[英]Finding last digits with modulo from a million digits int; need an explanation?
Lets say I have a random million digit number = x, such that: 假设我有一个随机的百万位数= x,这样:
len(str(x)) = 1000000
From looking at some explanations I can use x % (10 ** n) to find the last n digits. 通过查看一些解释,我可以使用x%(10 ** n)查找最后n个数字。 But I can't wrap my head around why that works.
但是我无法确定为什么这样做。
Such that if I wanted to find the last 11 digits of x my code would be: 这样,如果我想找到x的最后11位数字,我的代码将是:
x % (10 ** 11)
Could someone shed some light on this for me? 有人可以帮我一下吗?
Think of it this way: 这样想:
If you want to find the last n digits of a number, you can divide it by 10**n, and the remainder of the division will be the last n digits of the number. 如果要查找数字的最后n位数字,可以将其除以10 ** n,除法的其余部分将是数字的最后n位数字。 In mathematical terms, the last
n
digits of number x
are given by: 用数学术语来说,数字
x
的最后n
数字由下式给出:
x % (10 ** n)
In case you don't know, the modulo operator (%) divides two numbers and returns the remainder . 如果您不知道,模运算符(%)将两个数相除并返回余数 。
%
returns the remainder after division, just like in primary school when you learned long division. %
返回除法后的余数,就像在小学学习长除法时一样。 So 7%3 = 1
. 因此
7%3 = 1
。 I guess you know that **
is exponentiation, so 10**3 = 1000
. 我猜您知道
**
是幂运算,所以10**3 = 1000
。 Your example x % (10**11)
divides x
by 10**11
and tells you the remainder. 您的示例
x % (10**11)
将x
除以10**11
并告诉您余数。 That must be the digits left over once you take away the largest possible multiple of 10**11
, in other words the last n
digits. 一旦您拿走了
10**11
的最大可能倍数,换句话说就是最后n
数字,那一定是剩下的数字。
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