2019西安 B-Product (莫比乌斯反演 杜教筛)

Product

#include<bits/stdc++.h>
#define me(a,x) memset(a,x,sizeof(a))
#define sc scanf
#define itn int
#define IN freopen("in.txt","r",stdin);
#define OUT freopen("out.txt","w",stdout);
#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=1e6+5;
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;}
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;}
void exgcd(ll a,ll b,ll &x,ll &y){if(!b)x=1,y=0;else exgcd(b,a%b,y,x),y-=(a/b)*x;}
int mod;
int prime[N],tot=0;
int phi[N];
ll F[N];
bool vis[N];
void f_mod(ll &x){
    if(x>=mod)
    x-=x/mod*mod;
}
int f_2(ll n){
    ll ans=n*(n+1)/2;
    f_mod(ans);
    return ans;
}
void pre(){
    phi[1]=1;
    for(int i=2;i<N;i++){
        if(!vis[i])prime[++tot]=i,phi[i]=i-1;
        for(int j=1;j<=tot&&i*prime[j]<N;j++){
            vis[i*prime[j]]=1;
            if(i%prime[j]==0){
                phi[i*prime[j]]=phi[i]*prime[j];
                break;
            }else phi[i*prime[j]]=phi[i]*(prime[j]-1);
        }
    }
    for(int i=1;i<N;i++){
        for(int j=i;j<N;j+=i){
            ll x=1LL*i*phi[j/i];
            f_mod(x);
            F[j]+=x;
            f_mod(F[j]);
        }
        F[i]+=F[i-1];
        f_mod(F[i]);
    }
}
map<int,int>p;
map<int,bool>q;
int cal(int n){
    if(n<N)return F[n];
    if(q[n]) return p[n];
    ll ans=0;
    for(int i=1,last;i<=n;i=last+1){
        last=n/(n/i);
        int x=f_2(n/i);
        ll num=(1LL*f_2(last)-f_2(i-1)+mod)*x;
        f_mod(num);
        ans+=num;f_mod(ans);
    }
    for(int i=2,last;i<=n;i=last+1){
        last=n/(n/i);
        ll x=(1LL*last-i+1)*cal(n/i);
        f_mod(x);
        ans-=x;
        if(ans<0)ans=(ans%mod+mod)%mod;
    }
    q[n]=true;
    return p[n]=ans;
}
int main(){
    int n,m,p;
    sc("%d%d%d",&n,&m,&p);mod=p-1;
    pre();
    ll ans=0;
    for(int i=1,last;i<=n;i=last+1){
        last=n/(n/i);
        ll y=1LL*(n/i)*(n/i);
        f_mod(y);
        ll x=(1LL*cal(last)-cal(i-1)+mod)*y;
        f_mod(x);
        ans+=x;
        f_mod(ans);
    }
    printf("%lld\n",ksm(m,ans+mod,p));
}