2017多校训练Contest5: 1001 Rikka with Candies hdu6085

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<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

离线

把a压进一个bitset x1

从大到小枚举k，每次把大于k的b的所有倍数压入bitset x2里

然后直接(x1>>k)&x2求出当前k的答案

手写会好一点，用系统的bitset需要卡一下常数

#include<map>#include<cmath>#include<queue>#include<bitset>#include<vector>#include<cstdio>#include<string>#include<cstring>#include<cassert>#include<iostream>#include<algorithm>using namespace std;inline int read(){    int x=0;char ch=getchar();    while(ch<'0'||ch>'9'){ch=getchar();}    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}    return x;}inline bool cmp(int x,int y){    return x>y;}bitset<50051> x1,x2;int a[50051],b[50051];int ans[50051];int main(){    int T;    //scanf("%d",&T);    T=read();    while(T>0)    {        T--;        int n,m,q;        //scanf("%d%d%d",&n,&m,&q);        n=read();        m=read();        q=read();        int i,j;        int maxx=0;        for(i=1;i<=n;i++)        {            //scanf("%d",&a[i]);            a[i]=read();            maxx=max(maxx,a[i]);        }        for(i=1;i<=m;i++)        {            //scanf("%d",&b[i]);            b[i]=read();            maxx=max(maxx,b[i]);        }        sort(b+1,b+1+m,cmp);        x1.reset();        x2.reset();        for(i=1;i<=n;i++)            //x1[a[i]]=1;            x1.flip(a[i]);        int d=1;        b[0]=0;         memset(ans,0,sizeof(ans));        for(i=maxx;i>=0;i--)        {            while(d<=m&&b[d]>i)            {                for(j=0;j<=maxx;j+=b[d])                    //x2[j]=(x2[j]^1);                    x2.flip(j);                d++;                //x2.flip(0);            }            //x3=((x1>>i)&x2);            ans[i]=((x1>>i)&x2).count();        }        int x;        for(i=1;i<=q;i++)        {            //scanf("%d",&x);            x=read();            if(ans[x]&1)                puts("1");            else                puts("0");            //printf("%d\n",ans[x]%2);        }    }    return 0;}