Can you find it?

Time Limit: 10000/3000 MS (Java/Others)    Memory Limit: 32768/10000 K (Java/Others)
Total Submission(s): 35824    Accepted Submission(s): 8840

Problem Description
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.
 

Input
There 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.
 

Output
For 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
 

Author
wangye
 

Source
HDU 2007-11 Programming Contest

题目大意：先输入三列数字，分别标记为Ai，Bi，Ci，判断是否存在 Ai，Bj，Ck使得其相加为后面所输入的数。
思路：先将前两列的数相加，用一个数组储存，然后使用sort将其排序，最后二分查找。 
#include<stdio.h>
#include<algorithm>
using namespace std;
int L_[505], N_[505], M_[505];
int sum_[250005];
int main()
{
	int L, N, M;
	int S;
	int x,num=0,temp;
	while (~scanf("%d %d %d", &L, &N, &M))
	{
		num++;
		int k = 0;
		for(int i = 0; i < L; i++)
		{
			scanf("%d", &L_[i]);
		}
		for (int i = 0; i < N; i++)
		{
			scanf("%d", &N_[i]);
		}
		for (int i = 0; i < M; i++)
		{
			scanf("%d", &M_[i]);
		}
		for (int i = 0; i < N; i++)
		for (int j = 0; j < M; j++)
			sum_[k++] = N_[i] + M_[j];
		sort(sum_, sum_ + k);
		scanf("%d", &S);
		printf("Case %d:\n", num);
		while (S--)
		{
			int flag = 0;
			scanf("%d", &x);
			for (int i = 0; i < L; i++)
			{
				temp = x - L_[i];
				int left = 0,right=k-1;
				int mid;
				while (left <= right)
				{
					mid = (left + right) / 2;
					if (sum_[mid] > temp)
					{
						right=mid-1;
					}
					else if (sum_[mid]<temp)
					{
						left=mid+1;
					}
					else
					{
						flag = 1;
						break;
					}
				}
				if (flag==1)
				{
					printf("YES\n");
					break;
				}
			}
			if (flag == 0)
			{
				printf("NO\n");
			}
		}
	}
	return 0;
}