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

給三組整數，在下面每給一個整數，判斷從三組數各取一個相加與此整數是否可以相等。
直接暴力復雜度為O(N^3)，不可行。可以將前兩組枚舉，復雜度為O(N^2)，再排序，對於每一個目標數，枚舉第三組數，二分查找前兩組的和即可。

源代碼：
#include<iostream>
#include<algorithm>
#include<string.h>

using namespace std;

int a[505],b[505],c[505],d[250005];

int main()
{
	int L,M,N,S,s=1,x,y,z,flag;
	while(cin>>L>>M>>N)
	{
		y=0;
		flag=0;
		cout<<"Case "<<s++<<":"<<endl;
		memset(a,0,sizeof(a));
		memset(b,0,sizeof(b));
		memset(c,0,sizeof(c));
		memset(d,0,sizeof(d));
		for(int i=0;i<L;i++)
			cin>>a[i];
		for(int i=0;i<M;i++)
			cin>>b[i];
		for(int i=0;i<N;i++)
		{
			cin>>c[i];	
			for(int j=0;j<M;j++)
				d[y++]=c[i]+b[j];
		}	
		sort(d,d+y);
		int ans=unique(d,d+y)-d;
		/*cout<<ans<<" "<<y;
		for(int i=0;i<y;i++)
			cout<<d[i]<<endl;*/
		
		cin>>S;
		for(int i=0;i<S;i++)
		{
			cin>>x;
			flag=0;
			for(int j=0;j<L;j++)
			{
				z=x-a[j];
				if(binary_search(d,d+ans,z))
				{
					flag=1;
					break;
				}
			}
			if(flag)
				cout<<"YES"<<endl;
				else
					cout<<"NO"<<endl;
		}
	}
	
	return 0;
 } 
 

HDU - 2141 ： Can you find it?