python中的主要因素

我正在使用下面的代码寻找2500的素因子,但我的代码目前只打印2个,我不确定为什么会这样.

no = 2500
count = 0
# Finding factors of 2500
for i in range(1,no):
    if no%i == 0:
    # Now that the factors have been found, the prime factors will be determined  
        for x in range(1,no):
            if i%x==0: 
                count = count + 1
            """Checking to see if the factor of 2500, itself only has two factor implying it is prime"""  
                if count == 2:
                    print i

谢谢

解决方法:

使用sieve of eratosthenes首先生成素数列表：

 from math import sqrt
def sieve_of_eratosthenes(n):
    primes = range(3, n + 1, 2) # primes above 2 must be odd so start at three and increase by 2
    for base in xrange(len(primes)):
        if primes[base] is None:
           continue
        if primes[base] >= sqrt(n): # stop at sqrt of n
            break
        for i in xrange(base + (base + 1) * primes[base], len(primes), primes[base]):
            primes[i] = None
    primes.insert(0,2)
    sieve=filter(None, primes)
    return  sieve

def prime_factors(sieve,n):
    p_f = []
    for prime in sieve:
        while n % prime == 0:
            p_f.append(prime)
            n /= prime
    if n > 1:
        p_f.append(n)
    return p_f
sieve = sieve_of_eratosthenes(2500)
print prime_factors(sieve,2500)

标签：prime-factoring,python
来源： https://codeday.me/bug/20190928/1825203.html