问题
I'm having trouble understanding how to make this into a formula.
for (int i = 1; i <= N; i++) {
for (int j = 1; j <= N; j += i) {
I realize what happens, for every i++ you have 1 level of multiplication less of j.
i = 1, you get j = 1, 2, 3, ..., 100
i = 2, you get j = 1, 3, 5, ..., 100
I'm not sure how to think this in terms of Big-theta.
The total of j is N, N/2, N/3, N/4..., N/N (My conclusion)
How would be best to try and think this as a function of N?
回答1:
So your question can be actually reduced to "What is the tight bound for the harmonic series 1/1 + 1/2 + 1/3 + ... + 1/N?" For which the answer is log N (you can consider it as continuous sum instead of discrete, and notice that the integral of 1/N is log N)
Your harmonic series is the formula of the whole algorithm (as you have correctly concluded)
So, your sum:
N + N/2 + N/3 + ... + N/N = N * (1 + 1/2 + 1/3 + ... + 1/N) = Theta(N * log N)
So the tight bound for the algorithm is N*log N
See the [rigorous] mathematical proof here (see the "Integral Test" and "Rate of Divergence" part)
回答2:
Well, you can methodically use Sigma notation:
来源：https://stackoverflow.com/questions/18863422/asymptotic-analysis