Can you find it?

Time Limit : 10000/3000ms (Java/Other)   Memory Limit : 32768/10000K (Java/Other)

Total Submission(s) : 25   Accepted Submission(s) : 7

Problem 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

题意：分别有a,b,c三个数组，给出一个数，看这个数能不能再a,b,c,里面分别找出一个数求和得到

题解：先把两个数组加起来，另外一个数组用二分；

代码：

1 #include
       
         2 #include
        
          3 #include
         
           4 using namespace std; 5 __int64 D[1000010],A[510],B[510],C[510],L,N,M;//D数组要开大；；；  6 int erfen(__int64 *a,__int64 c,__int64 d){  7     int l,r,mid; 8     l=0;r=L*N-1;////  9     while(l<=r){10         mid=(l+r)/2;11         if(a[mid]+c
         
        
       

 stl:

1 #include
       
         2 #include
        
          3 #include
         
           4 using namespace std; 5 __int64 D[1000010],A[510],B[510],C[510],L,N,M;//D数组要开大；；；  6 /*int erfen(__int64 *a,__int64 c,__int64 d){  7     int l,r,mid; 8     l=0;r=L*N-1;  9     while(l<=r){10         mid=(l+r)/2;11         if(a[mid]+c
         
        
       

 map和set做全都MLE了，无奈了，别人的都对的；

set代码:

1 #include
       
         2 #include
        
          3 using namespace std; 4 const int MAXN=505; 5 int A[MAXN],B[MAXN],C[MAXN]; 6 int main(){ 7     int L,N,M,S,X,flot=0,ok; 8     while(~scanf("%d%d%d",&L,&N,&M)){
     set
         
          num; 9         for(int i=0;i
         
        
       

map代码：

1 #include
       
         2 #include