Question

我是一名Python新手，所以如果答案显而易见并且我对一些简单的事情一无所知，我道歉。

我正在尝试计算素数因子的数量，以便打印出来：“它有X因素”，但我正在撞墙 - 我不确定如何或最好的方式计算先前生成的列表的数量。

我也对我的Prime Number决心的一部分感到困惑。它通常效果很好，但是当我输入15,25,35等数字时，它们都回来说，当它们不是时，数字是素数。它也不是5的倍数。

感谢所有帮助！

我的代码：

# Main Program

def main():
 choice = "y"
 while choice.lower() == "y":
      print ("Prime Number Checker")
      print ()
      is_prime()
      print()
      choice = input("Try Again (y/n)?: ")
 choice = "n"
 while choice.lower() == "n":
      break

# Prime number determination and Output

def is_prime():
      num = int(input("Please enter an integer between 1 and 5,000: "))
      if num > 1:
           for i in range (2, num):
                if (num % i) != 0:
                          print()
                          print ("The factors of your number are:")
                          factors (num)
                          print (num, "is a Prime Number")
                          break
                elif (num % i) == 0:
                          print ()
                          print ("The factors of your number are:")
                          factors (num)
                          print (num, "is NOT a Prime Number")
                          numfacs (num)
                          break
      else:
           print ()
           print ("Number not in range.")
           print ()

# Factor Determination and Output

def factors(num):
 for i in range (1, num+1):
      if (num % i == 0):
           print (i)

# Number of Facotors and Output

def numfacs(num):
 for i in range (1, num+1):
      if (num % i == 0):
           list = []
           print ("Length: ", len(list))
           break

if __name__ == "__main__":
main()

Answer 1

您的问题正在发生，因为如果您评估的数字（例如15）不能被for循环的第一个元素（在这种情况下为数字2）分割，那么它将被评估为素数。要解决此问题，您可以在循环运行时添加标记（如果数字为素数）。如果它停止为素数（如果你可以将它除以另一个数字），则该标志变为False。

贝娄是我所说的一个有效例子。如果你不理解某些事情，请说出来。

# Main Program

def main():
 choice = "y"
 while choice.lower() == "y":
      print ("Prime Number Checker")
      print ()
      is_prime()
      print()
      choice = input("Try Again (y/n)?: ")
 choice = "n"
 while choice.lower() == "n":
      break

# Prime number determination and Output

def is_prime():
    flag=False
    num = int(input("Please enter an integer between 1 and 5,000: "))
    if num > 1:
        for i in range (2, num):
            if (num % i) != 0:
                flag=True
            elif (num % i) == 0:
                flag=False
                print ()
                print ("The factors of your number are:")
                factors (num)
                print (num, "is NOT a Prime Number")
                numfacs (num)
                break
        if flag == True :
            print(num,i,num % i)
            print()
            print ("The factors of your number are:")
            factors (num)
            print (num, "is a Prime Number")
    else:
        print ()
        print ("Number not in range.")
        print ()

# Factor Determination and Output

def factors(num):
 for i in range (1, num+1):
      if (num % i == 0):
           print (i)

# Number of Facotors and Output

def numfacs(num):
 for i in range (1, num+1):
      if (num % i == 0):
           list = []
           print ("Length: ", len(list))
           break

main()

注意：此行不适用于3.0版本的Python版本：

choice = input("Try Again (y/n)?: ")

而是使用它：

choice = str(raw_input("Try Again (y/n)?: "))

Python - 计算素因子

1 个答案: