【多校训练】hdu 6085 Rikka with Candies bitset

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 ith child has Ai dollars and the unit price of the ith 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 10dollars 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 ith child’s favorite candy is the jth 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

15 5 51 2 3 4 51 2 3 4 50 1 2 3 4

 

Sample Output

00001

 

题意：

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：

求a%b==k的数量  

->求a-k==b*j的数量

->可以用bitset优化成 ((a>>k)&bn).count()，bn为b的倍数。由于b一定要大于k，所以从大到小枚举。

具体看代码：

////  main.cpp//  1001////  Created by zc on 2017/8/30.//  Copyright © 2017年 zc. All rights reserved.//#include <iostream>#include<cstring>#include<cstring>#include<cmath>#include<bitset>#include<algorithm>using namespace std;const int N=55000;bitset<N>a,b,ans,bn;int main(int argc, const char * argv[]) {    int T;    scanf("%d",&T);    while(T--)    {        int n,m,q,t,mmax=0;        a.reset();b.reset();ans.reset();bn.reset();        scanf("%d%d%d",&n,&m,&q);        for(int i=0;i<n;i++)        {            scanf("%d",&t);            a.set(t);        }        for(int i=0;i<m;i++)        {            scanf("%d",&t);            b.set(t);            mmax=max(mmax,t);        }        for(int i=mmax;i>=0;i--)        {            ans[i]=((a>>i)&bn).count()&1;//a%b==k, b>k  ->  a-k==j*b, b>k   ->  (a>>k)&bn            if(b[i])            {                for(int j=0;j<N;j+=i)    bn.flip(j);            }        }        for(int i=0;i<q;i++)        {            scanf("%d",&t);            if(ans[t])  puts("1");            else    puts("0");        }    }}