Factors and Multiples

Factors and Multiples

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Statistics

Forum

Time Limit: 2 second(s)

Memory Limit: 32 MB

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 4or 5.

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

Input

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.

Output

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

Sample Input	Output for Sample Input
2 4 2 3 4 5 4 6 7 8 9 3 100 200 300 1 150	Case 1: 3 Case 2: 0

 题意：两个集合，删除元素使下一个集合没有上一个集合的倍数，问最少删除几个元素。匹配~匹配~匹配~匹配~匹配~匹配~匹配~匹配~匹配~匹配~匹配~匹配~匹配~匹配~匹配~匹配~匹配~匹配~匹配~匹配~匹配~匹配~匹配~匹配~匹配~匹配~匹配~匹配~匹配~匹配~匹配~匹配~匹配~匹配~匹配~匹配~匹配~匹配~匹配~匹配~匹配~匹配~匹配~匹配~匹配~匹配~匹配~匹配~匹配~匹配~匹配~匹配~匹配~匹配~匹配~匹配~匹配~匹配~匹配~匹配~匹配~匹配~匹配~匹配~匹配~匹配~匹配~匹配~匹配~

是猪么

#include<iostream>
#include<cstdio>
#include<cstring>

using namespace std;

#define N 110

int used[N], vis[N], n, m;
int maps[N][N];
int a[N], b[N];

int found(int x)
{
    for(int i = 0; i < m; i++)
    {
        if(maps[x][i] && !vis[i])
        {
            vis[i] = 1;
            if(used[i] == -1 || found(used[i]))
            {
                used[i] = x;
                return true;
            }
        }
    }
    return false;
}

int main()
{
    int t, k = 1;

    scanf("%d", &t);

    while(t--)
    {
        memset(used, -1, sizeof(used));
        memset(maps, 0, sizeof(maps));

        scanf("%d", &n);
        for(int i = 0; i < n; i++)
            scanf("%d", &a[i]);
        scanf("%d", &m);
        for(int j = 0; j < m; j++)
            scanf("%d", &b[j]);
        for(int i = 0; i < n; i++)
            for(int j = 0; j < m; j++)
            if(b[j] % a[i] == 0)
            maps[i][j] = 1;
        int cou = 0;
        for(int i = 0; i < n; i++)
        {
            memset(vis, 0, sizeof(vis));
            if(found(i))
                cou++;
        }
        printf("Case %d: %d
", k++, cou);
    }
    return 0;
}

好好的福利场被人家抢了~是不是傻，是不是猪，是不是~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

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