hdu2141

Can you find it?

Time Limit: 10000/3000 MS (Java/Others)    Memory Limit: 32768/10000 K (Java/Others)
Total Submission(s): 8566    Accepted Submission(s): 2237

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

 

思路：二分法，将Ai+Bj的值存入数组sum中，然后用二分法检测（X-sum[i]）是否等于Ck

 

代码：

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int a[510],b[510],c[510],sum[260000],k;
int main()
{
    int i,j,x,cnt=1,s,l,m,n,flag,tag,left,right,mid;
    while(scanf("%d%d%d",&l,&n,&m)!=EOF)
    {
        for(i=0; i<l; i++)
            scanf("%d",&a[i]);
        for(i=0; i<n; i++)
            scanf("%d",&b[i]);
        for(i=0; i<m; i++)
            scanf("%d",&c[i]);
        k=0;
        for(i=0; i<l; i++)
            for(j=0; j<n; j++)
                sum[k++]=a[i]+b[j];
        sort(sum,sum+k);
        scanf("%d",&s);
        printf("Case %d: ",cnt++);
        while(s--)
        {
            scanf("%d",&x);
            flag=0;
            for(i=0; i<m; i++)
            {
                int t=x-c[i];
                left=0;
                right=k-1;
                tag=0;
                while(left<=right)
                {
                    mid=(left+right)>>1;
                    if(sum[mid]>t)
                        right=mid-1;
                    else if(sum[mid]<t)
                        left=mid+1;
                    else
                    {
                        tag=1;
                        break;
                    }
                }
                if(tag)
                {
                    flag=1;
                    break;
                }
            }
            if(flag==1)
                printf("YES ");
            else
                printf("NO ");
        }
    }
    return 0;
}