>分治算法

题目：设有两个已排好序的数组，设计一个O(logn)时间算法，找出X和Y数组的2n个数的中位数。

#include <stdio.h>

void Mid_Num( int a[] , int b[] , int n )
{
    int n1 , left1 , left2 , right1 , right2  ;
    float m1 , m2 , m ;
    left1 = left2 = 0 ;
    right1 = right2 = n-1 ; 
    n1 = n;
    while( n1 > 2 )
    {
        if( n1 % 2 == 0 )       //if the number is not odd
        {   
            m1 = ( a[left1 + n1/2] + a[left1 + (n1 - 1)/2] ) / 2 ;  //the mid number is (a[n-1/2] + a[n/2])/2
            m2 = ( b[left2 + n1/2] + b[left2 + (n1 - 1)/2] ) /2 ;
            if( m1 > m2 )
            {
                right1 = left1 + ( n1 - 1 )/2 ;
                left2 = left2 + n1/2 ;
            }
            else
            {
                left1 = left1 + n1/2 ;
                                right2 = left2 + ( n1 - 1 )/2 ;
            }
            n1 = ( n1 + 1 )/2 ; 
        }
        else
        {
            m1 = a[left1 + n1/2] ;      //if the number is odd,the mid number is a[n/2]
                        m2 = b[left2 + n1/2] ;
                        if( m1 > m2 )
                        {
                                right1 = left1 + n1/2 ;
                                left2 = left2 + n1/2 ;
                        }
                        else
                        {
                                left1 = left1 + n1/2 ;
                                right2 = left2 + n1/2 ;
                        }
                        n1 = ( n1 + 1 )/2 ;
        }
//      printf( "m1 = %f,m2 = %f\n",m1,m2 ) ;
    }
//  printf( "a[left1] = %d,a[right1] = %d\nb[left2] = %d,b[right2] = %d\n ",a[left1],a[right1],b[left2],b[right2] ) ;

    if( n1 == 2 )   
    {
        if( a[right1] <= b[left2] )
            m = ( a[right1] + b[left2] )*1.0 /2 ;
        else if( a[left1] <= b[left2] )
        {
            if( a[right1] <= b[right2] )
                 m = ( a[right1] + b[left2] )*1.0 / 2 ;
            else
                m = ( b[left2] + b[right2] )*1.0 / 2 ;
        }   
        else if( a[left1] <= b[right2] )
        {
            if( a[right1] <= b[right2] )
                m = ( a[left1] + a[right1] )*1.0 / 2 ;
            else
                m = ( a[left1] + b[right2] )*1.0 / 2 ;
        }
    }
    printf( "m = %f\n",m );
}

int main( void )
{
    int i , n ;
    printf( "please input n:\n" ) ;
    scanf( "%d",&n ) ;
    int a[n] , b[n] ;
    printf( "please input %d number sort arry a:\n",n ) ;
    for( i = 0 ; i < n ; i++ )
        scanf( "%d",a+i ) ;
    printf( "please input %d number sort arry b:\n",n ) ;
        for( i = 0 ; i < n ; i++ )
                scanf( "%d",b+i ) ;
    Mid_Num( a , b , n ) ;
    return 0 ;

}

分治算法的基本思想：将大规模地问题分解为k个规模小的问题，且每个小问题与原问题相似。