O-can you find it? (dichotomy)

Description

Give 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 is many cases. Every data case was described as followed:in the first line there be three integers L, N, M, in the second line there was L integers represent the sequence A, in the third line there is N integers represent the sequences B, in the Forth line There is M integers represent the sequence C. The fifth line there was an integer s represents there was s integers X to be calculated. 1<=l, N, m<=500, 1<=s<=1000. All the integers is 32-integers.

Output

For each case, firstly you has to print the case number as the form ' case D: ', then for the S queries, and you calculate if T He formula can is 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 problem and analysis

Given three array a,b,c, each array has a number of numbers (<=500), and given a number s to determine if there is a s=a[i]+b[j]+c[k], there is a set of numbers on the output Yes, a group does not exist on the output of No.

Because there are only three 500-size arrays, I just started to write three for loop, time out, and then look at the topic more clearly, each set of test data given in S is <=1000, it is easy to timeout, and, the number can be negative, so I sort all sorts, also useless.

Helpless under direct Baidu, only to find that already AC people are first to add a array and b array of numbers into a ab[500x500] array, so that the equivalent of ab[i]+c[j]=s; and then deformation, ab[i]=c[j]+s; so I know what I'm going to do next, As long as in the AB array with the previous two-point search to see if C[j]+s can find out, seems to have had before this kind of data relations to convert the problem.

The code is as follows:

#include<cstdio>
#include<cstdlib>
int a[501],b[501],c[501],d[260000];int kNM;
IntCmp(Constvoid*a,Constvoid*b)
{
Return*(Int*) A-*(Int*) b;
}
IntMain(void)
{
int T, SMm=1;
While(scanf("%d%d%d", &k, &n, &m)!=eof)
{For(int I=0; I<k; I++)scanf("%d", &a[I]);
For(int I=0; I<n; I++)scanf("%d", &b[I]);
For(int I=0; I<m; I++)scanf("%d", &c[I]);
int cnt=0;
For(int I=0; I<n; I++)
For(Int J=0; j<m; j++)
D[CNT++]=b[I]+c[j];
Qsort(AK,sizeof(A[0]), CMP);
Qsort(dCnt,sizeof(d[0]), CMP);
scanf("%d", &t);
Printf("Case%d:\ n"Mm++);
While(t--)
{
int flag=0;
scanf("%d", &s);
For(int I=0; I<k; I++)
{
int TL=0Tr=cnt-1;
int TT=s-A[I];
While(TL<tr)
{
int mid= (TL+tr)/2;
If(d[TL]==tt|| D[TR]==tt|| D[Mid]==tt){Flag=1;Break;}
ElseIf(d[Mid]<tt) TL=mid+1;
else TR=mid-1;
}
If(Flag==1)Break;
}
If(Flag==1) puts< Span class= "Sh-symbol" > ( "YES"              else if (Flag==0puts ( "NO"           }
      }
     return 0
/span>

O-can you find it? (dichotomy)