杭电 2141 Can you find it? （二分法）

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

定义三个数组分别存a,b,c的值，再定义数组d存a和b所有可能的和，用给定的数减去c数组中的数得到k，用二分法在d数组找k值，若能找到输出yes，否则输出no。

#include<cstdio>
#include<algorithm>
using namespace std;
int a[505],b[505],c[505],d[505*505];
int main()
{
    int j,l,p,m,n,i,k,t,s,num,r,le,mid,key;
    num=0;
    while(scanf("%d %d %d",&l,&m,&n)!=EOF)
    {
        for(i = 0 ; i < l ; i++)
        {
            scanf("%d",&a[i]);
        }
        for(i = 0 ; i < m ; i++)
        {
            scanf("%d",&b[i]);
        }
        for(i = 0 ; i < n ; i++)
        {
            scanf("%d",&c[i]);
        }    
        p=0;
        for(i = 0 ; i < l ; i++)
        {
            for(j = 0 ; j < m ; j++)
            {
                d[p++]=a[i]+b[j];
            }
        }
        sort(c,c+n);
        sort(d,d+p);
        scanf("%d",&t);
        printf("Case %d:
",++num);
        while(t--)
        {
            scanf("%d",&s);
            key=0;
            for(i = 0 ; i < n ; i++)
            {
                k=s-c[i];
                le=0;
                r=p-1;                                                      
                while(le <= r)
                {
                    mid=(le+r)/2;
                    if(k == d[mid])
                    {
                        key=-1;
                        break;
                    }
                    if(k > d[mid])
                    {
                        le=mid+1;
                    }
                    else
                    {
                        r=mid-1;
                    }
                  }
                 if(key == -1) break;
            }
            if(key == -1) printf("YES
");
            else
                printf("NO
");
        }
    }
}