Can you find it?

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

给三个数组 每个数组取一个数 相加能否等于下面的例子

#include<cstdio>          
#include<algorithm>
using namespace std;
int a[501];
int b[501];
int c[501];
int d[505001];
int main()
{
	int L,n,m,t,q;
		int sum=1;
    while(scanf("%d%d%d",&L,&n,&m)!=EOF)
{
    		for(int i=1;i<=L;i++)
	{
		scanf("%d",&a[i]);
	}
		for(int i=1;i<=n;i++)
	{
		scanf("%d",&b[i]);
	}
		for(int i=1;i<=m;i++)
	{
		scanf("%d",&c[i]);
	}
	sort(a+1,a+L+1);
	sort(b+1,b+n+1);
	sort(c+1,c+m+1);
    int cnt=0;
		for(int i=1;i<=L;i++)
	{
		for(int j=1;j<=n;j++)
		{
			d[++cnt]=a[i]+b[j];
		}
	}
	sort(d+1,d+cnt+1);

	scanf("%d",&t);
	printf("Case %d:
",sum++);
	while(t--)
	{
	    int f=0;
		scanf("%d",&q);
		for(int i=1;i<=m;i++)
		{
			int l=1,r=cnt;
			while(l<=r)
			{
				int mid=(l+r)/2;
				if(d[mid]+c[i]==q)
				{
						f=1;
				        break;
				}
				if(d[mid]+c[i]<q)
				{
					l=mid+1;
				}
				else
				{
					r=mid-1;
				}
			}
			if(f==1)
			{
				  break;
			}
		}
		if(f==1)
		{
		printf("YES
");	
		}
		else
		{
			printf("NO
");
		}
	}
}
	return 0;
}