[杜教筛] 51Nod 1227 平均最小公倍数

来源：互联网发布：淘宝宝贝涨价被降权了编辑：程序博客网时间：2024/06/16 07:26

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<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+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=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;}

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