Can you find it?

Time Limit: 10000/3000 MS (Java/Others) Memory Limit: 32768/10000 K (Java/Others)
Total Submission(s): 37762 Accepted Submission(s): 9221

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

思路：首先将任意两个数组两两相加，保存到一个数组中，这个数组一定要开大一写，最好是500*500，排序，然后加这个新的数组和第三个数组相加，利用二分的思想进行查找。

#include<cstdio>

#include<algorithm>

#include<iostream>

using namespace std;

int al[],an[],am[],d[];

int main()

{

    int l,n,m,s,flag=,s1;

    while(~scanf("%d%d%d",&l,&n,&m))

    {

        for(int i=;i<=l;i++) scanf("%d",&al[i]);

        for(int i=;i<=n;i++) scanf("%d",&an[i]);

        for(int i=;i<=m;i++) scanf("%d",&am[i]);

        int k=;

        for(int i=;i<=l;i++){

            for(int j=;j<=n;j++)

                d[k++]=al[i]+an[j];

        }

        sort(d+,d+k);

        scanf("%d",&s);

        printf("Case %d:\n",flag++);

        while(s--)

        {

            bool sign = false;

            scanf("%d",&s1);

            for(int i=;i<=m;i++)

            {

                int left=,right=k;

                while(left<=right)

                {

                    int middle = (left+right)/;

                    if(d[middle]+am[i]==s1) {sign = true; break;}

                    if(d[middle]+am[i]<s1) left = middle+;

                    else right= middle -;

                }

            }

            if(sign) cout<<"YES"<<endl;

            else cout<<"NO"<<endl;

        }

    }

    return ;

}

巴特西

HDU 2007-11 Programming Contest

Can you find it?

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