Can you find it?

Time Limit: 10000/3000 MS (Java/Others)    Memory Limit: 32768/10000 K (Java/Others)
Total Submission(s): 17036    Accepted Submission(s):
4337

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

题意：给你三组数，每组数的个数分别为L N M再给你S个数x判断能否从这三组数中分别找到一个数使Ai+Bj+Ck = X.

题解：暴力肯定超时，所以要用二分，先将三组数中任意两组存入数组str[]中，然后再对每个x进行遍历

1、测试数据太弱，而且这样的一组测试数据本来就是个坑，好多不正确的情况都能得出正确答案

2、输出格式好坑，本来我以为是连续输出NO YES NO一直错没想到竟然是输一个数出一个结果

3、一定要注意给str[]数组排序（二分查找的条件是有序数列）

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
#define MAX 1100
bool cmp(int a,int b)
{
	return a<b;
}
int a[MAX];
int b[MAX];
int c[MAX];
int d,str[300000];
int main()
{
	int n,m,j,i;
	int k=1,l;
	while(scanf("%d%d%d",&l,&n,&m)!=EOF)
	{
		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]);
		int len=0;
		for(i=0;i<n;i++)
		{
			for(j=0;j<m;j++)
			{
				str[len++]=b[i]+c[j];
			}
		}
		sort(str,str+len,cmp);
		int ok;
		printf("Case %d:\n",k++);
		int s;
		scanf("%d",&s);
		for(i=0;i<s;i++)
		{
			scanf("%d",&d);
			ok=0;
			for(j=0;j<l;j++)
			{
				int left = 0,right = len,mid = 0;
				int goal = d - a[j];
				while(right >= left)
				{
					mid = (right + left) >> 1;
					if(str[mid] < goal)
					left = mid + 1;
					else if(str[mid] > goal)
					right = mid - 1;
					else
					{
						ok=1;
						break;
					}
				}
				if(ok)
					break;
			}
			if(ok)
				printf("YES\n");
			else
			    printf("NO\n");
		}
	}
	return 0;
}