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

题意：查找在三个数组中能否找到等于X的数。

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<math.h>
#include<algorithm>
#include<queue>
#include<string>
#include<map>
#include<stack>
#define maxn 505
#define inf 0x3f3f3f3f
#define ll long long
using namespace std;
ll a[maxn],b[maxn],c[maxn],x[maxn*maxn];
int main(){
    int l,n,m,t=0;
    while(~scanf("%d%d%d",&l,&n,&m)){
            memset(a,0,sizeof(a));
            memset(b,0,sizeof(b));
            memset(c,0,sizeof(c));
            memset(x,0,sizeof(x));
        for(int i=0;i<l;i++){
            cin>>a[i];
        }
        for(int j=0;j<n;j++)cin>>b[j];
        for(int i=0;i<m;i++)cin>>c[i];
        int nn;cin>>nn;
        sort(a,a+l);
        sort(b,b+n);
        sort(c,c+m);int p=0;
        for(int i=0;i<l;i++){
            for(int j=0;j<n;j++){
               x[p++]=a[i]+b[j];
            }
        }
        sort(x,x+p);int q;
        printf("Case %d:\n",++t);
        for(int i=0;i<nn;i++){
            cin>>q;
            int flag=0;
            for(int j=0;j<m;j++){
                int y=q-c[j];
                int ans=*lower_bound(x,x+p,y);
                if(ans==y){flag=1;cout<<"YES"<<endl;break;}
            }
            if(flag==0)cout<<"NO"<<endl;
        }
    }
}