ปลั๊กอินสำหรับไพธอนบอกให้ท่านเตรียมปัจจัย
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อันดับหนึ่ง factorization เป็นภาษาไพธอน
import math
def primeFactors(n):
# no of even divisibility
while n % 2 == 0:
print(2)
n = n / 2
# n reduces to become odd
for i in range(3, int(math.sqrt(n)) + 1, 2):
# while i divides n
while n % i == 0:
print(i)
n = n / i
# if n is a prime
if n > 2:
print(n)
primeFactors(256)ปลั๊กอินสำหรับไพธอนบอกให้ท่านเตรียมปัจจัย
# There is no quick way to calculate the prime factors of a number.
# In fact, prime factorization is so famously hard that it's what puts the "asymmetric" in asymmetric RSA encryption.
# That being said, it can be sped up a little bit by using divisibility rules, like checking if the sum of the digits is divisible by 3.
def factors(num):
ps = [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] # Primes from https://primes.utm.edu/lists/small/10000.txt. Primes can also be generated by iterating through numbers and checking for factors, or by using a probabilistic test like Rabin-Miller.
pdict = {}
for p in ps:
if p <= num:
while (num / p).is_integer():
if str(p) in pdict:
pdict[str(p)] += 1
else:
pdict[str(p)] = 1
num /= p
if num == 1: break
return pdict
# Returns a dictionary in the form {"base": "exponent"}หน้าคล้ายกัน
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