HDOJ 2141 Can you find it?（二分）

Can you find it?

Time Limit: 10000/3000 MS (Java/Others)    Memory Limit: 32768/10000 K (Java/Others)
Total Submission(s): 16442    Accepted Submission(s): 4178

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
   
   
   
   

 

ac代码：

#include<stdio.h>
#include<algorithm>
#include<iostream>
using namespace std;
int a[555],b[555],c[555],num[255555];//最好定义在外面吧
int main()
{
    int n,m,k,sum,t,max,i,j;
    int Case=1;
    while(scanf("%d%d%d",&n,&m,&k)!=EOF)
    {
        for(i=0;i<n;i++)
        scanf("%d",&a[i]);
        for(i=0;i<m;i++)
        scanf("%d",&b[i]);
        for(i=0;i<k;i++)
        scanf("%d",&c[i]);
        sort(c,c+k);
        int q=0;
        for(i=0;i<n;i++)//先把前两排的数加起来
        for(j=0;j<m;j++)
        num[q++]=a[i]+b[j];
        sort(num,num+q);//排序
        scanf("%d",&t);
        printf("Case %d:\n",Case++);
        while(t--)
     { 
        int bz=0;
        scanf("%d",&max);
        for(i=0;i<k;i++)
        {
            sum=max-c[i];
            int low=0;
            int high=q-1;
            int mid=(low+high)/2;
            while(low+1<high)//二分求
            {
                if(num[mid]>sum)
                high=mid;
                else if(num[mid]<sum)
                low=mid;
                else
                {
                    bz=1;
                    break;
                }
                mid=(low+high)/2;
            }
            if(num[low]==sum||num[high]==sum)
            {
                bz=1;
            }
            if(bz)
            {
            break;
            }
        }
        if(bz==0)
        printf("NO\n");
        else
        printf("YES\n");
     }
    }
    return 0;
}