Description: For a given set of data to find their maximum or minimum, the use of recursion and divided way of thinking to solve the problem. n = 2, the first comparison of two data elements can be found on the maximum and minimum yuan yuan. When n> 2 when n data elements can be divided into roughly equal halves, half of the? N/2? Data elements, while the other half there? N/2? Data elements. First, to identify their group were the largest and smallest yuan yuan, and then compare the two greatest element, you can have n elements of the greatest element to compare the two smallest element, you can get the smallest element of n-yuan .
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File list (Check if you may need any files):
lq\Debug\lq.exe
..\.....\lq.ilk
..\.....\lq.obj
..\.....\lq.pch
..\.....\lq.pdb
..\.....\StdAfx.obj
..\.....\vc60.idb
..\.....\vc60.pdb
..\lq.cpp
..\lq.dsp
..\lq.dsw
..\lq.ncb
..\lq.opt
..\lq.plg
..\ReadMe.txt
..\StdAfx.cpp
..\StdAfx.h
..\Debug
lq
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