hdu 2141:Can you find it?（数据结构，二分查找）

Can you find it?

Time Limit: 10000/3000 MS (Java/Others)    Memory Limit: 32768/10000 K (Java/Others)
Total Submission(s): 9180    Accepted Submission(s): 2401

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

 

Author

wangye

 

Source

HDU 2007-11 Programming Contest

 

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　　数据结构：二分（折半）查找。

　　这道题一开始还在纳闷怎么用二分查找，后来看人家的思路才发现需要用巧办法，做法是先将前两个数列相加产生sab数列，这个时候sab+c = x，那么sab= x-c，每次询问x时，就用x减去c中的所有数，依次在sab数列中查找。

　　没想到这样的方法，脑子还是太木，不甘心啊 >_<

　　本题代码：

 

 #include <iostream>
 #include <algorithm>
 using namespace std;
 int qn;    //sab数组总数
 int sab[];
 int binsearch(int q[],int n,int k)    //二分查找
 {
     int left=,right=n,mid;
     while(left<=right){
         mid = (left+right)/;
         if(q[mid]==k)
             return mid;
         if(q[mid]>k)
             right = mid - ;
         else
             left = mid + ;
     }
     return ;
 }
 int main()
 {
     int l,n,m;
     int count = ;
     while(cin>>l>>n>>m){
         qn = ;
         int A[],B[],C[];
         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];
         }
         for(int i=;i<=l;i++)
             for(int j=;j<=n;j++)
                 sab[qn++] = A[i] + B[j];    //产生sab数列
         sort(sab+,sab+qn-);    //对sab数列进行排序
         int s;
         cin>>s;
         cout<<"Case "<<count++<<":"<<endl;
         while(s--){
             int t;
             cin>>t;
             int i;
             for(i=;i<=m;i++){
                 int tt = t - C[i];
                 if(binsearch(sab,qn-,tt)){    //查找有没有 x-c
                     cout<<"YES"<<endl;
                     break;
                 }
             }
             if(i>m)
                 cout<<"NO"<<endl;
         }
     }
     return ;
 }

Freecode : www.cnblogs.com/yym2013