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[N];ll gcd(ll a,ll b){if(a%b==0)return b;elsereturn 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;elseans-=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;}