Question

如何根据乘法函数进行阶乘？多数民众赞成我所拥有的和我遇到的问题是它给了我0而不是5！= 120。

编辑NEWEEST：如何修复双因素来给我正确的数字？如果我做5，那它应该给我5 !! = 15但是它给了我12为什么呢？

消息来源

def multiply(num1,num2):

    sum_of_multiplication= 0
    for i in range(num2):
        sum_of_multiplication = add(sum_of_multiplication,num1)
    return sum_of_multiplication



def factorial(num1):
    factorial_num = num1
    for i in range(1,num1):
        num1 = multiply(factorial_num,num1)
        print(factorial_num)
    print(str(num1) + "!= " + str(factorial_num))



 def double_factorial(num1):
    double_factorial_num = 2
    for i in range(1,num1-2):
        double_factorial_num = multiply(double_factorial_num, i)
        print(double_factorial_num)
    print(str(num1) + "!!= " + str(double_factorial_num))

double_factorial = double_factorial(int(input("please enter your intger:")))

Answer 1

这是因为你在乘法中包含0。在范围函数中（在阶乘函数中），将1作为第一个参数使其看起来像range(1,num1)。这样，乘法将从1开始，而不是0。

Answer 2

由于factorial()问题导致range()略有偏差，这是一个很容易犯的错误，但您应该可以通过查看range()的文档来快速解决问题。但是，您的double_factorial()函数没有成功的希望，因为您未能预先处理关键问题，数字的奇偶校验（奇数或偶数）：

def add(a, b):
    return a + b

def multiply(a, b):

    my_sum = 0

    for _ in range(b):
        my_sum = add(my_sum, a)

    return my_sum

def factorial(number):
    product = 1

    for multiplicand in range(2, number + 1):
        product = multiply(product, multiplicand)

    print(str(number) + "!", "=", product)

def double_factorial(number):
    parity = number % 2 + 2  # start at 2 or 3

    product = 1

    for multiplicand in range(parity, number + 1, 2):
        product = multiply(product, multiplicand)

    print(str(number) + "!!", "=", product)

number = int(input("please enter your integer: "))

factorial(number)

double_factorial(number)

<强>输出

> python3 test.py
please enter your integer: 5
5! = 120
5!! = 15
>

如何使用基于另一个具有两个参数的函数的乘法来进行阶乘

2 个答案: