A

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 31 2 31 2 31 2 331410

Sample Output

Case 1:NOYESNO

刚开始看到这道题时大部分人会想到用for循环来完成，但是肯定会超时的，所以我们就想到用二分法来完成，这虽然是一道二分的简单题，把二分法学会会用其实是很困难的。该题还要注意到flag的用法。

#include <iostream>#include <cstdio>#include<algorithm>using namespace std;const int MAX=500+5;int arr[MAX*MAX];int k,flag;void bs(int x){    int left,right,mid;    left=0,right=k-1;          while(left<=right)    {        mid=(left+right)>>1;     //要比单纯的（left+right）/2快很多        if(arr[mid]>x)            right=mid-1;        else if(arr[mid]<x)            left=mid+1;        else        {            flag=1;            return;        }    }    return;}int main(){    int L,N,M,Case=1;    while(~scanf("%d%d%d",&L,&N,&M))    {        int a[MAX],b[MAX],c[MAX],T,x;        for(int i=0;i<L;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]);        k=0;        for(int i=0;i<L;i++)            for(int j=0;j<N;j++)            arr[k++]=a[i]+b[j];        //本题的二分主要用在这儿把数组a[i]的和fa数组b[i]又        sort(arr,arr+k);        scanf("%d",&T);        printf("Case %d:\n",Case++);        while(T--)        {            flag=0;            scanf("%d",&x);            for(int i=0;i<M;i++)            {                 bs(x-c[i]);                 if(flag)                 {                  printf("YES\n");                  break;                 }            }            if(!flag)            printf("NO\n");        }    }    return 0;}