知道數字是否為質數的優化方法？

Question

def print_prime_factors(number):
  # Start with two, which is the first prime
  factor = 2
  # Keep going until the factor is larger than the number
  while factor <= number:
    # Check if factor is a divisor of number
    if number % factor == 0:
      # If it is, print it and divide the original number
      print(factor)
      number = number / factor
    else:
      # If it's not, increment the factor by one
      factor += 1

      
  return "Done"

print_prime_factors(100)
# Should print 2,2,5,5
# DO NOT DELETE THIS COMMENT

我正在使用這種方式來檢查數字是否為質數。 還有其他快速方法嗎？

Answer 1

你問： Is there any other fast way? .

一種方法是大約。 通過消除所有偶數因子（ 2除外），將 function 的運行時間減半：

def print_prime_factors_odd(number):
  # Start with two, which is the first prime
  factor = 2
  addend = 1
  # Keep going until the factor is larger than the number
  while factor <= number:
    # Check if factor is a divisor of number
    if number % factor == 0:
      # If it is, print it and divide the original number
      print(factor)
      number = number / factor
    else:
      # If it's not, increment the factor by the addend
      factor += addend
      addend = 2    # Once factor becomes 3, then use odd factors after that

      
  print("Done")

Answer 2

標志 = 假

if num == 1: print(num, "is not a prime number") elif num > 1: # 檢查因數 for i in range(2, num): if (num % i) == 0: # if factor找到，設置flag為True flag = True # break out of loop break

# check if flag is True
if flag:
    print(num, "is not a prime number")
else:
    print(num, "is a prime number")