hdu 2141

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//这题的解题思路是将a b 中的元素每种相加可能的值都存放于一个数组中，如a[]={1,2,3};b={1,2} 则 ab[]={2,3,3,4,4,5};
//将ab[]中的元素进行升序排序
//将要查找的值分别减去c[]中的一个元素，然后再在ab[]中用二分查找的方式进行搜索，这样要搜索的值就有c[]元素的个数
//只要有一个值能在ab[]中查找成功，则输出 yes 反之输出 no

// 468 ms 1940 kb
#include <iostream>
#include <vector>
#include <algorithm>
#include <stdio.h>
using namespace std;

vector<int> ab;
int search(int search_num){

   int low=0;
   int high=ab.size()-1;
   int mid;
   while(low<=high){
      mid=(low+high)/2;

      if(search_num<ab[mid]) high=mid-1;

      if(search_num>ab[mid]) low=mid+1;

      if(search_num==ab[mid]) return 1;
   }
return 0;
}
int main(){
    int x;
    vector<int> a;
    vector<int> b;

int l,n,m;

    int s;
    int cases=0;
     a.reserve(500);
     b.reserve(500);
     ab.reserve(250000);

     int c[550];
while(cin>>l>>n>>m){

      a.clear();
      b.clear();
      ab.clear();
      for(int i=0;i<l;i++){
         scanf("%d",&x);
         a.push_back(x);
      }
      for(int i=0;i<n;i++){
         scanf("%d",&x);
         b.push_back(x);
      }
      for(int i=0;i<m;i++){
         scanf("%d",&c[i]);
      }
      for(int i=0;i<l;i++)
        for(int j=0;j<n;j++)
          ab.push_back(a[i]+b[j]);

      sort(ab.begin(),ab.end());

      cout<<"Case "<<++cases<<":"<<endl;
      cin>>s;
      for(int i=0;i<s;i++){
          scanf("%d",&x);
          int flag;
            for(int j=0;j<m;j++){

                flag=search(x-c[j]);
                if(flag) break;
            }
             if(flag)
                cout<<"YES"<<endl;
             else cout<<"NO"<<endl;
      }
}
return 0;
}

Can you find it?

Time Limit: 10000/3000 MS (Java/Others) Memory Limit: 32768/10000 K (Java/Others)
Total Submission(s): 3764 Accepted Submission(s): 939

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 31 2 31 2 31 2 331410