O - Can you find it?(二分查找)

O - Can you find it?

Time Limit:3000MS     Memory Limit:10000KB     64bit IO Format:%I64d & %I64u

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

//很简单的题目

运用二分的思想去查找 a 和 b 组合的数组，是否有等于要求的

 #include <iostream>
 #include <algorithm>
 #include <stdio.h>
 using namespace std;

 int a[];
 int b[];
 int c[];
 int sum[*];
 int x,y,z,num;//数组a，b，c的元素个数

 int find(int x)
 {
     int l=,r=num-;
     int mid;
     while (l<=r)
     {
         mid=(l+r)/;
         if (sum[mid]==x) return ;
         if (sum[mid]>x) r=mid-;
         if (sum[mid]<x) l=mid+;
     }
     return ;
 }

 int main()
 {
     int Case=;
     int i,j;
     while (scanf("%d%d%d",&x,&y,&z)!=EOF)
     {
         for(i=;i<x;i++)
             scanf("%d",&a[i]);
         for(i=;i<y;i++)
             scanf("%d",&b[i]);
         for(i=;i<z;i++)
             scanf("%d",&c[i]);
         sort(c,c+z);
         num=;
         for (i=;i<x;i++)
         {
             for (j=;j<y;j++)
             {
                 sum[num++]=a[i]+b[j];
             }
         }
         sort(sum,sum+num);
         int t,times;
         scanf("%d",&times);
         printf("Case %d:\n",++Case);
         while (times--)
         {
             scanf("%d",&t);
             for (i=;i<z;i++)
             {
                 if (find(t-c[i]))
                 {
                     printf("YES\n");
                     break;
                 }
             }
             if (i==z)
                 printf("NO\n");
         }
     }
     return ;
 }