Posted on 2008-03-30 22:26 
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算法 
			 
			
		 
		求解T(n)=T(ceil(n/2))+1
猜测解为O(lgn)
只需证T(n)<=clg(n-b)。于是T(n)<=clg(ceil(n/2-b))+1
                                                     <=clg(n/2-b+1)+1
                                                      ...
                                                       <=clg(n-b)
  
主方法:
形如T(n)=aT(n/b)+f(n),注意a>=1,b>1
比较f(n)和nlogba,则T(n)为较大者,如果f(n)=Q(nlogba),则T(n)=Q(nlogbalgn)