hdu 2141 Can you find it?

Can you find it?

Time Limit: 10000/3000 MS (Java/Others)    Memory Limit: 32768/10000 K (Java/Others)
Total Submission(s): 22306    Accepted Submission(s): 5624

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
   
   
   
   

 

Author

wangye

 

Source

HDU 2007-11 Programming Contest

比较典型的二分题，当然hash也是可行的；

#include <iostream>
#include <cstring>
#include <string>
#include <algorithm>
#include <cstdio>

using namespace std;
int L,M,N;
int a[505],b[505],c[505];
int sum[2500005];
bool binarysearch(int x,int cnt)
{
    int left=0,right=cnt-1,mid;
    while(left<=right)
    {
        mid=(left+right)/2;
        if(sum[mid]>x)
        {
            right=mid-1;
        }
        else if(sum[mid]<x)
        {
            left=mid+1;
        }
        else
        {
            return true;
        }
    }
    return false;
}
int main()
{
    int icase=0;
    while(scanf("%d %d %d",&L,&M,&N)!=EOF)
    {
        for(int i=0;i<L;++i)
        {
            scanf("%d",&a[i]);
        }
        for(int i=0;i<M;++i)
        {
            scanf("%d",&b[i]);
        }
        for(int i=0;i<N;++i)
        {
            scanf("%d",&c[i]);
        }
        int cnt=0;
        int S;
        scanf("%d",&S);
        for(int i=0;i<L;++i)
        {
            for(int j=0;j<M;++j)
            {
                sum[cnt++]=a[i]+b[j];
            }
        }
        printf("Case %d:\n",++icase);
        sort(sum,sum+cnt);
        int s;
        bool flag;
        for(int k=0;k<S;++k)
        {
            scanf("%d",&s);
            flag=0;
            for(int i=0;i<N;++i)
            {
                flag=binarysearch(s-c[i],cnt);
                if(flag) break;
            }
            if(flag) printf("YES\n");
            else printf("NO\n");
        }
    }
    return 0;
}