能否请您解释一下下面的线，一些很好的例子。
由n个阿左算术移 相当于乘以2 N （提供的值不 溢出）。
和
一个正确的算术 由二的补码值的n移位 相当于除以2 N 和 向负无穷大舍入。如果 二进制数被视为那些' 补充，那么同样右移 运算结果除以2 N 和向零舍入。
解决方案我会解释，我们是比较熟悉的基础发生了什么：10
在基地10，假设你有一个数N = 123。现在，你SHIFT这个数字向左K = 3个位置，以0填充清空位数所以，你得到X = 123000。
注意，X = N * 10 K 。
与基地2的情况是相似的。
例1（基数为10）|实施例2（基数为2） | N = 123 | N = 110101（53 10为基数） K = 3 | K = 2（在基体10） N'LT;＆LT; K = 123000 | N'LT;＆LT; K = 11010100（以10为基数212） | 10 ^ K = 1000 | 2 ^ K = 100（在基体2;在基座10 4） N * 10 ^ K = 123000 | N * 2 ^ K = 11010100（53 * 4 = 10为基212） |与右移的情况下仅仅是这个过程的一面镜子，也是在基地10例如类似，如果我在基地10有123456，我转变右3个位置，我得到123这千分之十二万三千四百五十六（整数除法），其中1000 = 10 3 。
Can you please explain the below lines, with some good examples.
A left arithmetic shift by n is equivalent to multiplying by 2n (provided the value does not overflow).
And:
A right arithmetic shift by n of a two's complement value is equivalent to dividing by 2n and rounding toward negative infinity. If the binary number is treated as ones' complement, then the same right-shift operation results in division by 2n and rounding toward zero.
解决方案
I will explain what happens in a base that we're more familiar with: 10.
In base 10, let's say you have a number N=123. Now, you "shift" this number to the left k=3 positions, filling the emptied digits with 0. So you get X=123000.
Note that X = N * 10k.
The case with base 2 is analogous.
Example 1 (base 10) | Example 2 (base 2) | N = 123 | N = 110101 (53 in base 10) k = 3 | k = 2 (in base 10) N << k = 123000 | N << k = 11010100 (212 in base 10) | 10^k = 1000 | 2^k = 100 (in base 2; 4 in base 10) N * 10^k = 123000 | N * 2^k = 11010100 (53 * 4 = 212 in base 10) |The case with right shift is simply a mirror of the process, and is also analogous in base 10. For example, if I have 123456 in base 10, and I "shift" right 3 positions, I get 123. This is 123456 / 1000 (integer division), where 1000 = 103.