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  • Factors and Multiples

    Factors and Multiples
    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

     1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 
 5 using namespace std;
 6 
 7 #define N 110
 8 
 9 int used[N], vis[N], n, m;
10 int maps[N][N];
11 int a[N], b[N];
12 
13 int found(int x)
14 {
15     for(int i = 0; i < m; i++)
16     {
17         if(maps[x][i] && !vis[i])
18         {
19             vis[i] = 1;
20             if(used[i] == -1 || found(used[i]))
21             {
22                 used[i] = x;
23                 return true;
24             }
25         }
26     }
27     return false;
28 }
29 
30 int main()
31 {
32     int t, k = 1;
33 
34     scanf("%d", &t);
35 
36     while(t--)
37     {
38         memset(used, -1, sizeof(used));
39         memset(maps, 0, sizeof(maps));
40 
41         scanf("%d", &n);
42         for(int i = 0; i < n; i++)
43             scanf("%d", &a[i]);
44         scanf("%d", &m);
45         for(int j = 0; j < m; j++)
46             scanf("%d", &b[j]);
47         for(int i = 0; i < n; i++)
48             for(int j = 0; j < m; j++)
49             if(b[j] % a[i] == 0)
50             maps[i][j] = 1;
51         int cou = 0;
52         for(int i = 0; i < n; i++)
53         {
54             memset(vis, 0, sizeof(vis));
55             if(found(i))
56                 cou++;
57         }
58         printf("Case %d: %d
", k++, cou);
59     }
60     return 0;
61 }

    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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    让未来到来 让过去过去
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  • 原文地址：https://www.cnblogs.com/Tinamei/p/4746900.html
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