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
 
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这题写的不爽，不写题解了，看代码吧
 1 #include <stdio.h>
 2 #include <string.h>
 3 #include <stdlib.h>
 4 #include <iostream>
 5 #include <algorithm>
 6 
 7 using namespace std;
 8 
 9 long long sum[250005];
10 int str1[505],str2[505],str3[505];
11 
12 int Table(int l,int n,int m)
13 {
14 
15     int cas=0;
16     for(int i=0; i<l ;i++)
17     {
18 
19         for(int j=0; j<n; j++)
20         {
21 
22             sum[cas++]=str1[i]+str2[j];
23         }
24     }
25     return cas;
26 }
27 
28 bool BSearch(long long sum[],int k,int cas)
29 {
30 
31     int left=0,right=cas-1;
32     while(left<=right)
33     {
34 
35         int mid=(left+right)>>1;
36         if(sum[mid] == k) return true;
37         else if(k < sum[mid]) right=mid-1;
38         else left = mid+1;
39     }
40     return false;
41 }
42 
43 int main()
44 {
45 
46     int l,m,n,i,j;
47     int ss=1;
48     while(scanf("%d%d%d",&l,&n,&m)!=EOF)
49     {
50 
51         memset(str1,0,sizeof(str1));
52         memset(str2,0,sizeof(str2));
53         memset(str3,0,sizeof(str3));
54 
55         for(i=0; i<l; i++)
56             scanf("%d",&str1[i]);
57         for(j=0; j<n; j++)
58             scanf("%d",&str2[j]);
59         for(i=0; i<m; i++)
60             scanf("%d",&str3[i]);
61         int k,tmp;
62         scanf("%d",&k);
63         int cas=Table(l,n,m);
64         sort(sum,sum+cas);
65         printf("Case %d:\n",ss++);
66         while(k--)
67         {
68                 scanf("%d",&tmp);
69                 for(i=0; i<m; i++)
70                 {
71 
72                     if(BSearch(sum,tmp-str3[i],cas))
73                     {
74 
75                         printf("YES\n");
76                          break;
77                     }
78                 }
79                 if(i>=m)printf("NO\n");
80         }
81         
82     }
83     return 0;
84 }
View Code