Hdu6085 Rikka with Candies（2017多校第5场）

Rikka with Candies

Time Limit: 7000/3500 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1180    Accepted Submission(s): 510

Problem Description

As we know, Rikka is poor at math. Yuta is worrying about this situation, so he gives Rikka some math tasks to practice. There is one of them:

There are  n  children and  m  kinds of candies. The  i th child has  Ai  dollars and the unit price of the  i th kind of candy is  Bi . The amount of each kind is infinity. 

Each child has his favorite candy, so he will buy this kind of candies as much as possible and will not buy any candies of other kinds. For example, if this child has  10  dollars and the unit price of his favorite candy is  4  dollars, then he will buy two candies and go home with  2  dollars left.

Now Yuta has  q  queries, each of them gives a number  k . For each query, Yuta wants to know the number of the pairs  (i,j)(1≤i≤n,1≤j≤m)  which satisfies if the  i th child’s favorite candy is the  j th kind, he will take  k  dollars home.

To reduce the difficulty, Rikka just need to calculate the answer modulo  2 .

But It is still too difficult for Rikka. Can you help her?

 

Input

The first line contains a number  t(1≤t≤5) , the number of the testcases. 

For each testcase, the first line contains three numbers  n,m,q(1≤n,m,q≤50000) . 

The second line contains  n  numbers  Ai(1≤Ai≤50000)  and the third line contains  m  numbers  Bi(1≤Bi≤50000) .

Then the fourth line contains  q  numbers  ki(0≤ki<maxBi)  , which describes the queries.

It is guaranteed that  Ai≠Aj,Bi≠Bj  for all  i≠j .

 

Output

For each query, print a single line with a single  01  digit -- the answer.

 

Sample Input

  

   1
5 5 5
1 2 3 4 5
1 2 3 4 5
0 1 2 3 4
  

 

Sample Output

  

   0
0
0
0
1
  

 

Source

2017 Multi-University Training Contest - Team 5

————————————————————————————————————

​ 题目的意思是存在两个长度分别为  n,m  的数组  A,B  。有  q  个询问，每个询问给出一个数字  k  ，可以得到使得  AimodBj=k  的种数。求该种数的奇偶性。

思路：打表出所有查询k，k<Bi,从最大的Bi开始枚举。如果a%b==k,那么(a-k)%b==0

所以我们考虑用bitset优化这个过程

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <cmath>
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <vector>
#include <bitset>

using namespace std;

#define LL long long
const int INF = 0x3f3f3f3f;
#define mod 10000007
#define mem(a,b) memset(a,b,sizeof a)

bitset<50005>bt,bi;

int b[50005],ans[500005];
int main()
{
    int a,T,m,n,k;;
    for(scanf("%d",&T); T--;)
    {
        bi.reset();
        bt.reset();
        scanf("%d%d%d",&n,&m,&k);
        for(int i=0; i<n; i++)
            scanf("%d",&a),bt.set(a);
        int mx=-1;
        for(int i=0; i<m; i++)
            scanf("%d",&b[i]),mx=max(mx,b[i]);
        sort(b,b+m);
        int cnt=m-1;
        for(int i=mx; i>=0; i--)
        {
            ans[i]=((bi<<i)&bt).count()&1;
            if(b[cnt]>=i&&cnt>=0)
            {
                for(int j=0; j<=mx; j+=b[cnt])
                    bi.flip(j);
                    cnt--;
            }
        }
        for(int i=0; i<k; i++)
        {
            scanf("%d",&a);
            printf("%d\n",ans[a]);
        }

    }
    return 0;
}