HDU2141——二分——Can you find it?

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

/*
把两个加在一起，和第三个二分
换下顺序就A了。。否则TLE

*/
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

const int MAX = 500 + 10;
int L[MAX], N[MAX], M[MAX];
int LN[1000100],Ln[1000100];
int l, n, m, q;

int main()
{
int cas = 0;
while(~scanf("%d%d%d", &l, &n, &m)){
for(int i = 1; i <= l; i++)
scanf("%d", &L[i]);
for(int i = 1; i <= n; i++)
scanf("%d", &N[i]);
for(int i = 1; i <= m; i++)
scanf("%d", &M[i]);
int cout = 1;
for(int i = 1; i <= l; i++){
for(int j = 1; j <= n; j++){
LN[cout++] = L[i] + N[j];
}
}
sort(LN + 1, LN + cout);
int cout1 = 1;
for(int i = 1; i < cout; i++){
if(LN[i] != LN[i+1]) ;
Ln[cout1++] = LN[i];
}
scanf("%d", &q);
cas++;
printf("Case %d:\n", cas);
int p;
for(int i = 1; i <= q; i++){
scanf("%d", &p);
int flag = 0;
for(int  j = 1; j <= m; j++){
int ll = 1, rr = cout1 - 1;
while(ll <= rr){
int mid = (ll + rr) >> 1;
if(Ln[mid] + M[j] == p){
//    printf("%d %d\n", mid, M[j]);
flag = 1;
break;
}
else if(Ln[mid] + M[j] > p)
rr = mid - 1;
else ll = mid + 1;
}
if(flag == 1) break;
}
if(flag == 1) printf("YES\n");
else printf("NO\n");
}
}
return 0;
}