hdu 2141 Can you find it?（二分查找变例）

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

 

初看很简单就是判断三个数加起来的和，但是三个for循环速度还是很慢的，所以细想要用到二分查找。

把函数改为：A+B=X-C，然后二分搜一下就可以了。

完全用的是二分查找的模板。

 #include <stdio.h>
 #include <math.h>
 #include <queue>
 #include <vector>
 #include <stack>
 #include <map>
 #include <string>
 #include <string.h>
 #include <algorithm>
 #include <iostream>
 using namespace std;
 #define K 505
 int LN[K*K];
 int binarysearch(int a[],int l,int r,int k){
     int mid;
     while(r-l>){
         mid=(r+l)/;
         if(a[mid]<=k)
             l=mid;
         else
             r=mid;
     }
     if(a[l]==k)
         return ;
     else
         return ;
 }//二分查找
 int main()
 {
     int i,j,count=,q;
     int L[K],N[K],M[K],s,n,m,l;
     while(~scanf("%d%d%d",&l,&n,&m)){
         int h=;
         for(i=;i<l;i++)
             scanf("%d",&L[i]);
         for(i=;i<n;i++)
             scanf("%d",&N[i]);
         for(i=;i<m;i++)
             scanf("%d",&M[i]);
         for(i=;i<l;i++)
           for(j=;j<n;j++)
              LN[h++]=L[i]+N[j];
          sort(LN,LN+h);
          scanf("%d",&s);
          printf("Case %d:\n",count++);
         for(i=;i<s;i++)
         {
             scanf("%d",&q);
             int p=;
             for(j=;j<m;j++)
             {
                 int a=q-M[j];
                 if(binarysearch(LN,,h,a))
                 {
                     printf("YES\n");
                     p=;
                     break;
                 }
             }
             if(!p)
               printf("NO\n");
         }
     }
     return ;
 }