Can you find it?（数组+二分hdu2141）

Can you find it?

Time Limit: 10000/3000 MS (Java/Others)    Memory Limit: 32768/10000 K (Java/Others) Total Submission(s): 8575    Accepted Submission(s): 2241

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

 

题意：给你三组数据，每组有L,N,M个数，每组取一个相加，问在他们的和数组里有没有x这个数？；

 

思路：数组+二分。。。。。

 

ps：http://acm.hdu.edu.cn/showproblem.php?pid=2141

 

详见代码：

#include<cstdio>

#include<cmath>

#include<cstdlib>

#include<iostream>

#include<algorithm>

using namespace std;



int cmp(int x,int y)

{

    return x<y;

}

__int64 num[300000];

int main()

{

    int S,ans,i,j,k,temp,t=1;

    int L,N,M,a[505],b[505],c[505],p;



    int left,right,middle;

    while(scanf("%d%d%d",&L,&N,&M)!=EOF)

    {

//        memset(num,0,sizeof(num));

        for(i=0;i<L;i++)

        {

            scanf("%d",&a[i]);

        }

//        sort(a,a+L,cmp);

        for(i=0;i<N;i++)

        {

            scanf("%d",&b[i]);

        }

//        sort(b,b+N,cmp);

        for(i=0;i<M;i++)

        {

            scanf("%d",&c[i]);

        }

//        sort(c,c+M,cmp);

        ans=0;

        for(i=0;i<L;i++)

            for(j=0;j<N;j++)

            {

                num[ans++]=a[i]+b[j];

            }

        sort(num,num+ans,cmp);

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

        scanf("%d",&S);

        while(S--)//for(i=0;i<S;i++)写成这样，，，那怪超时。。。顿时郁闷了！

        {

            temp=0;

            scanf("%d",&p);

            for(i=0;i<M;i++)

            {

                if(num[0]<=p-c[i]&&p-c[i]<=num[ans-1])

                {

                    left=0;

                    right=ans-1;



                    while(right-left>=0)

                    {

                        middle=(left+right)/2;



                        if(num[middle]>p-c[i])

                        {

                            right=middle-1;

                        }

                        else if(num[middle]<p-c[i])

                        {

                            left=middle+1;

                        }

                        else

                        {temp=1;break;}

                    }

                    

                }

                else

                    temp=0;

                if(temp)

                    break;

            }

            if(temp)

                printf("YES\n");

            else

                printf("NO\n");

        }

    }

    return 0;

}

/*

3 3 3

1 2 3

1 2 3

1 2 3

11

4

1

10

9

8

7

6

5

3

2

11

*/