CSP(NOIP)初赛时间复杂度分析整理

原文链接: https://www.luogu.org/discuss/show/155153

给定

T ( n ) = { 1 , n = 1 a T ( n b ) + f ( n ) , otherwise T(n)=\begin{cases}1,&n=1\\a T(\dfrac{n}{b})+f(n),&\text{otherwise}\end{cases}

试计算 T ( n ) T(n)

c c r i t = log b a c_{crit}=\log_ba

  • f ( n ) = O ( n c ) f(n)=O(n^c) c < c c r i t c<c_{crit} 则有 T ( n ) = Θ ( n c c r i t ) T(n)=\Theta(n^{c_{crit}})
  • k \exists k 使得 f ( n ) = n c c r i t log k n f(n)=n^{c_{crit}}\log^kn ,则有: T ( n ) = { Θ ( n c c r i t log k + 1 n ) , k > 1 Θ ( n c c r i t log log n ) , k = 1 Θ ( n c c r i t ) , k < 1 T(n)=\begin{cases}\Theta(n^{c_{crit}}\log^{k+1}n),&k>-1\\\Theta(n^{c_{crit}}\log\log n),&k=-1\\\Theta(n^{c_{crit}}),&k<-1\end{cases}
  • f ( n ) = Ω ( n c ) f(n)=\Omega(n^c) c > c c r i t c>c_{crit} 则有 T ( n ) = Θ ( f ( n ) ) T(n)=\Theta(f(n))

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转载自blog.csdn.net/kkkksc03/article/details/102616032