HDU 2141 Can you find it？

　　　　　　　　　　Can you find it?

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

#include<cstdio>
#include<algorithm>
using namespace std;

int p[3][505];
long long temp[250010];
int s[1005];

/*int search(int l,int r,int num)
{
    while(l<=r)
    {
        int mid=(l+r)/2;
        if(temp[mid]==num)
        return mid;
        else if(temp[mid]>num)
        r=mid-1;
        else
        l=mid+1;
    }
    return l;
}*/

int main()
{
    //freopen("in.txt","r",stdin);
    int a,b,c,i,j,k,n;
    int m=0;
    while(scanf("%d%d%d",&a,&b,&c)!=EOF)
    {
        printf("Case %d:\n",++m); 
        for(i=0;i<a;i++)
        scanf("%d",&p[0][i]);
        for(i=0;i<b;i++)
        scanf("%d",&p[1][i]);
        for(i=0;i<c;i++)
        scanf("%d",&p[2][i]);
        scanf("%d",&n);
        for(i=0;i<n;i++)
        scanf("%d",&s[i]);
        int k=-1;
        for(i=0;i<b;i++)
        for(j=0;j<c;j++)
        temp[++k]=p[1][i]+p[2][j];
        sort(temp,temp+k);
        for(j=0;j<n;j++)
        {
            int target;
            int ans;
            for(i=0;i<a;i++)
            {
                target=s[j]-p[0][i]; 
                int ans=lower_bound(temp,temp+k,target)-temp;
                //ans=search(0,k,target);
                if(temp[ans]==target) 
                break;
            }
            if(i!=a)
            printf("YES\n");
            else
            printf("NO\n");    
        }
    } 
    
}