2 k-v kv型
2 k-widths k宽度
2 k coatings 双组分涂料
computer y 2 k 计算机年
y 2 k problems yk问题
2 k polyurethane 双组分聚氨酯
y 2 k solutions yk问题
t-2 k collector tk捕收剂
atlas 2 k language atlask语言
the y 2 k problem yk问题
To get the information in 1k blocks, use the -k command-line option, as shown in Listing 2.
要获得以1 k块为单位的信息,可以使用- k命令行选项,如清单2所示。
或者是1/2,k,x的平方。
So we get 2gl times the sine of theta mu k*cos 2 and the whole thing to the power one-half.
所以我们得到2gl*sinθ,θ,minus,mu,of,k,times,the,cosine,of,theta,所有项除以。
And if you do so, you will end up with 1.312 mega joules per mole for this quantity K.
这样做之后,对于K常量你就能得到2,1。312百万焦耳每摩尔。
Yeah. You're jumping slightly ahead of me, but basically, I'm done when this is equal to 1 right? Because I get down to the base case, so I'm done when b u is over 2 to the k is equal to 1, and you're absolutely right, that's when k is log base 2 of b.
因为这就是最基本的情况了,因此当b/2的k次方等于1的时候就停止了1,你说的太对了,就是k等于b的以2为底的对数的时候,你们都坐的挺靠后的啊,我不知道是不是我讲的不太明白?
And if I pull it out one more level, 12k it's 12 plus 12 plus t of b over 8, 12 k because I'll have 12 of those to add up, plus t of b over 2 to the k.
总结一下也就是说,在k步以后,总步数应该是,那这种情况什么时候才能停止呢?,才能到达最基本的情况呢?
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