hdu 6085 Rikka with Candies (set计数)

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

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

 

给你一个长度为n的a数组,m的b数组,再给你q个询问,每个询问为一个k,问在a b 数组中满足 a[i]%b[j]=k的i,j有多少对,答案对2取模.保证a  b 中的数组不会重复.

答案对2取模的话就提示你在用bitset来加速处理.我们先对询问排一个序,b数组排一个序.我们先用一个aa的bitset存下a数组中出现的数字

bb存b[j]的倍数,因为如果[i]%b[j]=k,那么b[j]应该是>k的.然后对已每一个询问不断更新bb也就是不断加上下一个较小的b[j]

对于每一个询问的结果我们只需要(((aa>>q[i].k)&bb).count())&1;这样就是答案了

代码如下:

 #include <bits/stdc++.h>
 using namespace std;
 const int BufferSize=<<;
 char buffer[BufferSize],*head,*tail;
 inline char Getchar() {
     if(head==tail) {
         int l=fread(buffer,,BufferSize,stdin);
         tail=(head=buffer)+l;
     }
     return *head++;
 }
 inline int read() {
     int x=,f=;char c=Getchar();
     for(;!isdigit(c);c=Getchar()) if(c=='-') f=-;
     for(;isdigit(c);c=Getchar()) x=x*+c-'';
     return x*f;
 }
 inline void write(int x)
 {
     if(x>=)write(x/);
     putchar(x%+'');
 }
 const int maxn = ;
 int n,m,qu;
 int a[maxn],b[maxn];
 int ret[maxn];
 bitset<maxn> aa,bb;
 struct query
 {
     int k,id;
 }q[maxn];
 bool cmp (query q1,query q2)
 {
     return q1.k>q2.k;
 }
 int main()
 {
     //freopen("de.txt","r",stdin);
     int T;
     T=read();
     while (T--){
         n=read(),m=read(),qu=read();
         aa.reset();
         bb.reset();
         for (int i=;i<=n;++i){
             a[i]=read();
             aa.set(a[i]);
         }
         for (int i=;i<=m;++i)
             b[i]=read();
         sort(b+,b++m);
         for (int i=;i<=qu;++i)
             q[i].k=read(),q[i].id=i,ret[i]=;
         sort(q+,qu++q,cmp);
         int now = m+;
         for (int i=;i<=qu;++i){
             while (now->=&&b[now-]>q[i].k){
                 now--;
                 for (int j=;j<=;j+=b[now])//枚举b[now]的倍数,从0开始
                     bb[j]=bb[j]^;
             }
             ret[q[i].id] =(((aa>>q[i].k)&bb).count())&;
         }
         for (int i=;i<=qu;++i)
             write(ret[i]),putchar();
     }
     return ;
 }

加上快速读入3400ms左右,时限是3500...