How many integers can you find HDU - 1796

 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

7

题意：给你一个数n，一个含m个元素的集合，求小于n的数中是m中某个元素的倍数有多少个。

题目分析：这是一道简单的容斥原理的题目，下面根据这道题给出容斥原理的三种实现方法： dfs，队列数组，二进制；

注意：这道题求的是小于n的数中有多少符合条件的，不包含n本身；

二进制实现：

#include<cstdio>
#include<iostream>
#include<vector>
using namespace std;
vector<int>x;
int gcd(int a,int b)
{
    if(b==0)
        return a;
    return gcd(b,a%b);
}
int solve(int n)
{
     int ans=0;
    for(int i=1;i<(1<<x.size());i++)
    {
        int lcm=1,cnt=0;
        for(int j=0;j<x.size();j++)
          if(i&(1<<j))
          {
              cnt++;
              int g=gcd(lcm,x[j]);
              lcm=lcm/g*x[j];
          }
          if(cnt&1)
            ans+=n/lcm;
          else
            ans-=n/lcm;
    }
    return ans;
}
int main()
{
    int n,m;
    int temp;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        n--;
        x.clear();
        for(int i=1;i<=m;i++)
        {
           scanf("%d",&temp);
           if(temp)  ///不能为0
           x.push_back(temp);
        }
        cout<<solve(n)<<endl;
    }
    return 0;
}

深搜的两种实现：

1.用数组存储元素

#include<cstdio>
#include<iostream>
#include<vector>
using namespace std;
int get_lcm(int a,int b)
{
    int x=a;
    int y=b;
    while(b)
    {
        int t=a;
        a=b;
        b=t%b;
    }
    return x/a*y;
}

long long ans;
int b[20],k,n;
void dfs(int i,int cnt,int lcm)///i代表数组的下标
{                              ///cnt代表几个元素的最小公倍数   
    lcm=get_lcm(lcm,b[i]);      ///lcm代表最小公倍数
    if(cnt&1)
        ans+=n/lcm;
    else
        ans-=n/lcm;
    for(int j=i+1;j<k;j++)
        dfs(j,cnt+1,lcm);
}
int main()
{
    int m;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        n--;
        ans=k=0;
        for(int i=1;i<=m;i++)///k记录非0的个数
        {
            scanf("%d",b+k);///给这个数组赋的值不能为0；
            if(!b[k++]) k--;///这种赋值方法有趣
        }
        for(int i=0;i<k;i++)
            dfs(i,1,1);
        cout<<ans<<endl;
    }
    return 0;
}

2.用vector实现

#include<cstdio>
#include<iostream>
#include<vector>
using namespace std;
int get_lcm(int a,int b)///获得最小公倍数
{
    int x=a;
    int y=b;
    while(b)
    {
        int t=a;
        a=b;
        b=t%b;
    }
    return x/a*y;
}

vector<int>x;
int ans;
int n;
void dfs(int i,int cnt,int lcm)///i代表下标
{                              ///cnt代表几个数
    lcm=get_lcm(x[i],lcm);     ///lcm代表最小公倍数
    if(cnt&1)
        ans+=n/lcm;
    else
        ans-=n/lcm;
    for(int j=i+1;j<x.size();j++)
        dfs(j,cnt+1,lcm);
}
int main()
{
    int m;
    int temp;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        n--;
        x.clear();
        ans=0;
        for(int i=1;i<=m;i++)
        {
            scanf("%d",&temp);
            if(temp)
                x.push_back(temp);
        }
        for(int i=0;i<x.size();i++)
            dfs(i,1,1);
            cout<<ans<<endl;
    }
    return 0;
}