Can you find it?
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
1 #include<cstdio> 2 #include<algorithm> 3 using namespace std; 4 5 int p[3][505]; 6 long long temp[250010]; 7 int s[1005]; 8 9 /*int search(int l,int r,int num) 10 { 11 while(l<=r) 12 { 13 int mid=(l+r)/2; 14 if(temp[mid]==num) 15 return mid; 16 else if(temp[mid]>num) 17 r=mid-1; 18 else 19 l=mid+1; 20 } 21 return l; 22 }*/ 23 24 int main() 25 { 26 //freopen("in.txt","r",stdin); 27 int a,b,c,i,j,k,n; 28 int m=0; 29 while(scanf("%d%d%d",&a,&b,&c)!=EOF) 30 { 31 printf("Case %d: ",++m); 32 for(i=0;i<a;i++) 33 scanf("%d",&p[0][i]); 34 for(i=0;i<b;i++) 35 scanf("%d",&p[1][i]); 36 for(i=0;i<c;i++) 37 scanf("%d",&p[2][i]); 38 scanf("%d",&n); 39 for(i=0;i<n;i++) 40 scanf("%d",&s[i]); 41 int k=-1; 42 for(i=0;i<b;i++) 43 for(j=0;j<c;j++) 44 temp[++k]=p[1][i]+p[2][j]; 45 sort(temp,temp+k); 46 for(j=0;j<n;j++) 47 { 48 int target; 49 int ans; 50 for(i=0;i<a;i++) 51 { 52 target=s[j]-p[0][i]; 53 int ans=lower_bound(temp,temp+k,target)-temp; 54 //ans=search(0,k,target); 55 if(temp[ans]==target) 56 break; 57 } 58 if(i!=a) 59 printf("YES "); 60 else 61 printf("NO "); 62 } 63 } 64 65 }