hdu 1796 How many integers can you find(容斥原理)

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#include <iostream>
#include <cstring>
#include <vector>
using namespace std;
typedef long long ll;
const int N=1e6+20;
ll n,m,num,ans;
int fac
;
ll gcd(ll a,ll b)
{
if(a%b==0)
return b;
else
return gcd(b,a%b);
}
ll dfs(int id,int flag,ll LCM)
{
LCM=fac[id]/gcd(fac[id],LCM)*LCM;//flag个数的LCM
if(flag%2)
ans+=n/LCM;
else
ans-=n/LCM;
for(int i=id+1;i<=num;i++)
{
dfs(i,flag+1,LCM);
}
}
int main()
{
while(cin>>n>>m)
{
ans=0;
num=0;
for(int i=0;i<m;i++)
{
ll x;
cin>>x;
if(x)
{
fac[++num]=x;
}
}
n--; //小于n
//能被fac[i]整除的为集合pi 则小于N&&能被任意一个fac[i]整除的个数为 p1Up2..Upm
//p1Up2=n/p1+n/p2-n/LCM(p1,p2)..
for(int i=1;i<=num;i++)
{
dfs(i,1,fac[i]);
}
cout<<ans<<endl;

}
return 0;
}