洛谷 P2231 [HNOI2002]跳蚤

https://www.luogu.org/problemnew/show/P2231

题意相当于：
有n个位置a[1..n]，每个位置可以填[1,m]中任一个整数，问共有多少种填法满足gcd(a[1],a[2],..,a[n],m)=1

可以反演一下

 #include<cstdio>
 #include<algorithm>
 #include<cstring>
 #include<vector>
 #include<cmath>
 using namespace std;
 #define fi first
 #define se second
 #define mp make_pair
 #define pb push_back
 typedef long long ll;
 typedef unsigned long long ull;
 typedef pair<int,int> pii;
 const ll N=;
 ll prime[N+],len;
 bool nprime[N+];
 ll ans;
 ll poww(ll a,ll b)
 {
     ll bs=a,an=;
     for(;b;b>>=,bs*=bs)
         if(b&)
             an*=bs;
     return an;
 }
 ll getmu(ll x)
 {
     ll i,ans=,ed=sqrt(x+0.5);
     bool fl;
     for(i=;i<=len&&prime[i]<=ed;i++)
     {
         fl=;
         while(x%prime[i]==)
         {
             if(fl)    return ;
             fl=;
             x/=prime[i];
             ans*=(-);
         }
     }
     if(x!=)    ans*=(-);
     return ans;
 }
 ll n,m;
 void work(ll d)
 {
     ans+=poww(m/d,n)*getmu(d);
 }
 int main()
 {
     ll i,j;
     for(i=;i<=N;i++)
     {
         if(!nprime[i])    prime[++len]=i;
         for(j=;j<=len&&i*prime[j]<=N;j++)
         {
             nprime[i*prime[j]]=;
             if(i%prime[j]==)    break;
         }
     }
     scanf("%lld%lld",&n,&m);
     ll ed=sqrt(m+0.5);
     for(i=;i<=ed;i++)
         if(m%i==)
         {
             work(i);
             if(m/i!=i)    work(m/i);
         }
     printf("%lld",ans);
     return ;
 }