hdu 2141 二分查找，三个数相加

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

 

#include<iostream>
#include<stdio.h>
#include<string>
#include<string.h>
#include<algorithm>

using namespace std;
int a[505],b[505],c[505];
int sab[250005];
bool Bi_Search(int a[],int n,int b)
{
    int l=0;
    int r=n-1;
    int mid;
    while(l<=r)
    {
        mid=(l+r)>>1;
        if(a[mid]==b)
            return 1;
        else if(a[mid]>b)
            r=mid-1;
        else
            l=mid+1;
    }
    return 0;
}
int main()
{
    int L,M,N;
        int i,j,k;
        int s;
        int q,con=1;

    while(cin>>L>>M>>N)
    {
        for( i=0;i<L;i++)
            cin>>a[i];
        for( j=0;j<M;j++)
            cin>>b[j];
        for( k=0;k<N;k++)
            cin>>c[k];
        for(k=0,i=0;i<L;i++)
            for(j=0;j<M;j++)
            {
                sab[k++]=a[i]+b[j];
            }
        sort(sab,sab+k);

        cin>>s;
        cout<<"Case "<<con++<<":"<<endl;
        while(s--)
        {
            cin>>q;
            for(j=0;j<N;j++)
            {
                if(Bi_Search(sab,k,q-c[j]))
                   break;
            }
            if(j==N)
                cout<<"NO"<<endl;
            else
                cout<<"YES"<<endl;
        }
    }
    return 0 ;
}