Rikka with Candies

Time Limit: 7000/3500 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 553 Accepted Submission(s): 232

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 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 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

#include<bits/stdc++.h>
using namespace std;
const int N = 5e4+100;
bitset<N> a,b[2],c,d;
int num[N];
int n,m,q;
int mx,mxm;
bool vis[N];
void dfs(bitset<N>&now,bitset<N> tmp,int x){
    now ^= (now >> (x/2));
    tmp >>= (x/2);
    now &= tmp;
    if(vis[x/2]) {b[0] ^= now;vis[x/2] = false;}
    x/=2;
    if(x%2 == 0){
        dfs(now,tmp,x);
    }
}
int main(){
    int T;
    cin >> T;
    while(T--){
        scanf("%d %d %d",&n,&m,&q);
        memset(vis,false,sizeof(vis));
        a.reset();//原陣列
        b[0].reset();//ans
        b[1].reset();//每次個j
        c.reset();//去除後面一段
        mx = 0;//a陣列最大值
        mxm = 0;//b陣列最大值
        for(int i= 1;i <= n;i ++){
            int now;
            scanf("%d",&now);
            mx = max(now,mx);
            a.set(now);
        }
        for(int i= 1; i<= m;i ++) {
            int now;
            scanf("%d",&now);
            mxm = max(mxm,now);
            vis[now] = true;
        }
        for(int i= 0;i <= mxm;i ++) c.set(i);
        for(int i= mxm;i >= mxm/2;i --){
            c.reset(i);
            d = a;
            b[1].reset();
            int nn = 0;
            while(nn <= mx){
                b[1] ^= d;
                d >>= i;
                nn += i;
            }
            b[1] &= c;
            if(vis[i]) {b[0] ^= b[1];vis[i] = false;}
            if(i%2 == 0)
                dfs(b[1],c,i);
        }
        for(int i = 1;i <= q;i ++){
            int now;
            scanf("%d",&now);
            printf("%d\n",b[0][now]==true?1:0);
        }


    }
    return 0;
}