Amortized Analysis 均摊分析

转自 https://zxi.mytechroad.com/blog/sp/amortized-analysis/

 

 

 

证明：1+2+4+8+...+n = 2n-1

a1=1，项数k=log2(n)+1

=等比数列求和=a1*(q^n-1)/(q-1)

①：Sn=a1+a2+...+an

②：q*Sn=q*a1+q*a2+...+q*an = a2+a3+...+an+q*an

①-②：(1-q)*Sn=(a1-q*an)=(a1-q*a1*q^(n-1))=a1-a1*q^n

-> Sn=a1(1-q^n)/(1-q)

所以本为题的和

= 1*(1-2^(log2(n)+1))/(1-2)

= 2^log2(n) * 2^1 - 1

= n * 2 -1

= 2n-1