hdu 2141

                                            Can you find it?

Time Limit: 10000/3000 MS (Java/Others) Memory Limit: 32768/10000 K (Java/Others)

Total Submission(s): 24458 Accepted Submission(s): 6192

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

题目大意就是给你三个数组，然后再给一个数X,在这三个数组中分别找出一个数，是否存在使得Ai+Bj+Ck = X的数。

Code:

<span style="font-size:18px;">

#include <iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
bool binary_search1(__int64 D[],__int64 key,int len,__int64 temp)
{
    int begin=0;
    int end=len-1;
    int mid;
    while(begin<=end)
    {
        mid=((end-begin)>>1)+begin;
        if(temp+D[mid]==key)
            return true;
        else if(temp+D[mid]<key)
            begin=mid+1;
        else
            end=mid-1;
    }
    return false;
}
int main()
{
    int cas=1,L,M,N;
    __int64 A[505],B[505],C[505],D[505*505];//D数组的大小为n*n
    while(~scanf("%d%d%d",&L,&N,&M))
    {
        for(int i=0;i<L;i++)
            scanf("%I64d",&A[i]);
        for(int i=0;i<N;i++)
            scanf("%I64d",&B[i]);
        for(int i=0;i<M;i++)
            scanf("%I64d",&C[i]);
        int k=0;
        for(int i=0;i<N;i++){//用一个新的数组去保存B数组+C数组的值
            for(int j=0;j<M;j++){
                D[k++]=B[i]+C[j];
            }
        }
        sort(D,D+k);//排序
        int s;
        scanf("%d",&s);
        printf("Case %d:\n",cas++);
        while(s--)
        {
            __int64 key;
            int flag=0;
            scanf("%I64d",&key);
            for(int i=0;i<L;i++)//枚举A数组中的每一个数
            {
                if(binary_search1(D,key,k,A[i]))//然后再D数组中进行二分查找
                {
                    printf("YES\n");
                    flag=1;
                    break;
                }
            }
            if(!flag)
                printf("NO\n");
        }
    }
    return 0;
}</span>