hdu-2141

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：

 #include<bits/stdc++.h>
 using namespace std;
 
 int a[],b[],c[];
 int num[];
 
 int bin(int q[],int x,int y){
     int left=,right=x,mid;
     while(left<=right){
         mid=(left+right)/;
         if(q[mid]==y)
         return ;
         if(q[mid]>y)
         right=mid-;
         else
         left=mid+;
     }
     return ;
 }
 
 int main(){
     int l,n,m,s,ans=;
     while(cin>>l>>n>>m){
     ans++;
     for(int i=;i<l;i++){
         cin>>a[i];
     }
     for(int i=;i<n;i++){
         cin>>b[i];
     }
     for(int i=;i<m;i++){
         cin>>c[i];
     }
     cin>>s;
     int x,cnt=;
 
     for(int i=;i<l;i++){
         for(int j=;j<n;j++){
             num[cnt++]=a[i]+b[j];
         }
     }
     sort(num,num+cnt);
     printf("Case %d:\n",ans);
     while(s--){
         cin>>x;
         int flag=;
         for(int i=;i<m;i++){
             int temp=x-c[i];
             if(bin(num,cnt-,temp)){
                 cout<<"YES"<<endl;
                 flag=;
                 break;
             }
         }
         if(!flag)
         cout<<"NO"<<endl;
     }
 }
     return ;
 }