hdu 2141 （二分）

链接：http://acm.hdu.edu.cn/showproblem.php?pid=2141

Can you find it?

Time Limit: 10000/3000 MS (Java/Others)    Memory Limit: 32768/10000 K (Java/Others)
Total Submission(s): 11503    Accepted Submission(s): 3021

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

 

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

这题写的不爽，不写题解了，看代码吧

 #include <stdio.h>
 #include <string.h>
 #include <stdlib.h>
 #include <iostream>
 #include <algorithm>

 using namespace std;

 long long sum[];
 int str1[],str2[],str3[];

 int Table(int l,int n,int m)
 {

     int cas=;
     for(int i=; i<l ;i++)
     {

         for(int j=; j<n; j++)
         {

             sum[cas++]=str1[i]+str2[j];
         }
     }
     return cas;
 }

 bool BSearch(long long sum[],int k,int cas)
 {

     int left=,right=cas-;
     while(left<=right)
     {

         int mid=(left+right)>>;
         if(sum[mid] == k) return true;
         else if(k < sum[mid]) right=mid-;
         else left = mid+;
     }
     return false;
 }

 int main()
 {

     int l,m,n,i,j;
     int ss=;
     while(scanf("%d%d%d",&l,&n,&m)!=EOF)
     {

         memset(str1,,sizeof(str1));
         memset(str2,,sizeof(str2));
         memset(str3,,sizeof(str3));

         for(i=; i<l; i++)
             scanf("%d",&str1[i]);
         for(j=; j<n; j++)
             scanf("%d",&str2[j]);
         for(i=; i<m; i++)
             scanf("%d",&str3[i]);
         int k,tmp;
         scanf("%d",&k);
         int cas=Table(l,n,m);
         sort(sum,sum+cas);
         printf("Case %d:\n",ss++);
         while(k--)
         {
                 scanf("%d",&tmp);
                 for(i=; i<m; i++)
                 {

                     if(BSearch(sum,tmp-str3[i],cas))
                     {

                         printf("YES\n");
                          break;
                     }
                 }
                 if(i>=m)printf("NO\n");
         }

     }
     return ;
 }