打印范圍內的素數

Question

k=int(input())
res=[2]
for i in range(2,k+1):
    if i%2==0:
       continue 
    else:
      for j in range(2,i):
          if i%j==0 or j%2==0 :
              break
      else:
             res.append(i)
print(res)

此代碼用於在給定的數字范圍內查找素數。 我試圖運行代碼，但列表只有第 2 個。誰能告訴我發生了什么？ 如果我刪除j%2==0它正在工作。 我只想知道我的錯誤。

Answer 1

您應該使用當前的結果來加速您的過程。 您只需要測試素數的整除性。 但是你正在建立一個質數列表。 所以使用它！

k=int(input())
primes=[]
for i in range(2,k+1):
    if all(i%p!=0 for p in primes):
        primes.append(i)

您還可以通過僅選擇不如其他人建議的 sqrt(i) 的素數元素來改進。

import math
k=int(input())
primes=[]
for i in range(2,k+1):
    j=math.sqrt(i)
    if all(i%p!=0 for p in primes if p<=j):
        primes.append(i)

Answer 2

在您的內部循環j變量從值 2 開始，然后您有一個 if 語句始終為True因為j%2==0始終為2%2==0始終為True ，因此您總是從第一步break內部for循環迭代

您可以使用：

import math 

k=int(input())
res=[]
for i in range(2, k+1):
    for x in range(2, int(math.sqrt(i) + 1)):
        if i % x == 0 :
            break
    else:
        res.append(i)
# k = 20

輸出：

[2, 3, 5, 7, 11, 13, 17, 19]

為了有效地生成素數，您可以使用埃拉托色尼篩法：

# 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

k=int(input())

def gen_primes(k):
    gp = _gen_primes()

    p = next(gp)
    while p < k:
        yield p
        p = next(gp)

res = list(gen_primes(k))

Answer 3

您的代碼有一個問題，在內部循環中 or 條件不正確，正如@kederrac 所強調的那樣。 您不需要 j%2==0 因為 j 總是從 2 開始，而i%j==0已經涵蓋了條件

k=int(input())
res=[2]
for i in range(2,k+1):
    if i%2==0:
       continue 
    else:
      for j in range(2,i):
          if i%j==0 :
              break
      else:
             res.append(i)
print(res)