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Can you find it?

Time Limit: 10000/3000 MS (Java/Others)    Memory Limit: 32768/10000 K (Java/Others)
Total Submission(s): 29184    Accepted Submission(s): 7286

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

题意：给出三个数列ABC，从ABC各取一个整数求是否能够等于X，是则输出YES，否NO。

三组500直接查找会超时

可将A+B+C=X转换为A+B=X-C

之后即可将等式两边看做两个整体进行二分查找。

AC代码：

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 
 4 int a[510],b[510],c[510];
 5 int num[250010];
 6 
 7 int bin(int q[],int x,int y){
 8     int left=0,right=x,mid;
 9     while(left<=right){
10         mid=(left+right)/2;
11         if(q[mid]==y)
12         return 1;
13         if(q[mid]>y)
14         right=mid-1;
15         else
16         left=mid+1;
17     }
18     return 0;
19 }
20 
21 int main(){
22     int l,n,m,s,ans=0;
23     while(cin>>l>>n>>m){
24     ans++;
25     for(int i=0;i<l;i++){
26         cin>>a[i];
27     }
28     for(int i=0;i<n;i++){
29         cin>>b[i];
30     }
31     for(int i=0;i<m;i++){
32         cin>>c[i];
33     }
34     cin>>s;
35     int x,cnt=0;
36     
37     for(int i=0;i<l;i++){
38         for(int j=0;j<n;j++){
39             num[cnt++]=a[i]+b[j];
40         }
41     }
42     sort(num,num+cnt);
43     printf("Case %d:\n",ans);
44     while(s--){
45         cin>>x;
46         int flag=0;
47         for(int i=0;i<m;i++){
48             int temp=x-c[i];
49             if(bin(num,cnt-1,temp)){
50                 cout<<"YES"<<endl;
51                 flag=1;
52                 break;
53             }
54         }
55         if(!flag)
56         cout<<"NO"<<endl;
57     }
58 }
59     return 0;
60 } 

hdu-2141

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原文地址：http://www.cnblogs.com/Kiven5197/p/6711106.html