输入10个数，逆序输出（折半）。
求详解，百度上都是折半查找法

折半排序法、也就是二分排序法。

快速排序  时间复杂度O(nlogn)

具体点额

折半也叫二分.....具体的就是在数组中查找一个数。其查找的前提条件是数组的值按照从大到小或者从小到大排列
比如a[10]={1,2,3,4,5,6,7,8,9,10}中查找a
令begin=0,end=9,mid=(begin+end)/2
if(a[mid]>a) end=mid;
else if(a[mid]<a) begin=mid;
就这样循环知道找到或者退出为止

#include < stdio.h >
#include < stdlib.h >
#define N 15
void main ( )
{
    int i, number, top, bott, mid, loca, a[N], flag = 1, sign;
    char c;
    printf ( "enter data:\n" );
    scanf ( "%d", &a[0] );
    i = 1;
    while ( i < N )
    {
        scanf ( "%d", &a[i] );
        if ( a[i] >= a[i - 1])
            i++;
        else
            printf ( "enter this data again:\n" );
    }
    printf ( "\n" );
    for ( i = 0;i < N; i++ )
        printf ( "%d", a[i] );
    printf ( "\n" );
    while ( flag )
    {
        printf ( "input number to look for : ");
        scanf ( "%d", &number );
        sign = 0;
        top = 0;
        bott = N - 1;
        if (  ( number < a[0]) || ( number > a[N-1]) )
            loca = -1;
        while ( ( !sign ) && ( top <= bott ) )
        {
            mid = ( bott + top )/2;
            if ( number == a[mid] )
            {
                loca = mid;
                printf ( "Has found %d ,its position is %d\n", number, loca + 1 );
                sign = 1;
            }
            else if ( number < a[mid] )
                bott = mid -1;
            else
                top = mid + 1;
        }
        if ( !sign ||loca == -1 )
            printf ( "can not find %d.\n",number );
        printf ( "continue or not ( Y/N )?" );
        scanf ( "%c",&c );
        if ( c == 'N' || c == 'n' )
            flag = 0;
    }
    system ( "pause" );
}

这是个找到一个和已知数组相同的程序，用的正是，折半排序法，前两天写的一个程序，给你看看

好复杂。。我水平不够。。还是谢谢你了，我先参悟下

你这个是折半查找吧