Can you find it?（hdu 2141 二分查找）

Can you find it?

Time Limit: 10000/3000 MS (Java/Others)    Memory Limit: 32768/10000 K (Java/Others)
Total Submission(s): 19416    Accepted Submission(s): 4891

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

 

lower_bound提交就会出错，binary_search()也可以过

 #include <iostream>
 #include <cstring>
 #include <cstdio>
 #include <algorithm>
 #include <set>
 using namespace std;
 int a[],b[],c[],x;
 __int64 su[*];
 int l,n,m,s,u;
 int d=,sum;
 bool solve(int x)
 {
     int i,k;
     int ans;
     for(i=;i<m;i++)
     {
         ans=x-c[i];
         int l=,r=u-,mid,e;
         while(l<=r)
         {
             mid=(l+r)/;
             if(su[mid]==ans)    return ;
             else if(su[mid]>ans)    r=mid-;
             else l=mid+;
         }
     }
     return ;
 }
 int main()
 {
     int i,j;
     freopen("in.txt","r",stdin);
     while(scanf("%d%d%d",&l,&n,&m)!=EOF)
     {
         for(i=;i<l;i++)    scanf("%d",&a[i]);
         for(i=;i<n;i++)    scanf("%d",&b[i]);
         for(i=;i<m;i++)    scanf("%d",&c[i]);
         scanf("%d",&s);
         u=;
         for(i=;i<l;i++)
             for(j=;j<n;j++)
                 su[u++]=a[i]+b[j];
         sort(su,su+u);
         printf("Case %d:\n",d++);
         for(i=;i<s;i++)
         {
             scanf("%d",&x);
             if(solve(x))    printf("YES\n");
             else     printf("NO\n");
         }
     }
 }