[杜教筛] 51Nod 1227 平均最小公倍数

糖老师博客传送门：http://blog.csdn.net/skywalkert/article/details/50500009

跟最小公倍数类似

求phi·id的前缀和 把phi卷上一个1

#include
#include
#include
#include
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+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 S;

inline int Sum(ll n){
  if (n<=maxn) return sum[n];
  if (S.find(n)!=S.end()) return S[n];
  int tem=sum2(n);
  ll l,r;
  for (l=2;l*l<=n;l++) (tem+=P-l%P*Sum(n/l)%P)%=P;
  for (ll t=n/l;l<=n;l=r+1,t--)
    r=n/t,(tem+=P-sum1(l,r)*Sum(t)%P)%=P;
  return S[n]=tem;
}

inline int Solve(ll n){
  int tem=0; ll l,r;
  for (l=1;l*l<=n;l++) (tem+=Sum(n/l)%P)%=P;
  for (ll t=n/l;l<=n;l=r+1,t--)
    r=n/t,(tem+=(r-l+1)%P*Sum(t)%P)%=P;
  tem=(ll)tem*inv2%P;
  tem=(n%P*inv2%P+tem)%P;
  return tem;
}

int main(){
  ll l,r;
  freopen("t.in","r",stdin);
  freopen("t.out","w",stdout);
  Pre();
  scanf("%lld%lld",&l,&r);
  printf("%d\n",(Solve(r)+P-Solve(l-1))%P);
  return 0;
}