打印范围内的素数

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)

This code is for finding prime numbers in a given range of numbers. I tried to run the code but the list is having only number 2. Can anyone tell me what is happening? If I remove j%2==0 it's working. I just want to know my mistake.

Answer 1

You should use your current result to accelerate your process. You only need to test divisibility by primes. But you are building a list of primes. So use it !

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

You can also improve by selecting only prime elements which are inferior to sqrt(i) like others suggested.

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

in your inner loop j variable starts from value 2 and then you have an if statement that is always True because j%2==0 is always 2%2==0 which is always True , so you always break from the first step of inner for loop iteration

you can use:

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

output:

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

for efficient prime generation, you can use the Sieve of Eratosthenes:

# 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

Your code had one issue, in the inner loop the or condition is incorrect, as highlighted by @kederrac. You don't need the j%2==0 as j always start from 2 and i%j==0 already covers the condition

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)