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 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

 

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：

考虑枚举k,此时只有bj>k,ai>=k,才可能满足ai%bj=k 
bbi={i的>k的因子属于b[1...m]的个数} 
此时余数为k的情况数=∑ni=1bbai−k 
考虑用bitset存放a,bitsetA>>k等价于a数组全部元素减去k,丢弃<0的元素 
bitsetBB[i]=bbi & 1 
ans[k]=(bitsetA>>k) & bitsetBB.count() & 1

求解bb,对每个bi,枚举其倍数,O(N∑1bi) 
最坏情况下变为N∗调和级数=O(N∑Ni=11N)约等于O(N∗ln(N+1)) 
枚举k,对bitsetA和bitsetBB进行and操作O(N232) 
所以复杂度大概O(N232+N∗ln(N+1))

#include <iostream>#include <algorithm>#include <cstdio>#include <cmath>#include <cstring>#include <string>#include <string.h>#include <map>#include <set>#include <queue>#include <deque>#include <list>#include <bitset>#include <stack>#include <stdlib.h>#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;bitset<maxn>a,b,ans;bitset<maxn>bb;//b的倍数，bb［i］＝1:有奇数个y满足i％b［y］＝＝0void 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<maxn;j+=i)                bb.flip(j);    }}int main(){    ios::sync_with_stdio(false);    int T;    scanf("%d",&T);//cin>>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<n;i++)        {            int x;            scanf("%d",&x);//cin>>x;            a.set(x);        }        for(int i=0;i<m;i++)        {            int x;            scanf("%d",&x);//cin>>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"<<endl;            else puts("0");//cout<<"0"<<endl;        }    }    return 0;}