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
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
题意：
有N个人，每个人拥有Ai的钱，有M种物品，每个物品的单位价值是Bj，每种物品的数量都是无限的，我们现在有一个查询X，表示询问存在多少对购买方案对，使得Ai%Bj==X.
输出方案数%2的结果。
分析：
我们开两个bitset，tmp里边存的是拥有i的钱的人是否存在，如果存在，该位为1，否则为0.ans里边存的是价格为i的物品是否存在，我们等比例放大物品的购买倍数即可。
然后我们预处理答案暴力跑一下。
代码：
#include<stdio.h>
#include<bitset>
#include<string.h>
#include<iostream>
using namespace std;
int b[50050];
int q[50050];
int vis[50050];
int output[50050];
bitset<50050>tmp,ans,temp,a;
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
memset(vis,0,sizeof(vis));
tmp.reset();///初始化，将所有的二进制位都设为0
ans.reset();
int n,m,qq,mx=0;
scanf("%d%d%d",&n,&m,&qq);
for(int i=1; i<=n; i++)
{
int x;
scanf("%d",&x);
tmp.set(x);///将tmp中x处的二进制位设为1
a.set(x);
}
for(int i=1; i<=m; i++)
{
scanf("%d",&b[i]);
mx=max(mx,b[i]);
vis[b[i]]=1;
}
for(int i=1; i<=qq; i++)
scanf("%d",&q[i]);
for(int i=mx; i>=0; i--)
{
temp=ans&(tmp>>i);///tmp表示的二进制数进行移位
//printf("i===%d temp.count()==%d\n",i,temp.count());
output[i]=temp.count()%2;///count()计算的是二进制中1的个数
if(vis[i]==1)///有这个价格的东西
{
for(int j=0; j<=mx; j+=i)///依次寻找当前这个价格的倍数
ans.flip(j);///把二进制的第j位取反
}
}
for(int i=1; i<=qq; i++)
printf("%d\n",output[q[i]]);
}
return 0;
}