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         else16         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