[杜教筛] 51Nod 1238 最小公倍数之和 V3

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比最大公约数之和要难搞一些

吉丽博客传送门：http://jiruyi910387714.is-programmer.com/posts/195270.html

这道题两个要点

首先

∑1<=i<=n [(n,i)==1]*i == 1/2*([n==1]+n*phi(n))

还有就是对于phi(i)*i*i也就是phi·id·id的前缀和

我们单独卷一下phi

(phi*1)·id·id=id·id·id

#include<cstdio>#include<cstdlib>#include<algorithm>#include<tr1/unordered_map>using namespace std;using namespace std::tr1;typedef long long ll;const int P=1000000007;const int inv2=(P+1)/2;const int inv3=(P+1)/3;const int maxn=5e6;int prime[(int)1e6],num;int vst[maxn+5],phi[maxn+5],sum[maxn+5];inline void Pre(){  phi[1]=1;  for (int i=2;i<=maxn;i++){    if (!vst[i]) phi[i]=i-1,prime[++num]=i;    for (int j=1;j<=num && (ll)i*prime[j]<=maxn;j++){      vst[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<=maxn;i++) sum[i]=((ll)phi[i]*i%P*i%P+sum[i-1])%P;}inline ll sum1(ll n){ return (n+1)%P*(n%P)%P*inv2%P;}inline ll sum2(ll n){ return (n%P)*((n+1)%P)%P*((2*n+1)%P)%P*inv2%P*inv3%P; }inline ll sum3(ll n){ return sum1(n)*sum1(n)%P; }inline ll sum1(ll l,ll r){ return (sum1(r)+P-sum1(l-1))%P; }inline ll sum2(ll l,ll r){ return (sum2(r)+P-sum2(l-1))%P; }inline ll sum3(ll l,ll r){ return (sum3(r)+P-sum3(l-1))%P; }unordered_map<ll,int> S;inline int Sum(ll n){  if (n<=maxn) return sum[n];  if (S.find(n)!=S.end()) return S[n];  int tem=sum3(n);  ll l,r;  for (l=2;l*l<=n;l++) (tem+=P-l%P*(l%P)%P*Sum(n/l)%P)%=P;  for (ll t=n/l;l<=n;l=r+1,t--)    r=n/t,(tem+=P-sum2(l,r)*Sum(t)%P)%=P;  return S[n]=tem;}int main(){  ll n; int Ans=0,tem=0;  freopen("t.in","r",stdin);  freopen("t.out","w",stdout);  Pre();  scanf("%lld",&n);  Ans=sum1(n)*inv2%P;  ll l,r;  tem=0;  //for (l=1;l<=n;l++) {  //  (tem+=(ll)l*l%P*phi[l]%P*sum1(n/l))%=P;  //printf("%d %d\n",tem,l);  //}  for (l=1;l*l<=n;l++) (tem+=(ll)l*l%P*phi[l]%P*sum1(n/l)%P)%=P;  for (ll t=n/l;l<=n;l=r+1,t--)    r=n/t,(tem+=(Sum(r)+P-Sum(l-1))%P*sum1(t)%P)%=P;  (Ans+=(ll)tem*inv2%P)%=P;  Ans=(Ans*2%P+P-sum1(n))%P;  printf("%d\n",Ans);  return 0;}

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