HDUOJ 2141 Can you find it? (二分查找)

Give you three sequences of numbers A, B, C, then we give you a number X. Now you need to calculate if you can find the three numbers Ai, Bj, Ck, which satisfy the formula Ai+Bj+Ck = X. 

InputThere are many cases. Every data case is described as followed: In the first line there are three integers L, N, M, in the second line there are L integers represent the sequence A, in the third line there are N integers represent the sequences B, in the forth line there are M integers represent the sequence C. In the fifth line there is an integer S represents there are S integers X to be calculated. 1<=L, N, M<=500, 1<=S<=1000. all the integers are 32-integers. 
OutputFor each case, firstly you have to print the case number as the form "Case d:", then for the S queries, you calculate if the formula can be satisfied or not. If satisfied, you print "YES", otherwise print "NO". 
Sample Input

3 3 3
1 2 3
1 2 3
1 2 3
3
1
4
10

Sample Output

Case 1:
NO
YES
NO

文章大概意思是：第一行L,M,N，第二行L个数，第三行M个数，第四行N个数，第五行一个S，表示接下来有S次询问Ai，是否能在L,M,N中各找一个数相加等于Ai。

这道题我看完第一个想法就是暴力解决，直接算出来L,M,N相加的所以结果，时间复杂度为O（n^3），然后用二分查找，我试了一下，结果超时了，空间超限。

第二种思想就是半暴力解决，算出L,M相加的结果SUM，然后用Ai减去，在N中查找，看是否能找到。（SUMk+Nj=Ai）时间复杂度为O(N^2+lgN)，本来想着会WA，结果AC了，算是水过呀。。。。

具体代码如下：

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<algorithm>
using namespace std;
int BinarySearch(int a[],int x,int low,int high)//二分查找
{
	if(low>high)
		return -1;
	int mid=(low+high)/2;
	if(x==a[mid])
		return mid;
	if(x<a[mid])
		return BinarySearch(a,x,low,mid-1);
	return BinarySearch(a,x,mid+1,high);
}
int main()
{
	int temp=1,l,m,n;
	while(scanf("%d%d%d",&l,&m,&n)!=EOF)
	{
		int i;
		int *a,*b,*c;
		a=(int*)malloc(sizeof(int)*l);
		b=(int*)malloc(sizeof(int)*m);
		c=(int*)malloc(sizeof(int)*n);
		for(i=0;i<l;i++)
			cin>>a[i];
		for(i=0;i<m;i++)
			cin>>b[i];
		for(i=0;i<n;i++)
			cin>>c[i];
		
		int s;
		cin>>s;
		int *h;
		h=(int *)malloc(sizeof(int)*s);
		for(i=0;i<s;i++)
			cin>>h[i];
		int *p;//储存最终结果
		p=(int *)malloc(sizeof(int)*(m*n));
		int j,k,q=0;
		for(j=0;j<m;j++)
			for(k=0;k<n;k++)
				p[q++]=b[j]+c[k];		
		sort(p,p+q);//升序排序
		printf("Case %d:\n",temp);
		temp++;
		int flag;
		for(i=0;i<s;i++)
		{
			for(flag=0,j=0;j<l;j++)
			{
				if(BinarySearch(p,h[i]-a[j],0,m*n-1)!=-1)
				{
					flag=1;
					break;
				}					
			}
			if(flag==1)
				cout<<"YES"<<endl;
			else
				cout<<"NO"<<endl;			
		}
	}
	return 0;
}