hdoj-2141-Can you find it?

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 
     
     
     
     

 

题意很容易理解，很水的一道模拟题，给你L个数A，N个数B，M个数C，然后一个X，求Ai+Bj+Ck = X，我拿到题的第一瞬间想到暴力，可是O(N3)，1<=L, N, M<=500, 1<=S<=1000.是肯定会爆了，于是把第一个数组合第二个数组的所有组合线求出来，再将第三个数组排序，然后二分的去找，就ac了~~~

#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;

int num[250005];

bool tree(int a[],int x,int y)
{
    bool f=false;
    int lf,rig,mid;
    lf=0;
    rig=x-1;
    mid=(lf+rig)/2;
    //printf("1\n");
    while(lf<=rig)
    {
        //printf("1\n");
        if(a[mid]>y)
            {
                rig=mid-1;
            }
            else if(a[mid]<y)
            {
                lf=mid+1;
            }
        else{
            f=true;
            break;
        }
        mid=(lf+rig)/2;
    }
    return f;
}

int main()
{
    int ll[505],mm[505],nn[505];
    int l,n,m,s;
    int Cas=0;
    bool flag;
    while(scanf("%d%d%d",&l,&n,&m)!=EOF)
    {
        memset(ll,0,sizeof(ll));
        memset(nn,0,sizeof(nn));
        memset(mm,0,sizeof(mm));
        for(int i=0;i<l;i++)
            scanf("%d",&ll[i]);
        for(int i=0;i<n;i++)
            scanf("%d",&nn[i]);
        for(int i=0;i<m;i++)
            scanf("%d",&mm[i]);
        int k=0;
        for(int i=0;i<l;i++)
            for(int j=0;j<n;j++)
        {
            num[k++]=ll[i]+nn[j];
        }
        sort(num,num+k);
            Cas++;
            printf("Case %d:\n",Cas);
        scanf("%d",&s);
        while(s--)
        {
            flag=0;
            int x;
            scanf("%d",&x);
            for(int i=0;i<m;i++)
            {
                if(tree(num,k,x-mm[i]))
                    {
                        flag=1;
                        break;
                    }
            }
            if(flag) printf("YES\n");
            else printf("NO\n");
        }
    }
    return 0;
}