[51nod1238] 最小公倍数之和 V3（杜教筛）

题面

传送门

题解

懒了……这里写得挺好的……

//minamoto
#include<bits/stdc++.h>
#define R register
#define ll long long
#define IT map<ll,int>::iterator
#define fp(i,a,b) for(R int i=a,I=b+1;i<I;++i)
#define fd(i,a,b) for(R int i=a,I=b-1;i>I;--i)
#define go(u) for(int i=head[u],v=e[i].v;i;i=e[i].nx,v=e[i].v)
using namespace std;
const int N=6e6+5,P=1e9+7,inv2=500000004,inv6=166666668;
bitset<N>vis;int p[N],phi[N],g[N],m,sqr;ll n;map<ll,int>mp;
inline int calc(R int x){return (1ll*x*(x+1)>>1)%P;}
inline int calc2(R int x){return 1ll*x*(x+1)%P*((x<<1)+1)%P*inv6%P;}
inline int calc3(R int x){x=calc(x);return 1ll*x*x%P;}
void init(int n){
    phi[1]=g[1]=1;
    fp(i,2,n){
        if(!vis[i])p[++m]=i,phi[i]=i-1,g[i]=1ll*phi[i]*i%P*i%P;
        for(R int j=1,k;j<=m&&1ll*i*p[j]<=n;++j){
            vis[k=i*p[j]]=1;
            if(i%p[j]==0){phi[k]=phi[i]*p[j],g[k]=1ll*phi[k]*k%P*k%P;break;}
            phi[k]=phi[i]*(p[j]-1),g[k]=1ll*g[i]*g[p[j]]%P;
        }
    }
    fp(i,2,n)(g[i]+=g[i-1])%=P;
}
int G(ll n){
    if(n<=sqr)return g[n];
    IT it=mp.find(n);
    if(it!=mp.end())return it->second;
    int res=calc3(n%P),las=1,now;
    for(ll i=2,j;i<=n;i=j+1)
        j=n/(n/i),now=calc2(j%P),res=(res-1ll*(now-las+P)*G(n/i)%P+P)%P,las=now;
    return mp[n]=res;
}
int main(){
//  freopen("testdata.in","r",stdin);
    scanf("%lld",&n),init(sqr=N-5);
    int res=0,las=0,now=0;
    for(R ll i=1,j;i<=n;i=j+1)
        j=n/(n/i),now=calc(j%P),(res+=1ll*(now-las+P)*G(n/i)%P)%=P,las=now;
    printf("%d\n",res);
    return 0;
}

原文地址：https://www.cnblogs.com/bztMinamoto/p/10437280.html