HDU 5212-Code (莫比乌斯反演)

HDU 5212-Code

#include<bits/stdc++.h>
#define me(a,x) memset(a,x,sizeof(a))
#define IN freopen("in.txt","r",stdin);
#define OUT freopen("out.txt","w",stdout);
#define sc scanf
#define itn int
#define STR clock_t startTime = clock();
#define END clock_t endTime = clock();cout << double(endTime - startTime) / CLOCKS_PER_SEC *1000<< "ms" << endl;
using namespace std;
const int N=10000+225;
const long long mod=10007;
const long long mod2=998244353;
const int oo=0x7fffffff;
const int sup=0x80000000;
typedef long long ll;
typedef unsigned long long ull;
template <typename it>void db(it *begin,it *end){while(begin!=end)cout<<(*begin++)<<" ";puts("");}
template <typename it>
string to_str(it n){string s="";while(n)s+=n%10+'0',n/=10;reverse(s.begin(),s.end());return s;}
template <typename it>int o(it a){cout<<a<<endl;return 0;}
inline ll mul_64(ll x,ll y,ll c){return (x*y-(ll)((long double)x/c*y)*c+c)%c;}
inline ll ksm(ll a,ll b,ll c){ll ans=1;for(;b;b>>=1,a=a*a%c)if(b&1)ans=ans*a%c;return ans;}
inline void exgcd(ll a,ll b,ll &x,ll &y){if(!b)x=1,y=0;else exgcd(b,a%b,y,x),y-=x*(a/b);}
int prime[N],tot=0,vis[N],mu[N];
ll f[N],h[N];
ll g(int n,int T){
    ll ans=0;
    for(int i=1;i<=n;i++){
        ans+=f[i*T];
    }
    return ans*ans%mod;
}
int main(){
    mu[1]=1;
    for(int i=2;i<N;i++){
        if(!vis[i])prime[++tot]=i,mu[i]=-1;
        for(int j=1;j<=tot&&i*prime[j]<N;j++){
            vis[i*prime[j]]=1;
            if(i%prime[j]==0){
                mu[i*prime[j]]=0;
                break;
            }else mu[i*prime[j]]=-mu[i];
        }
    }
    for(int i=1;i<N;i++){
        for(int j=i;j<N;j+=i){
            h[j]+=1LL*mu[j/i]*(i*i%mod-i);
            h[j]=(h[j]%mod+mod)%mod;
        }
    }
    int n;while(cin>>n){
        me(f,0);
        for(int i=1,x;i<=n;i++)sc("%d",&x),f[x]++;
        n=10000;
        ll ans=0;
        for(int i=1;i<=n;i++){
            ans+=g(n/i,i)*h[i]%mod;
            if(ans>=mod)ans-=ans/mod*mod;
        }
        printf("%lld\n",ans);
    }
}