hdu 6085 （bitset区间取值）

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 nn children and mm kinds of candies. The iith child has AiAi dollars and the unit price of the iith kind of candy is BiBi. 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 1010 dollars and the unit price of his favorite candy is 44 dollars, then he will buy two candies and go home with 22 dollars left. 

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

To reduce the difficulty, Rikka just need to calculate the answer modulo 22. 

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

Input

The first line contains a number t(1≤t≤5)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)n,m,q(1≤n,m,q≤50000). 

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

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

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

Output

For each query, print a single line with a single 0101 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

给出数组A和B，问多少对 A%B==ki。数据大小 5e4 考虑n^2/32*log 可以知道的是取模运算。

一个数%B 和%2B 结果相等，考虑到对ki的贡献，只有比ki大的B才会对ki有贡献，我们考虑遍历B.从大到小，然后不断更新B存在的位置(log),然后不断计算这样a%b(比当前b大）==ki的个数，这个只要用减法就可以做到，利用bitset的位运算，bx&(a>>i).count()&1,更方便快捷

#include <bits/stdc++.h>
using namespace std;
const int N = 50000+5;
bitset<N> a,b,ans,bx;

void solve(int x)
{
	bx.reset();
	for(int i=x;i>=0;i--)
	{
		ans[i]=(bx&(a>>i)).count()&1;
		if(b[i])
		{
			for(int j=0;j<N;j+=i)
				bx.flip(j);
		}
	}
}

int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		a.reset(),b.reset();
		int n,m,q,x,maxt=0;
		scanf("%d%d%d",&n,&m,&q);
		for(int i=1;i<=n;i++)
		{
			scanf("%d",&x);
			a.set(x);
		}
		for(int i=1;i<=m;i++)
		{
			scanf("%d",&x);
			b.set(x);
			maxt=max(x,maxt);
		}
		ans.reset();
		solve(maxt);
		while(q--)
		{
			int x;
			scanf("%d",&x);
			puts(ans[x]?"1":"0");
		}
	}
}