O(n^k)比O(n^k')对全部k, k' >= 0 and k' > k
O(n^2)会比O(n^2*logn)
请注意，您只能忽略常量，任何涉及输入大小的内容都不能被忽略。
因此，复杂度为T(n)=n^2 + n^2logn会是两者中更糟糕的一个，即O(n^2logn).
Little-oh
宽松地说，“小哦”是有保证的上限。是的，这叫小，而且限制比较多。
n^2 = O(n^k) for k >= 2 but n^2 = o(n^k) for k > 2
实际上，它是Big-Oh这占据了大部分的风头。
What about
T(n)= n^2.001 + n^2logn?
We have n2.001 = n2*n0.001 and n2 * log(n).
To settle the question, we need to figure out what would eventually be bigger, n0.001 or log(n).
It turns out that a function of the form nk with k > 0 will eventually take over log(n) for a sufficiently large n.
Same is the case here, and thus T(n) = O(n2.001).
Practically though, log(n) will be larger than n0.001.
(103300)0.001 < log(103300) (1995.6 < 3300), and the sufficiently large n in this case would be just around 103650, an astronomical number.
Worth mentioning again, 103650. There are 1082 atoms in the universe.
