You will be given two sets of integers. Let's call them set A and set
B. Set A contains n elements and set
B contains m elements. You have to remove
k₁ elements from set A and k₂ elements from set
B so that of the remaining values no integer in set B is a multiple of any integer in set
A. k₁ should be in the range
[0, n] and k₂ in the range [0, m].

You have to find the value of (k₁ + k₂) such that
(k₁ + k₂) is as low as possible. P is a multiple of
Q if there is some integer K such that
P = K * Q.

Suppose set A is {2, 3, 4, 5} and set
B is {6, 7, 8, 9}. By removing 2 and
3 from A and 8 from B, we get the sets
{4, 5} and {6, 7, 9}. Here none of the integers
6, 7 or 9 is a multiple of 4 or
5.

So for this case the answer is 3 (two from set
A and one from set B).

Input starts with an integer T (≤ 50), denoting the number of test cases.

The first line of each case starts with an integer n followed by
n positive integers. The second line starts with m followed by
m positive integers. Both n and m will be in the range
[1, 100]. Each element of the two sets will fit in a 32 bit signed integer.

For each case of input, print the case number and the result.

2

4 2 3 4 5

4 6 7 8 9

3 100 200 300

1 150

Case 1: 3

Case 2: 0

Problem Setter: Sohel Hafiz

Special Thanks: Jane Alam Jan



给了两个集合A，B，分别有n，m个数，从A取k1个数，B取k2个数，使得b[ j ]%a[ i ]==0的情况不存在

刚开始以为可以暴力的，但是后来发现暴力真的是挺麻烦，把图画出来之后会发现，其实就是最小点覆盖，二分图性质：最小点覆盖=最大匹配，匈牙利算法跑一次

#include<cstdio>

#include<cstring>

#include<cmath>

#include<vector>

#include<algorithm>

using namespace std;

vector<int>map[200];

int used[200],pipei[200],a[200],b[200];

int n,m;

int find(int x)

{

	for(int i=0;i<map[x].size();i++)

	{

		int y=map[x][i];

		if(!used[y])

		{

			used[y]=1;

			if(pipei[y]==-1||find(pipei[y]))

			{

				pipei[y]=x;

				return 1;

			}

		}

	}

	return 0;

}

int main()

{

	int t,k=1;

	scanf("%d",&t);

	while(t--)

	{

		memset(a,0,sizeof(a));

		memset(b,0,sizeof(b));

		memset(pipei,-1,sizeof(pipei));

		scanf("%d",&n);

		for(int i=0;i<n;i++)

		{

			scanf("%d",&a[i]);

			map[i].clear();

		}

		scanf("%d",&m);

		for(int i=0;i<m;i++)

		scanf("%d",&b[i]);

		for(int i=0;i<n;i++)

		{

			for(int j=0;j<m;j++)

			{

				if(b[j]%a[i]==0)

				{

					map[i].push_back(j);

				}

			}

		}

		int sum=0;

		for(int i=0;i<n;i++)

		{

			memset(used,0,sizeof(used));

			sum+=find(i);

		}

		printf("Case %d: %d\n",k++,sum);

	}

	return 0;

}