Question

我正在上课的作业要求我们制作一个程序，通过重复减法计算整数除法（就像乘法相当于重复加法）。到目前为止，除了用户想要计算商的次数之外，一切都很好。

由于某种原因，程序总是运行一次，而不是我提示程序在for循环中执行的迭代/输入次数。

我可以得到一些帮助吗？我是新功能的，所以我确定程序不是可读的，但我需要弄清楚我的get_num_of_inputs（）函数是什么问题和/或num_of_inputs变量。

谢谢！：）

def main():
    num_of_inputs = get_num_of_inputs()
    numerator, denominator = get_input()
    quotient = calc_quotient(num_of_inputs, numerator, denominator)
    display_output(numerator, denominator, quotient)

def get_num_of_inputs():
    return int(input("What is the number of inputs?: "))

def get_input():
    numerator = int(input("\nEnter the numerator: "))
    denominator = int(input("Enter the denominator (must not be zero): "))
    return numerator, denominator

def calc_quotient(num_of_inputs, numerator, denominator):

    if denominator == 0:
        return print("Error: invalid input")

    if numerator == 0:
        return 0

    if denominator == 1:
        return numerator

    if denominator == -1:
        return -numerator

    numerator_abs = abs(numerator)
    denominator_abs = abs(denominator)

    for iterations in range(num_of_inputs):
        #case 1
        if numerator_abs > denominator_abs:
            result = 5
            acc = numerator_abs
            quotient = 0

            while result > 0:
                acc = acc - denominator_abs
                result = acc
                quotient += 1

            if result < 0:
                return quotient - 1

            if (numerator > 0 and denominator > 0) or (numerator < 0 and denominator < 0):
                return quotient

            if numerator_abs % denominator_abs == 0 and ((numerator > 0 and denominator < 0) \
                or (numerator < 0 and denominator > 0)):
                return -quotient

            if numerator_abs % denominator_abs > 0 and ((numerator > 0 and denominator < 0) \
                or (numerator < 0 and denominator > 0)):
                return -quotient - 1
        #case 2
        if denominator_abs > numerator_abs:

            if (numerator > 0 and denominator > 0) or (numerator < 0 and denominator < 0):
                return 0

            if (numerator > 0 and denominator < 0) or (numerator < 0 and denominator > 0):
                return -1
        #case 3
        if numerator_abs == denominator_abs:

            if (numerator > 0 and denominator > 0) or (numerator < 0 and denominator < 0):
                return 1

            if (numerator > 0 and denominator < 0) or (numerator < 0 and denominator > 0):
                return -1


def display_output(numerator, denominator, quotient):
    print("\nThe quotient of", numerator, "and", denominator, "is:", quotient)

main()

Answer 1

你的逻辑有点纠结。无需将num_of_inputs传递给calc_quotient函数。相反，您需要创建循环for次的num_of_inputs循环。在循环内部，您将得到一个新的分子和分母，调用calc_quotient，然后输出结果。

在下面的代码中，我简化了calc_quotient功能。

def calc_quotient(numerator, denominator):
    if denominator == 0:
        raise ValueError("denominator can't be zero!")

    sign = -1 if numerator * denominator < 0 else 1
    numerator = abs(numerator)
    denominator = abs(denominator)

    quotient = 0
    while numerator >= denominator:
        numerator -= denominator
        quotient += 1
    return sign * quotient

def get_num_of_inputs():
    return int(input("What is the number of inputs?: "))

def get_input():
    numerator = int(input("\nEnter the numerator: "))
    denominator = int(input("Enter the denominator (must not be zero): "))
    return numerator, denominator

def display_output(numerator, denominator, quotient):
    print("\nThe quotient of", numerator, "and", denominator, "is:", quotient)

def main():
    num_of_inputs = get_num_of_inputs()
    for i in range(num_of_inputs):
        numerator, denominator = get_input()
        quotient = calc_quotient(numerator, denominator)
        display_output(numerator, denominator, quotient)


if __name__ == '__main__':
    main()

函数中的迭代问题

1 个答案: