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:
",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
");
            else
                printf("NO
");
        }
    }
    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
*/