HDU 1241 Can you find it?

/*
--------------------------------------------
    stratege : 二分查找
    basic step : 
            int find (int min, int max, int x) // min是已排列的数组的下限，max为上限，x为所要查找的值
            {
                int mid ;
                while (min <= max)
                {
                    mid = (min + max) / 2 ;    //折半查找，顾名思义，从中间元素开始进行对比
                    if (ab[mid] == x)   return 1 ;
                    if (ab[mid] > x)    max = mid - 1 ;
                    if (ab[mid] < x)    min = mid + 1 ;
                }
                return 0 ;
            }
    URL : http://acm.hdu.edu.cn/showproblem.php?pid=2141   Can you find it?
******************************************************************
    Problem : 2141 ( Can you find it? )     Judge Status : Accepted
    RunId : 5330614    Language : G++    Author : a312745658
******************************************************************
------------------------------------------------------------------
*/

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <algorithm>
using namespace std ;

const int MAXN = 505 ;

int L, N, M ;
int a[MAXN], b[MAXN], c[MAXN] ;
int ab[250005] ;
int S, n ;

int cmp (const void *a, const void *b)
{
    return *(int *)a - *(int *)b ;
}

int find (int min, int max, int x)
{
    int mid ;
    while (min <= max)
    {
        mid = (min + max) / 2 ;
        if (ab[mid] == x)   return 1 ;
        if (ab[mid] > x)    max = mid - 1 ;
        if (ab[mid] < x)    min = mid + 1 ;
    }
    return 0 ;
}

int main()
{
    int i, j, k ;
    int cas = 1 ;
    int num ;

    while (scanf ("%d%d%d", &L, &N, &M) != EOF)
    {
        num = 0 ;
        for (i = 0; i < L; i ++)
            scanf ("%d", &a[i]) ;
        for (i = 0; i < N; i ++)
            scanf ("%d", &b[i]) ;
        for (i = 0; i < M; i ++)
            scanf ("%d", &c[i]) ;

        scanf ("%d", &S) ;

        for (i = 0; i < L; i ++) //将2个数组整合成1个数组，使其为2个数组中各个不同元素的和
            for (j = 0; j < N; j ++)
                ab[num++] = a[i] + b[j] ;
        qsort (c, M, sizeof(c[0]), cmp) ;   
        qsort (ab, num, sizeof (ab[0]), cmp) ; //注意：需对数组进行排序 
        printf ("Case %d:\n", cas ++) ;

        while (S -- && scanf ("%d", &n))
        {
            bool flag = false ;

            for (i = 0; i < M; i ++)
            {
                if (find (0, num-1, n-c[i])) //n-c[i], 即为ab[i]中的值，对ab[i]进行折半查找
                {
                    flag = true ;
                    break ;
                }
            }
            if (flag)
                printf  ("YES\n") ;
            else
                printf ("NO\n") ;
        }
    }
    return 0 ;
}