2017多校联合第五场1001/hdu6085Rikka with Candies(bitset)

Rikka with Candies

Time Limit: 7000/3500 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1389    Accepted Submission(s): 606


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)(1in,1jm)  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(1t5) , the number of the testcases. 

For each testcase, the first line contains three numbers  n,m,q(1n,m,q50000)

The second line contains  n  numbers  Ai(1Ai50000)  and the third line contains  m  numbers  Bi(1Bi50000) .

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

It is guaranteed that  AiAj,BiBj  for all  ij .
 

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
 

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Statistic |  Submit |  Discuss |  Note
题意:A数组n个数,B数组m个数,q个查询, 每次给出一个k,询问有多少对(i,j), 使得Ai % Bj = k, 输出对数对模2的值  

题解:

k,bj>k,ai>=k,ai%bj=k  
bbi={i>kb[1...m]}  
k=ni=1bbaik  
bitseta,bitsetA>>kak,<0  
bitsetBB[i]=bbi  &  1  
ans[k]=(bitsetA>>k)  &  bitsetBB.count()  &  1

bb,bi,,O(N1bi)  
N=O(NNi=11N)O(Nln(N+1))  
k,bitsetAbitsetBBandO(N232)  
O(N232+Nln(N+1))

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#define lowbit(x) (x&-x)
#define e exp(1.0)
//ios::sync_with_stdio(false);
//    auto start = clock();
//    cout << (clock() - start) / (double)CLOCKS_PER_SEC;
typedef long long ll;
typedef long long LL;
using namespace std;
int read()
{
    int res = 0, ch, flag = 0;
    
    if((ch = getchar()) == '-')             //判断正负
        flag = 1;
    
    else if(ch >= '0' && ch <= '9')           //得到完整的数
        res = ch - '0';
    while((ch = getchar()) >= '0' && ch <= '9' )
        res = res * 10 + ch - '0';
    
    return flag ? -res : res;
}

const int maxn=50000+10;
bitseta,b,ans;
bitsetbb;//b的倍数,bb[i]=1:有奇数个y满足i%b[y]==0
void solve(int maxk)
{
    bb.reset();
    ans.reset();
    for(int i=maxk;i>=0;--i)//枚举k
    {
        ans[i]=(bb&(a>>i)).count()&1;
        if(b[i])//枚举b[i]的倍数
            for(int j=0;j>T;
    int n,m,q;
    while(T--)
    {
        scanf("%d%d%d",&n,&m,&q);//cin>>n>>m>>q;
        a.reset();
        b.reset();
        int maxk=0;
        for(int i=0;i>x;
            a.set(x);
        }
        for(int i=0;i>x;
            b.set(x);
            maxk=max(maxk,x);
        }
        solve(maxk);
        while(q--)
        {
            int x;
            scanf("%d",&x);//cin>>x;
            if(ans[x])puts("1");//cout<<"1"<


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