HDU How many integers can you find 容斥

How many integers can you find

Time Limit: 12000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4249 Accepted Submission(s): 1211

Problem Description

Now you get a number N, and a M-integers set, you should find out how many integers which are small than N, that they can divided exactly by any integers in the set. For example, N=12, and M-integer set is {2,3}, so there is another set {2,3,4,6,8,9,10}, all the integers of the set can be divided exactly by 2 or 3. As a result, you just output the number 7.

Input

There are a lot of cases. For each case, the first line contains two integers N and M. The follow line contains the M integers, and all of them are different from each other. 0<N<2^31,0<M<=10, and the M integer are non-negative and won’t exceed 20.

Output

For each case, output the number.

Sample Input

12 2 2 3

Sample Output

题意：给n个数字，最大不会超过20的非负数，0忽略它可以。给你一个数字M，

问1-M-1中，有多少个数字能被这n数字中任何一个整除（只要满足其中一个能整除就行）。统计个数输出。

思路：容斥，简单容斥。一开始做zoj的一道题，果断数据太水，方法是不对的也能ac。

原来的思路是这样的，对n个数字，筛选掉ai倍数的数字，然后就容斥，但是明显这样的数据有问题

4 6，这样容斥后得到的结果是4 6 -24，不对的，应该是4 6 -12，所以应该是 4 6 -（4*6）/gcd（4，6）

略坑略坑。

 1 #include<iostream>
 2 #include<stdio.h>
 3 #include<cstring>
 4 #include<cstdlib>
 5 using namespace std;
 6 
 7 bool Hash[22];
 8 int f[22],len,qlen;
 9 __int64 Q[5002];
10 
11 int gcd(int a,int b)
12 {
13     if(a<0)a=-a;
14     if(b<0)b=-b;
15     if(b==0)return a;
16     int r;
17     while(b)
18     {
19         r=a%b;
20         a=b;
21         b=r;
22     }
23     return a;
24 }
25 void solve(__int64 m)
26 {
27     qlen = 0;
28     Q[0]=-1;
29     for(int i=1;i<=len;i++)
30     {
31         int k=qlen;
32         for(int j=0;j<=k;j++)
33         Q[++qlen]=-1*(Q[j]*f[i]/gcd(Q[j],f[i]));
34     }
35     __int64 sum = 0;
36     for(int i=1;i<=qlen;i++)
37     sum = sum+m/Q[i];
38     printf("%I64d
",sum);
39 }
40 int main()
41 {
42     int m,x;
43     __int64 n;
44     while(scanf("%I64d%d",&n,&m)>0)
45     {
46         n=n-1;
47         memset(Hash,false,sizeof(Hash));
48         for(int i=1;i<=m;i++)
49         {
50             scanf("%d",&x);
51             Hash[x]=true;
52         }
53         for(int i=1;i<=20;i++)
54         {
55             if(Hash[i]==true)
56             for(int j=i+i;j<=20;j=j+i)
57             if(Hash[j]==true) Hash[j]=false;
58         }
59         len = 0;
60         for(int i=1;i<=20;i++)if(Hash[i]==true) f[++len]=i;
61         solve(n);
62     }
63     return 0;
64 }