Can you find it?

Time Limit: 10000/3000 MS (Java/Others)    Memory Limit: 32768/10000 K (Java/Others)

Total Submission(s): 17556    Accepted Submission(s): 4439

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

 

 

Author

wangye

 

 

Source

 

 

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//二分法， 查找的是有序数组的下标，  

1 #include 
      
        2 #include 
       
         3 using namespace std; 4 int a[550], b[550], c[550]; __int64 d[250050]; 5 int main() 6 { 7     int l,m ,n, cas=1; 8     while(cin >> l >> m >> n) 9     {10         11         int i, j, k, t=0;12         for(i=0; i
        
         > a[i];14         for(i=0; i
         
          > b[i];16         for(i=0; i
          
           > c[i];18         for(i=0; i
           
            > sum;28 int ok = 0;29 for(i=0; i
            
             =sum) //32 {33 int l=0, r=t-1, mid;34 while(r-l >= 0)35 {36 mid = (l+r)/2;37 if(c[i] + d[mid] > sum) r = mid -1;38 else if(d[mid] + c[i] < sum) l = mid + 1;39 else40 { ok = 1; break;} 41 }42 if(ok) break;43 }44 }45 if(ok)46 printf("YES\n");47 else48 printf("NO\n");49 }50 }51 return 0;52 }
            
           
          
         
        
       
      

//感觉二分法做的确实有点懵。