Can you find it?——［二分查找］

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

 

解题思路：

　　首先我们考虑这个问题：给定两个序列A，B，和确定的数x，问是否存在i，j使满足A［i］＋B［j］＝x的？最快的方法是枚举A，然后在B中二分查找

x－A。现在回到这个问题，这道题给了三组序列A，B，C如何查找呢？我们不妨将A，C两数组合并成新数组LN，LN中每个元素都是Ai+Bj的和。然后枚举B，在LN中二分查找x－B。

＊这里有个细节：由于x为32位整数，当A+B>INT32_MAX时，可以不用加入LN.

代码如下：

 #include <iostream>
 #include <cstdio>
 #include <algorithm>
 #include <time.h>
 using namespace std;
 #define clock__ printf("%f\n",double(clock())/CLOCKS_PER_SEC);
 #define maxn 500
 #define INT_32_MAX ((1<<31)-1)
 typedef long long LL;
 int A[maxn+],B[maxn+],C[maxn+];
 int L,M,N,S;
 int LN[maxn*maxn+];
 int h;

 bool search_LN(int a,int b,int x){
     int len=b-a;
     int mid=a+len/;
     if(len==){
         if(LN[a]==x) return true;
         else return false;
     }
     if(LN[mid]==x) return true;
     else if(LN[mid]>x) return search_LN(a, mid, x);
     else return search_LN(mid, b, x);
 }

 int main() {

     int T=;
     while(scanf("%d%d%d",&L,&M,&N)==){
         printf("Case %d:\n",++T);
         for(int i=;i<L;i++)
             scanf("%d",&A[i]);
         for(int i=;i<M;i++)
             scanf("%d",&B[i]);
         for(int i=;i<N;i++)
             scanf("%d",&C[i]);
         h=;
         for(int i=;i<L;i++)
             for(int j=;j<N;j++){
                 if(LL(A[i])+C[j]<=INT_32_MAX)
                 LN[h++]=A[i]+C[j];
             }
         sort(LN, LN+h);
         scanf("%d",&S);
         for(int i=;i<=S;i++){
             int x;
             scanf("%d",&x);
             bool ok=;
             for(int j=;!ok&&j<M;j++){
                 if(search_LN(,h, x-B[j]))
                     ok=;
             }
             if(ok)printf("YES\n");
             else printf("NO\n");
         }
     }
     //clock__
     return ;
 }