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