计算A[i]+B[j]+C[k]=x
转换为A[i]+B[i]=x-C[k] ，这样在A[i]+B[i]中二分查找，x-C[k] ，不会超时。
题目来源：
http://acm.hdu.edu.cn/showproblem.php?pid=2141
Can you find it?
Time Limit: 10000/3000 MS (Java/Others) Memory Limit: 32768/10000 K (Java/Others)
Total Submission(s): 9548 Accepted Submission(s): 2523
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<iostream> 2 #include<stdio.h> 3 #include<string> 4 #include<string.h> 5 #include<algorithm> 6 7 using namespace std; 8 int a[505],b[505],c[505]; 9 int sab[250005]; 10 bool Bi_Search(int a[],int n,int b) 11 { 12 int l=0; 13 int r=n-1; 14 int mid; 15 while(l<=r) 16 { 17 mid=(l+r)>>1; 18 if(a[mid]==b) 19 return 1; 20 else if(a[mid]>b) 21 r=mid-1; 22 else 23 l=mid+1; 24 } 25 return 0; 26 } 27 int main() 28 { 29 int L,M,N; 30 int i,j,k; 31 int s; 32 int q,con=1; 33 34 while(cin>>L>>M>>N) 35 { 36 for( i=0;i<L;i++) 37 cin>>a[i]; 38 for( j=0;j<M;j++) 39 cin>>b[j]; 40 for( k=0;k<N;k++) 41 cin>>c[k]; 42 for(k=0,i=0;i<L;i++) 43 for(j=0;j<M;j++) 44 { 45 sab[k++]=a[i]+b[j]; 46 } 47 sort(sab,sab+k); 48 49 cin>>s; 50 cout<<"Case "<<con++<<":"<<endl; 51 while(s--) 52 { 53 cin>>q; 54 for(j=0;j<N;j++) 55 { 56 if(Bi_Search(sab,k,q-c[j])) 57 break; 58 } 59 if(j==N) 60 cout<<"NO"<<endl; 61 else 62 cout<<"YES"<<endl; 63 } 64 } 65 return 0 ; 66 }