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): 33248 Accepted Submission(s): 8241

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 Input3 3 31 2 31 2 31 2 331410Sample OutputCase 1:NOYESNOAuthorwangyeSourceHDU 2007-11 Programming Contest

#include<stdio.h>#include<iostream>#include<algorithm>#include<math.h>using namespace std;int main(){    long long l,m,n,cnt=1;    while(cin>>l>>m>>n)    {    long long a[510],b[510],c[510],num[250010];    for(int i=0;i<l;i++)//输入三个数组     cin>>a[i];    for(int i=0;i<m;i++)    cin>>b[i];    for(int i=0;i<n;i++)    cin>>c[i];    int k=0;    for(int i=0;i<l;i++)        for(int j=0;j<m;j++)            num[k++]=a[i]+b[j];//前两个数组两两相加得到新数组     sort(num,num+k);//从小到大排序     int s;    cin>>s;    cout<<"Case "<<cnt++<<":"<<endl;//输出第几个测试实例     while(s--)    {        long long x;        cin>>x;        bool st =false;        for(int i=0;i<n;i++)//第三个数组和新数组mid相加判断         {        int l=0,r=k-1;        while(l<=r)        {            int mid=(l+r)>>1;            if(num[mid]+c[i]==x)            {                st=true;                break;            }            else if(num[mid]+c[i]<x)                l=mid+1;            else                r=mid-1;        }           if(st)            break;        }        if(st)        cout<<"YES"<<endl;        else        cout<<"NO"<<endl;    }       }    return 0;}