HDOJ-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): 23679    Accepted Submission(s): 5982

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
  
   
   
  
   
   


  
   
   
  
   
   

    
    
   直接三个循环必定会超时，所以用二分算法即可，a+b各种情况算出来，然后x-c二分查找就行了。
  
   
   
  
   
   

    
    
   看了半天，原来j写成i了。
  
   
   
  
   
   

    
    
   代码如下：
  
   
   
  
   
   

   
    
    
#include <cstdio>
#include <cstring>
#include <math.h>
#include <algorithm>
#include <iostream> 
using namespace std;
int l,m,n,t,flag,sum[510*510],a[510],b[510],c[510];
void cmp(int x)
{
    int mid,z=0,y=t-1;
    while(z<=y)
    {
        mid=(z+y)>>1;
        if(x>sum[mid]) z=mid+1;
        else if(x<sum[mid]) y=mid-1;
        else {flag=1;return ;}
    }
    return ;
}
int main()
{
	int i,j,cnt=1,x,s;
    while(scanf("%d%d%d",&l,&m,&n)!=EOF)
    {
        for(i=0;i<l;i++)
            scanf("%d",&a[i]);
        for(i=0;i<m;i++)
            scanf("%d",&b[i]);
        for(i=0;i<n;i++)
            scanf("%d",&c[i]);
        t=0;
        for(i=0;i<l;i++)
            for(j=0;j<m;j++)
                sum[t++]=a[i]+b[j];
        sort(sum,sum+t);
        scanf("%d",&s);
        printf("Case %d:\n",cnt++);
        while(s--)
        {
            scanf("%d",&x);
            flag=0;
            for(i=0;i<n;i++)
            {
                cmp(x-c[i]);
                if(flag)
                {
                    printf("YES\n");
                    break;
                }
            }
            if(!flag) printf("NO\n");
        }
    }
    return 0;
}