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：

 

 1 #include <iostream>

 2 #include <algorithm>

 3 using namespace std;  4 int qn;    //sab数组总数 

 5 int sab[250005];  6 int binsearch(int q[],int n,int k)    //二分查找 

 7 {  8     int left=1,right=n,mid;  9     while(left<=right){ 10         mid = (left+right)/2; 11         if(q[mid]==k) 12             return mid; 13         if(q[mid]>k) 14             right = mid - 1; 15         else

16             left = mid + 1; 17  } 18     return 0; 19 } 20 int main() 21 { 22     int l,n,m; 23     int count = 1; 24     while(cin>>l>>n>>m){ 25         qn = 1; 26         int A[501],B[501],C[501]; 27         for(int i=1;i<=l;i++){ 28             cin>>A[i]; 29  } 30         for(int i=1;i<=n;i++){ 31             cin>>B[i]; 32  } 33         for(int i=1;i<=m;i++){ 34             cin>>C[i]; 35  } 36         for(int i=1;i<=l;i++) 37             for(int j=1;j<=n;j++) 38                 sab[qn++] = A[i] + B[j];    //产生sab数列 

39         sort(sab+1,sab+qn-1);    //对sab数列进行排序 

40         int s; 41         cin>>s; 42         cout<<"Case "<<count++<<":"<<endl; 43         while(s--){ 44             int t; 45             cin>>t; 46             int i; 47             for(i=1;i<=m;i++){ 48                 int tt = t - C[i]; 49                 if(binsearch(sab,qn-1,tt)){    //查找有没有 x-c 

50                     cout<<"YES"<<endl; 51                     break; 52  } 53  } 54             if(i>m) 55                 cout<<"NO"<<endl; 56  } 57  } 58     return 0; 59 }

 

Freecode : www.cnblogs.com/yym2013