1
while one == 1:
    # do stuff

你可以簡單地說：

while 1:
    #do stuff

Answer 15

類似於 user107745，但使用“全部”而不是雙重否定（可讀性更強，但我認為性能相同）：

import math
[x for x in xrange(2,10000) if all(x%t for t in xrange(2,int(math.sqrt(x))+1))]

基本上它迭代 (2, 100) 范圍內的 x 並只選擇那些沒有 mod == 0 的范圍內的所有 t (2,x)

另一種方法可能只是在我們進行時填充素數：

primes = set()
def isPrime(x):
  if x in primes:
    return x
  for i in primes:
    if not x % i:
      return None
  else:
    primes.add(x)
    return x

filter(isPrime, range(2,10000))

Answer 16

import time

maxnum=input("You want the prime number of 1 through....")

n=2
prime=[]
start=time.time()

while n<=maxnum:

    d=2.0
    pr=True
    cntr=0

    while d<n**.5:

        if n%d==0:
            pr=False
        else:
            break
        d=d+1

    if cntr==0:

        prime.append(n)
        #print n

    n=n+1

print "Total time:",time.time()-start

Answer 17

如果你想找到一個范圍內的所有質數，你可以這樣做：

def is_prime(num):
"""Returns True if the number is prime
else False."""
if num == 0 or num == 1:
    return False
for x in range(2, num):
    if num % x == 0:
        return False
else:
    return True
num = 0
itr = 0
tot = ''
while itr <= 100:
    itr = itr + 1
    num = num + 1
    if is_prime(num) == True:
        print(num)
        tot = tot + ' ' + str(num)
print(tot)

只需添加while its <=和您的范圍編號。
輸出：
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

Answer 18

這似乎是家庭作業，所以我會給出一個提示而不是詳細的解釋。 如果我假設錯誤，請糾正我。

當你看到一個偶數除數時，你做得很好。

但是，只要您看到一個不能整除的數字，就會立即打印“計數”。 例如，2 不能整除為 9。但這並不能使 9 成為質數。 您可能希望繼續下去，直到您確定范圍內沒有數字匹配為止。

（正如其他人所回答的那樣，Sieve 是一種更有效的方法……只是想幫助您了解為什么此特定代碼沒有按照您的意願行事）

Answer 19

您需要確保所有可能的除數不會平均划分您正在檢查的數字。 在這種情況下，只要其中一個可能的除數不能均勻地整除該數字，您就可以隨時打印正在檢查的數字。

此外，您不想使用 continue 語句，因為當您已經發現該數字不是素數時， continue 只會導致它檢查下一個可能的除數。

Answer 20

使用發電機：

def primes(num):
    if 2 <= num:
        yield 2
    for i in range(3, num + 1, 2):
        if all(i % x != 0 for x in range(3, int(math.sqrt(i) + 1))):
            yield i

用法：

for i in primes(10):
    print(i)

2、3、5、7

Answer 21

您可以以相當優雅的方式使用列表推導式創建素數列表。 取自這里：

>>> noprimes = [j for i in range(2, 8) for j in range(i*2, 50, i)]
>>> primes = [x for x in range(2, 50) if x not in noprimes]
>>> print primes
>>> [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47]

Answer 22

這是我見過的找到 n 以內的素數的最快方法。 在 1.2 秒內達到 100m。 它使用純 python 沒有依賴

def primes(lim):
  if lim<7:
    if lim<2: return []
    if lim<3: return [2]
    if lim<5: return [2, 3]
    return [2, 3, 5]
  n = (lim-1)//30
  m = n+1
  BA = bytearray
  prime1 = BA([1])*m
  prime7 = BA([1])*m
  prime11 = BA([1])*m
  prime13 = BA([1])*m
  prime17 = BA([1])*m
  prime19 = BA([1])*m
  prime23 = BA([1])*m
  prime29 = BA([1])*m
  prime1[0] = 0
  i = 0
  try:
    while 1:
      if prime1[i]:
        p = 30*i+1
        l = i*(p+1)
        prime1[l::p] = BA(1+(n-l)//p)
        l += i*6
        prime7[l::p] = BA(1+(n-l)//p)
        l += i*4
        prime11[l::p] = BA(1+(n-l)//p)
        l += i*2
        prime13[l::p] = BA(1+(n-l)//p)
        l += i*4
        prime17[l::p] = BA(1+(n-l)//p)
        l += i*2
        prime19[l::p] = BA(1+(n-l)//p)
        l += i*4
        prime23[l::p] = BA(1+(n-l)//p)
        l += i*6
        prime29[l::p] = BA(1+(n-l)//p)
      if prime7[i]:
        p = 30*i+7
        l = i*(p+7)+1
        prime19[l::p] = BA(1+(n-l)//p)
        l += i*4+1
        prime17[l::p] = BA(1+(n-l)//p)
        l += i*2+1
        prime1[l::p] = BA(1+(n-l)//p)
        l += i*4
        prime29[l::p] = BA(1+(n-l)//p)
        l += i*2+1
        prime13[l::p] = BA(1+(n-l)//p)
        l += i*4+1
        prime11[l::p] = BA(1+(n-l)//p)
        l += i*6+1
        prime23[l::p] = BA(1+(n-l)//p)
        l += i*2+1
        prime7[l::p] = BA(1+(n-l)//p)
      if prime11[i]:
        p = 30*i+11
        l = i*(p+11)+4
        prime1[l::p] = BA(1+(n-l)//p)
        l += i*2
        prime23[l::p] = BA(1+(n-l)//p)
        l += i*4+2
        prime7[l::p] = BA(1+(n-l)//p)
        l += i*2
        prime29[l::p] = BA(1+(n-l)//p)
        l += i*4+2
        prime13[l::p] = BA(1+(n-l)//p)
        l += i*6+2
        prime19[l::p] = BA(1+(n-l)//p)
        l += i*2+1
        prime11[l::p] = BA(1+(n-l)//p)
        l += i*6+2
        prime17[l::p] = BA(1+(n-l)//p)
      if prime13[i]:
        p = 30*i+13
        l = i*(p+13)+5
        prime19[l::p] = BA(1+(n-l)//p)
        l += i*4+2
        prime11[l::p] = BA(1+(n-l)//p)
        l += i*2+1
        prime7[l::p] = BA(1+(n-l)//p)
        l += i*4+1
        prime29[l::p] = BA(1+(n-l)//p)
        l += i*6+3
        prime17[l::p] = BA(1+(n-l)//p)
        l += i*2+1
        prime13[l::p] = BA(1+(n-l)//p)
        l += i*6+3
        prime1[l::p] = BA(1+(n-l)//p)
        l += i*4+1
        prime23[l::p] = BA(1+(n-l)//p)
      if prime17[i]:
        p = 30*i+17
        l = i*(p+17)+9
        prime19[l::p] = BA(1+(n-l)//p)
        l += i*2+1
        prime23[l::p] = BA(1+(n-l)//p)
        l += i*4+3
        prime1[l::p] = BA(1+(n-l)//p)
        l += i*6+3
        prime13[l::p] = BA(1+(n-l)//p)
        l += i*2+1
        prime17[l::p] = BA(1+(n-l)//p)
        l += i*6+3
        prime29[l::p] = BA(1+(n-l)//p)
        l += i*4+3
        prime7[l::p] = BA(1+(n-l)//p)
        l += i*2+1
        prime11[l::p] = BA(1+(n-l)//p)
      if prime19[i]:
        p = 30*i+19
        l = i*(p+19)+12
        prime1[l::p] = BA(1+(n-l)//p)
        l += i*4+2
        prime17[l::p] = BA(1+(n-l)//p)
        l += i*6+4
        prime11[l::p] = BA(1+(n-l)//p)
        l += i*2+1
        prime19[l::p] = BA(1+(n-l)//p)
        l += i*6+4
        prime13[l::p] = BA(1+(n-l)//p)
        l += i*4+2
        prime29[l::p] = BA(1+(n-l)//p)
        l += i*2+2
        prime7[l::p] = BA(1+(n-l)//p)
        l += i*4+2
        prime23[l::p] = BA(1+(n-l)//p)
      if prime23[i]:
        p = 30*i+23
        l = i*(p+23)+17
        prime19[l::p] = BA(1+(n-l)//p)
        l += i*6+5
        prime7[l::p] = BA(1+(n-l)//p)
        l += i*2+1
        prime23[l::p] = BA(1+(n-l)//p)
        l += i*6+5
        prime11[l::p] = BA(1+(n-l)//p)
        l += i*4+3
        prime13[l::p] = BA(1+(n-l)//p)
        l += i*2+1
        prime29[l::p] = BA(1+(n-l)//p)
        l += i*4+4
        prime1[l::p] = BA(1+(n-l)//p)
        l += i*2+1
        prime17[l::p] = BA(1+(n-l)//p)
      if prime29[i]:
        p = 30*i+29
        l = i*(p+29)+28
        prime1[l::p] = BA(1+(n-l)//p)
        l += i*2+1
        prime29[l::p] = BA(1+(n-l)//p)
        l += i*6+6
        prime23[l::p] = BA(1+(n-l)//p)
        l += i*4+4
        prime19[l::p] = BA(1+(n-l)//p)
        l += i*2+2
        prime17[l::p] = BA(1+(n-l)//p)
        l += i*4+4
        prime13[l::p] = BA(1+(n-l)//p)
        l += i*2+2
        prime11[l::p] = BA(1+(n-l)//p)
        l += i*4+4
        prime7[l::p] = BA(1+(n-l)//p)
      i+=1
  except:
    pass
  RES = [2, 3, 5]
  A = RES.append
  ti=0
  try:
    for i in range(n):
      if prime1[i]:
        A(ti+1)
      if prime7[i]:
        A(ti+7)
      if prime11[i]:
        A(ti+11)
      if prime13[i]:
        A(ti+13)
      if prime17[i]:
        A(ti+17)
      if prime19[i]:
        A(ti+19)
      if prime23[i]:
        A(ti+23)
      if prime29[i]:
        A(ti+29)
      ti+=30
  except:
    pass
  if prime1[n] and (30*n+1)<=lim:
    A(30*n+1)
  if prime7[n] and (30*n+7)<=lim:
    A(30*n+7)
  if prime11[n] and (30*n+11)<=lim:
    A(30*n+11)
  if prime13[n] and (30*n+13)<=lim:
    A(30*n+13)
  if prime17[n] and (30*n+17)<=lim:
    A(30*n+17)
  if prime19[n] and (30*n+19)<=lim:
    A(30*n+19)
  if prime23[n] and (30*n+23)<=lim:
    A(30*n+23)
  if prime29[n] and (30*n+29)<=lim:
    A(30*n+29)
  return RES

from time import time
n = time()
print (primes(100000000)[-1])
print (time() - n)

Answer 23

如果你想直接計算素數怎么辦：

def oprime(n):
counter = 0
b = 1
if n == 1:
    print 2
while counter < n-1:
    b = b + 2
    for a in range(2,b):
        if b % a == 0:
            break
    else:
        counter = counter + 1
        if counter == n-1:
            print b

Answer 24

def genPrimes():
    primes = []   # primes generated so far
    last = 1      # last number tried
    while True:
        last += 1
        for p in primes:
            if last % p == 0:
                break
        else:
            primes.append(last)
            yield last

Answer 25

對我來說，以下解決方案看起來簡單且易於遵循。

import math

def is_prime(num):

    if num < 2:
        return False

    for i in range(2, int(math.sqrt(num) + 1)):
        if num % i == 0:
            return False

return True

Answer 26

def check_prime(x):
    if (x < 2): 
       return 0
    elif (x == 2): 
       return 1
    t = range(x)
    for i in t[2:]:
       if (x % i == 0):
            return 0
    return 1

Answer 27

• continue 語句看起來是錯誤的。

• 你想從 2 開始，因為 2 是第一個質數。

• 您可以編寫“while True:”以獲得無限循環。