麻省理工学院公开课程代码，用于在 Python 中查找第 1000 个素数不起作用？

Question

I am just starting out on programming and my first personal assignment is to find the 1000th prime number. I think this code is correct, but it lists the first 1000 odd numbers instead. Any insight?

count = 1
num = 3
prime = [2]

while count <= 1000:
    for x in range(2, num + 1):
        if num % x == 0 and num == x:
            prime.append(num)
            num += 2
    count += 1

print(prime[1000])

Answer 1

I just wrote this one. It will ask you how many prime number user wants to see, in this case it will be 1000. Feel free to use it :).

# p is the sequence number of prime series
# n is the sequence of natural numbers to be tested if prime or not
# i is the sequence of natural numbers which will be used to devide n for testing
# L is the sequence limit of prime series user wants to see
p=2;n=3
L=int(input('Enter the how many prime numbers you want to see: '))
print ('# 1  prime is 2')
while(p<=L):
    i=2
    while i<n:
        if n%i==0:break
        i+=1
    else:print('#',p,' prime is',n); p+=1
    n+=1 #Line X

#when it breaks it doesn't execute the else and goes to the line 'X' 

Answer 2

import math

def is_prime(a):
    for i in xrange(2, int(math.sqrt(a)+1)):
        if a%i == 0:
            return False
    return True

primes=[]
count = 2
while len(primes) < 1000:
    if is_prime(count):
        primes.append(count)
    count += 1

print(primes)`