如何编写一个函数，只将整数除以一个均匀分配的数字？ [英] How do I write a function that will only divide an integer by a number that evenly divides into it?

问题描述

divide():
    ran1 = randint(0, 15) #dividend
    ran2 = randint(1, 15) #divisor
    while ran1 % ran2 == 0: #Can I close this? Is there a better way?
        divi = int(ran1 / ran2)
    print('What is ' + str(ran1) + ' / ' + str(ran2) + '?')

这是我的尝试，但它不起作用。

我想要的结果如4/2或20/5

不是4/3或17/4。

This is my attempt, but it doesn't work.
I want results like 4/2 or 20/5
not 4/3 or 17/4.

推荐答案

扭转问题。从两个数字开始相乘，得到第三个数字。现在你有了所有三个值（其中两个均匀划分第三个）并可以随意做任何事情。

Reverse the problem. Start with two numbers that multiply together to get a third. Now you have all three values (two of them dividing the third evenly) and can do anything you want with them.

引用：

而ran1％ran2 == 0：#Can我关闭了吗？有更好的方法吗？

while ran1 % ran2 == 0: #Can I close this? Is there a better way?

使用模运算符（％）的表达式是正确的。 而是没有意义的。



[更新]

试试这段代码

The expression using the modulo operator (%) is correct. The while is pointless.

[update]
Try this code

from random import randint


for x in range(0,100):
  ran1 = randint(0, 15) 
  ran2 = randint(1, 15) 

  if ran1 % ran2 == 0:
    print(str(ran2) + " divide exactly " + str(ran1))
  else:
    print(str(ran2) + " doesn't divide exactly " + str(ran1))

[/ update]

[/update]

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