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>1){
        mid=(r+l)/2;
        if(a[mid]<=k)
            l=mid;
        else
            r=mid;
    }
    if(a[l]==k)
        return 1;
    else
        return 0;
}//二分查找
int main()
{
    int i,j,count=1,q;
    int L[K],N[K],M[K],s,n,m,l;
    while(~scanf("%d%d%d",&l,&n,&m)){
        int h=0;
        for(i=0;i<l;i++)
            scanf("%d",&L[i]);
        for(i=0;i<n;i++)
            scanf("%d",&N[i]);
        for(i=0;i<m;i++)
            scanf("%d",&M[i]);
        for(i=0;i<l;i++)
          for(j=0;j<n;j++)
             LN[h++]=L[i]+N[j];
         sort(LN,LN+h);
         scanf("%d",&s);
         printf("Case %d:\n",count++);
        for(i=0;i<s;i++)
        {
            scanf("%d",&q);
            int p=0; 
            for(j=0;j<m;j++)
            {
                int a=q-M[j]; 
                if(binarysearch(LN,0,h,a)) 
                {
                    printf("YES\n");
                    p=1;
                    break;
                }
            }
            if(!p) 
              printf("NO\n");
        }
    }
    return 0;
}