程序有效但没有产生正确的输出？

Question

我制作了一个程序，可以找到低于 n 数的素数。

这是代码：-

import time

n = 10000

start = time.time()

primes = [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]

if n > 1000000:

    for j in range(1,n+1):

        for i in range(0,len(primes)):

            if j % primes[i] == 0:

                break

        else:

            primes[:25].append(j)

    print primes

else:

    for j in range(1,n+1):

        for i in range(0,len(primes[:25])):

            if j % primes[i] == 0:

                break

        else:

            primes[:25].append(j)

    print primes

end = time.time() - start

print end

我知道该代码的工作原理是找到前 1000 个所需的时间，而 100 万个素数则大不相同。

但它只打印最多 1000 的素数，即它不打印附加列表。我哪里错了？

Answer 1

复制和更改您的代码来测试它似乎是一个坏主意，因为这两个版本会分开。相反，修改初始条件并运行相同的代码：import timestart = time.time()n = 10000000primes = [ \&nbsp; &nbsp; 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, \&nbsp; &nbsp; 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, \&nbsp; &nbsp; 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, \&nbsp; &nbsp; 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283, 293, 307, 311, \&nbsp; &nbsp; 313, 317, 331, 337, 347, 349, 353, 359, 367, 373, 379, 383, 389, 397, 401, \&nbsp; &nbsp; 409, 419, 421, 431, 433, 439, 443, 449, 457, 461, 463, 467, 479, 487, 491, \&nbsp; &nbsp; 499, 503, 509, 521, 523, 541, 547, 557, 563, 569, 571, 577, 587, 593, 599, \&nbsp; &nbsp; 601, 607, 613, 617, 619, 631, 641, 643, 647, 653, 659, 661, 673, 677, 683, \&nbsp; &nbsp; 691, 701, 709, 719, 727, 733, 739, 743, 751, 757, 761, 769, 773, 787, 797, \&nbsp; &nbsp; 809, 811, 821, 823, 827, 829, 839, 853, 857, 859, 863, 877, 881, 883, 887, \&nbsp; &nbsp; 907, 911, 919, 929, 937, 941, 947, 953, 967, 971, 977, 983, 991, 997, \]# for testing:n = 10000primes = primes[:25]for j in range(3, n + 1, 2):&nbsp; &nbsp; for prime in primes:&nbsp; &nbsp; &nbsp; &nbsp; if j % prime == 0:&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; break&nbsp; # skip composites and primes in list&nbsp; &nbsp; else:&nbsp; # no break&nbsp; &nbsp; &nbsp; &nbsp; primes.append(j)print primesend = time.time() - startprint end即使这不仅仅是为了测试，您也可以将其转换为条件：if n <= 1000000:&nbsp; &nbsp; &nbsp;primes = primes[:25]并且仍然避免代码重复。我只想在最短的时间内找到质数如果是这种情况，难道您不应该只检查列表中的素数直到 的平方根j而不再检查吗？当然，您还需要更改主要列表管理逻辑。