Negative 1 plus 0 should add up to negative 1, if in fact, we're correct for the c n anion.
负一加上零应该等于负一,如果是这样,我们对于氰离子的结果就是正确的。
to the n, every value in the 1 bit vector we looked at last time is either 0 or 1. So it's a binary n number of n bits, 2 to the n.
从2到n,我们上次看到的,位向量的每个值不是0就是,所以它是n,比特的二进制数,从2到。
So when we count, generally, again, we start from zero, we go to N minus 1.
当我们计数时,一般的,再次强调,我们从0开始,到N-1结束。
If n = 1, x/n can take on only two values: 0 or 1.
当n=1时,x/n的值只可能等于0和1
Well, this question mark becomes an O, this question mark becomes an O, and then the loop terminates 0 1 2 because it's iterating from zero to N so that's zero, 1, 2 and the length of the string is 3 so the loop terminates, but I remember that I needed to have this special sentinel value so I'm just going to put it there manually.
嗯,这个问号变成了,这个问号变成了0,然后循环结束了,以为迭代从0到N,那就是,字符串的长度是3,然后循环就结束了,但是我记得我需要这个特殊的标记值,所以需要手动的加上它。
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