HDU 2141 Can you find it? （二分）

Can you find it?

Time Limit: 10000/3000 MS (Java/Others)    Memory Limit: 32768/10000 K (Java/Others)
Total Submission(s): 22628    Accepted Submission(s): 5724

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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题解：给你三组A、B、C的值和S值，问是否能找到Ai、Bj、Ck，使得Ai+Bj+Ck = S。
           二分，先将a数组和b数组，再相加成一个数组ab[500*500]。这样就相当于ab[i] + c[j] = s。
           再变形一下， ab[i] = s - c[j].
         只要在ab数组中用二分查找是否存在s-c[j]就可以了。

AC代码：

#include<iostream>
#include<algorithm>
#include<stdio.h>
using namespace std;
int a[505],b[505],c[505];
int ab[250005];
bool Binary_Search( int a, int l , int R )
{
	//二分
   if( l > R )
     return false;
   int mid = ( l + R ) / 2;
    if (ab[mid] == a )
    return true;
   else if( ab[mid] > a)
    Binary_Search( a, l, mid-1);
   else
    Binary_Search( a, mid+1, R);
}
int main()
{
 	int l,n,m;
 	int i,j,k;
 	int s,sum,cnt = 1;
 	while( cin >> l >> n >> m )
 	{
 	 	
		for( i = 0; i < l; i++)
  		 cin >> a[i];
  		 
 		 for( i = 0; i < n; i++)
  		 cin >> b[i];
  		 
 		 for( i = 0; i < m; i++)
   		 cin >> c[i];
   		
 		 for(k=0, i=0;i<l; i++)
  		 {
  		 	for( j=0; j<n; j++)
            {
            	ab[k++] = a[i] + b[j];
			}
		   }
      	sort( ab, ab + k );
 		 cin >> s;
  		cout << "Case " << cnt++ << ":"<<"\n";
 		 for( i = 0; i < s; i++ )
 		 {
  			 cin >> sum;
   			for( j = 0; j < m; j++)
    		if ( Binary_Search(sum-c[j] , 0 , k-1) ) //查找是否有满足的和
    		 break;
  			 if( j == m)
   			 puts("NO"); 
  			 else
    			puts("YES");
  		}
	}
 		return 0;
}