# 言之凿凿

Try harder.

2018.09.11

## $$T(n) = aT(\frac{n}{b}) + O(n^d)$$

$$\log_{b}a > d \quad \Rightarrow \quad O(n^{\log_{b}a})$$ $$\log_{b}a < d \quad \Rightarrow \quad O(n^{d})$$ $$\log_{b}a = d \quad \Rightarrow \quad O(n^{d}*\log{n})$$

$$\log_{b}a > d \quad \Rightarrow \quad O(n^{\log_{b}a})$$ $$\log_{b}a < d \quad \Rightarrow \quad O(n^{d})$$ $$\log_{b}a = d \quad \Rightarrow \quad O(n^{d}*\log{n})$$