loj#6491. zrq 学反演

题意:求\(\sum_{i_1=1}^m\sum_{i_2=1}^m...\sum_{i_n=1}^mgcd(i_1,i_2,...i_n)\)
题解:\(\sum_{d=1}^md\sum_{i_1=1}^m...\sum_{i_n=1}^m[(i_1,...i_n)==d]\)
\(=\sum_{d=1}^md\sum_{i_1=1}^{\lfloor \frac{m}{d} \rfloor}...\sum_{i_n=1}^{\lfloor \frac{m}{d} \rfloor}\sum_{t|i_1...t|i_n}^m\mu(t)\)
\(=\sum_{d=1}^md\sum_{t=1}^{\lfloor \frac{m}{d} \rfloor}t*{\lfloor \frac{m}{dt} \rfloor}^n\)
\(=\sum_{x=1}^m{\lfloor \frac{m}{x} \rfloor}^n\sum_{d|x}d*\mu(\frac{x}{d})\)
\(=\sum_{x=1}^m{\lfloor \frac{m}{x} \rfloor}^n*\phi(x)\)
杜教筛即可

//#pragma GCC optimize(2)
//#pragma GCC optimize(3)
//#pragma GCC optimize(4)
//#pragma GCC optimize("unroll-loops")
//#pragma comment(linker, "/stack:200000000")
//#pragma GCC optimize("Ofast,no-stack-protector")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#include<bits/stdc++.h>
#define fi first
#define se second
#define db long double
#define mp make_pair
#define pb push_back
#define pi acos(-1.0)
#define ll long long
#define vi vector<int>
#define mod 998244353
#define ld long double
//#define C 0.5772156649
#define ls l,m,rt<<1
#define rs m+1,r,rt<<1|1
#define pll pair<ll,ll>
#define pil pair<int,ll>
#define pli pair<ll,int>
#define pii pair<int,int>
//#define cd complex<double>
#define ull unsigned long long
//#define base 1000000000000000000
#define fin freopen("a.txt","r",stdin)
#define fout freopen("a.txt","w",stdout)
#define fio ios::sync_with_stdio(false);cin.tie(0)
inline ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
inline void sub(ll &a,ll b){a-=b;if(a<0)a+=mod;}
inline void add(ll &a,ll b){a+=b;if(a>=mod)a-=mod;}
template<typename T>inline T const& MAX(T const &a,T const &b){return a>b?a:b;}
template<typename T>inline T const& MIN(T const &a,T const &b){return a<b?a:b;}
inline ll qp(ll a,ll b){ll ans=1;while(b){if(b&1)ans=ans*a%mod;a=a*a%mod,b>>=1;}return ans;}
inline ll qp(ll a,ll b,ll c){ll ans=1;while(b){if(b&1)ans=ans*a%c;a=a*a%c,b>>=1;}return ans;}

using namespace std;

const double eps=1e-8;
const ll INF=0x3f3f3f3f3f3f3f3f;
const int N=10000000+10,maxn=200000+10,inf=0x3f3f3f3f;

ull phi[N];
map<ull,ull>phii;
int prime[N],cnt;
bool mark[N];
void init()
{
    phi[1]=1;
    for(int i=2;i<N;i++)
    {
        if(!mark[i])prime[++cnt]=i,phi[i]=i-1;
        for(int j=1;j<=cnt&&i*prime[j]<N;j++)
        {
            mark[i*prime[j]]=1;
            if(i%prime[j]==0)
            {
                phi[i*prime[j]]=phi[i]*prime[j];
                break;
            }
            phi[i*prime[j]]=phi[i]*phi[prime[j]];
        }
    }
    for(int i=1;i<N;i++)phi[i]+=phi[i-1];
}
ull getphi(ull n)
{
    if(n<N)return phi[n];
    if(phii.find(n)!=phii.end())return phii[n];
    ull ans;
    if(n&1)ans=(n+1)/2*n;
    else ans=n/2*(n+1);
    for(ull i=2,j;i<=n;i=j+1)
    {
        j=n/(n/i);
        ans-=(j-i+1)*getphi(n/i);
    }
    return phii[n]=ans;
}
inline ull qq(ull a,ull b){ull ans=1;while(b){if(b&1)ans=ans*a;a=a*a,b>>=1;}return ans;}
int main()
{
    init();ull n,m;
    scanf("%llu%llu",&n,&m);
    ull ans=0;
    for(ull i=1,j;i<=m;i=j+1)
    {
        j=m/(m/i);
        ans+=(getphi(j)-getphi(i-1))*qq(m/i,n);
    }
    printf("%llu\n",ans);
    return 0;
}
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