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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


题目大意：输入3个数组，3个数组中的元素相加，判断是否能得到x；
题目的输入输出有点恶心人，，，写的时候弄的我晕 ，，哇了好几次
思路 ：一开始想到的是暴力枚举，，但是肯定会TLE 看了一下大佬们的博客，，用二分法方便一点  就是让A+B构成一个新的数组sum，x-c[i]构成一个新的数组cc，然后在sum中查找是否存在cc中的元素有的话返回YES否则返回NO
AC代码：(本人不太擅长二分所以代码质量不高)

#include<iostream>
#include<algorithm>
using namespace std;
int a[501];
int b[501];
int c[501];
int sum[500*500+1];
int pos;
int judge(int x){
    
    if(x<sum[0]||x>sum[pos-1]) return 0;
    
    int low=0,high=pos-1;
    while(low<=high){
        int mid=(high+low)/2;
//        cout<<mid<<endl;
        if(sum[mid]>x){
            high=mid-1;
        }
        else if(sum[mid]<x) low=mid+1;
        else {
            return 1;
        }
    }
    return 0;
}

int main()
{
    int l,m,n,ll=0;
    while(cin>>l>>m>>n)
    {
        ll++;
        for(int i=0;i<l;i++)
            cin>>a[i];
        for(int j=0;j<m;j++)
            cin>>b[j];
        for(int k=0;k<n;k++)
            cin>>c[k];
            
        pos=0;
        for(int i=0;i<l;i++)
            for(int j=0;j<m;j++){
                sum[pos++]=a[i]+b[j];
            }
        sort(sum,sum+pos);
        
        
        int xx;
        cin>>xx;
        
        printf("Case %d:
",ll);
        
        for(int i=1;i<=xx;i++){
            int x,flag=0;
            cin>>x;            
            for(int i=0;i<n;i++){
                if(judge(x-c[i])){
                    flag=1;
                    break;
                }
            }
            
            if(flag)
                printf("YES
");
            else printf("NO
");
            
        }
    }
return 0;
}