在python中逐步打印解决方案的计算器

Question

我儿子正试着做一个计算器来帮助他完成很多页的家庭作业。我知道简单的解决办法是告诉他关于wolfram的事情，但我认为这可能是一个有趣的项目。我需要一些帮助，如何迭代其余的数字和打印解决方案的一步一步的格式。到目前为止，这就是他所拥有的：

#  Variables
X = input("input the first number with space between digits:  ")
Y = int(input("input the second number:  "))
Xn = X.split(" ")

if int(Xn[0]) >= Y:  # if Y fits in X the do a simple division and return the remainder  

    Xy1 = int(Xn[0])
    fit0 = int(Xy1 / Y)
    rem0 = Xy1 - fit0 * Y

    print(" It fits " + str(fit0), "times ", "    the remainder is:  " + str(rem0))

else:  # if Y does not fit in X the  add the first two strings of X  and convert them to integers then do a simple 
    # division and return the remainder 

    Xy0 = (int(Xn[0] + Xn[1])) 
    fit = int(Xy0 / Y)
    rem = Xy0 - fit * Y
    print(" It fits " + str(fit), "times ", "    the remainder is:  " + str(rem))

Answer 1

在这里，我做了一个例子，一步一步地打印除法。

我硬编码了x(用空格分隔的数字除数)和除数。您只需将其更改为包含来自用户的输入。

x = "1 4 8 7"
divisor = 5

# convert x to a list of integers
x = [int(i) for i in  x.split(" ")]

dividend = x[0]
i = 0   # index of the current dividend digit
quotient = ""

while i < len(x)-1:
    i += 1
    quotient += str(dividend // divisor)
    remainder = dividend % divisor
    print(f"{dividend} / {divisor} -> {dividend // divisor} (remainder {remainder})")
    dividend = remainder * 10 + x[i]
    
quotient += str(dividend // divisor)
remainder = dividend % divisor

print(f"{dividend} / {divisor} -> {dividend // divisor} (remainder {remainder})")
print(f"Result: {quotient} (remainder {remainder})")

其结果是：

1 / 5 -> 0 (remainder 1)
14 / 5 -> 2 (remainder 4)
48 / 5 -> 9 (remainder 3)
37 / 5 -> 7 (remainder 2)
Result: 297 (remainder 2)

Answer 2

我想我误解了这个问题..。为什么不使用浮子呢？

x = float(input("input the first number:  "))
y = float(input("input the second number:  "))
print(f" It fits {x//y} times , the remainder is:  {x/y-x//y}")