51nod1238 最小公倍数之和 V3

题意:求(sum_{i=1}^nsum_{j=1}^nfrac{i*j}{gcd(i,j)})
题解:先枚举gcd,(sum_{d=1}^ndsum_{i=1}^{lfloor frac{n}{d} floor}sum_{j=1}^{lfloor frac{n}{d} floor}i*j[(i,j)==1])
(=sum_{d=1}^nd2*sum_{i=1}^{lfloor frac{n}{d} floor}sum_{j=1}^{i}i*j[(i,j)==1]-1)
(=sum_{d=1}^nd2*sum_{i=1}^{lfloor frac{n}{d} floor}isum_{j=1}^{i}j[(i,j)==1]-1)
(=sum_{d=1}^nd2*sum_{i=1}^{lfloor frac{n}{d} floor}i*frac{i*phi(i)+[i==1]}{2}-1)
(=sum_{d=1}^ndsum_{i=1}^{lfloor frac{n}{d} floor}i^2*phi(i))
假设(f(i)=i^2*phi(i),S(n)=sum_{i=1}^nf(i))
由杜教筛(g(1)S(n)=sum_{i=1}^ng*f-sum_{d=2}g(d)S(lfloor frac{n}{d} floor))
考虑(g(n)=n^2),g和f狄利克雷卷积(=sum_{d|n}d^2phi(d){frac{n}{d}}^2=n^3)
那么(S(n)=sum_{i=1}^n n^3-sum_{d=2}d^2S(lfloor frac{n}{d} floor))

//#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 double
#define mp make_pair
#define pb push_back
#define pi acos(-1.0)
#define ll long long
#define vi vector<int>
#define mod 1000000007
#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 Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
#define fin freopen("a.txt","r",stdin)
#define fout freopen("a.txt","w",stdout)
#define fio ios::sync_with_stdio(false);cin.tie(0)
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 void add(ll &a,ll b){a+=b;if(a>=mod)a-=mod;}
inline void sub(ll &a,ll b){a-=b;if(a<0)a+=mod;}
inline ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
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=5000000+10,maxn=3000000+10,inf=0x3f3f3f3f;

int prime[N],cnt,phi[N];
bool mark[N];
ll f[N],inv2=qp(2,mod-2),inv6=qp(6,mod-2);
map<ll,ll>phii;
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]*(prime[j]-1);
        }
    }
    for(int i=1;i<N;i++)
    {
        f[i]=1ll*i*i%mod*phi[i]%mod;
        add(f[i],f[i-1]);
    }
}
ll getf(ll n)
{
    if(n<N)return f[n];
    if(phii.find(n)!=phii.end())return phii[n];
    ll ans=n%mod*((n+1)%mod)%mod*inv2%mod;ans=ans*ans%mod;
    for(ll i=2,j;i<=n;i=j+1)
    {
        j=n/(n/i);
        ll tj=j%mod,ti=i%mod;
        ll te=tj*(tj+1)%mod*(2ll*tj+1)%mod*inv6%mod-(ti-1)*ti%mod*(2ll*ti-1)%mod*inv6%mod;
        te=(te%mod+mod)%mod;
        sub(ans,te*getf(n/i)%mod);
    }
    return phii[n]=ans;
}
int main()
{
    init();
    ll n,ans=0;scanf("%lld",&n);
    for(ll i=1,j;i<=n;i=j+1)
    {
        j=n/(n/i);
        add(ans,(j-i+1)%mod*((i+j)%mod)%mod*inv2%mod*getf(n/i)%mod);
    }
    printf("%lld
",ans);
    return 0;
}
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