51Nod 最小公倍数之和V3

这题公式真tm难推……为了这题费了我一个草稿本……

woc……在51Nod上码LaTeX码了两个多小时……

一开始码完了前半段，刚码完后半段突然被51Nod吃了，重新码完后半段之后前半段又被吃了，吓得我赶紧换Notepad++接着写……

有的细节懒得再码了，这么一坨LaTeX估计也够你们看了……

\begin{equation}
ans=\sum_{i=1}^n\sum_{j=1}^n [i,j]\\
=2\sum_{i=1}^n\sum_{j=1}^i [i,j]-\frac{n(n+1)}2\\
Let\space s(n)=\sum_{i=1}^n\sum_{j=1}^i [i,j],f(n)=\sum_{i=1}^n [i,n]\\
f(n)=\sum_{i=1}^n [i,n]\\
=\sum_{i=1}^n\frac{in}{(i,n)}\\
=n\sum_{i=1}^n\frac i{(i,n)}\\
=n\sum_{d|n}\sum_{i=1}^n[(i,n)=d]\frac i d\\
=n\sum_{d|n}\sum_{i=1}^{\frac n d}[(i,\frac n d)=1]i\\
=n\sum_{d|n}\sum_{i=1}^{d}[(i,d)=1]i\\
=n\sum_{d|n}\frac{\phi(d)d+[d=1]}2\\
=n\frac{1+\sum_{d|n}\phi(d)d}2\\
s(n)=\sum_{i=1}^n f(i)\\
=\frac{\sum_{i=1}^n i(1+\sum_{d|i}\phi(d)d)}2\\
=\frac{\sum_{i=1}^n i+\sum_{i=1}^n i\sum_{d|i}\phi(d)d}2\\
=\frac{\frac{n(n+1)}2+\sum_{i=1}^n i\sum_{d|i}\phi(d)d}2\\
=\frac{\frac{n(n+1)}2+\sum_{d=1}^n\phi(d)d\sum_{d|i}i}2\\
=\frac{\frac{n(n+1)}2+\sum_{d=1}^n\phi(d)d^2\sum_{i=1}^{\lfloor\frac n d\rfloor}i}2\\
=\frac{\frac{n(n+1)}2+\sum_{i=1}^n i\sum_{d=1}^{\lfloor\frac n i\rfloor}\phi(d)d^2}2\\
ans=2s(n)-\frac{n(n+1)}2\\
=\sum_{i=1}^n i\sum_{d=1}^{\lfloor\frac n i\rfloor}\phi(d)d^2\\
Let \space h(d)=\phi(d)d^2,g(n)=\sum_{d=1}^nh(d)\\
n=\sum_{d|n}\phi(d)\\
n^3=\sum_{d|n}\phi(d)n^2\\
=\sum_{d|n}\phi(d)d^2(\frac n d)^2\\
=\sum_{d|n}h(d)(\frac n d)^2\\
\sum_{i=1}^n i^3=\sum_{i=1}^n\sum_{d|i}h(d)(\frac i d)^2\\
=\sum_{d=1}^n h(d)\sum_{d|i}(\frac i d)^2\\
=\sum_{d=1}^n h(d)\sum_{i=1}^{\lfloor\frac n d \rfloor}i^2\\
=\sum_{i=1}^n i^2\sum_{d=1}^{\lfloor\frac n i\rfloor}h(d)\\
=\sum_{i=1}^n i^2 g(\lfloor\frac n i\rfloor)\\
g(n)=\sum_{i=1}^n i^3-\sum_{i=2}^ni^2 g(\lfloor\frac n i\rfloor)
\end{equation}
然后就是杜教筛的形式了，上杜教筛即可
\begin{equation}
\sum_{i=1}^n i^3=(\frac{n(n+1)}2)^2\\
\sum_{i=1}^n i^2=\frac{n(n+1)(2n+1)}6\\
ans=\sum_{i=1}^n i g(\lfloor\frac n i\rfloor)
\end{equation}
在外面套上一层分块不会影响复杂度，利用g的定义，筛出$\phi$之后即可算出较小的g，较大的g直接杜教筛算即可，总复杂度$O(n^{\frac 2 3})$

贴两份代码(虽然Python2的代码用Python2和Pypy2交都过不去......)：

 '''
 h(i)=phi(d)*d^2
 g(i)=sum{h(j)|1<=j<=i}
 g(n)=sum{i^3|1<=i<=n}-sum{i^2*g(n/i)|2<=i<=n}
 线筛预处理一部分g，大一些的部分直接上杜教筛即可
 s_3(n)=s_1(n)^2,s_2(n)=n(n+1)(2n+1)/6
 '''
 p=1000000007
 table_size=8000000
 def get_table(n):
     global phi
     notp=[False for i in xrange(n+1)]
     prime=[]
     cnt=0
     phi[1]=1
     for i in xrange(2,n+1):
         if not notp[i]:
             prime.append(i)
             cnt+=1
             phi[i]=i-1
         for j in xrange(cnt):
             if i*prime[j]>n:
                 break
             notp[i*prime[j]]=True
             if i%prime[j]:
                 phi[i*prime[j]]=phi[i]*(prime[j]-1)
             else:
                 phi[i*prime[j]]=phi[i]*prime[j]
                 break
     for i in xrange(2,n+1):
         phi[i]=phi[i]*i*i%p
         phi[i]=(phi[i]+phi[i-1])%p
 def s1(n):
     return (n*(n+1)>>1)%p
 def s2(n):
     return (n*(n+1)*((n<<1)+1)>>1)/3%p
 def S(n):
     if n<table_size:
         return phi[n]
     elif hashmap.has_key(n):
         return hashmap[n]
     ans=n*(n+1)/2
     ans*=ans
     ans%=p
     i=2
     while i<=n:
         last=n/(n/i)
         #print 'last=%d'%last
         ans-=(s2(last)-s2(i-1))*S(n/i)%p
         ans%=p
         i=last+1
     if ans<0:
         ans+=p
     hashmap[n]=ans
     return ans
 n=input()
 hashmap=dict()
 table_size=min(table_size,n)
 phi=[0 for i in xrange(table_size+1)]
 get_table(table_size)
 #print 'table OK'
 ans=0
 i=1
 while i<=n:
     last=n/(n/i)
     ans+=S(n/i)*(s1(last)-s1(i-1))%p
     ans%=p
     i=last+1
 print ans

 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<ext/pb_ds/assoc_container.hpp>
 #include<ext/pb_ds/hash_policy.hpp>
 #define s1(n) ((long long)(n)%p*(((n)+1)%p)%p*inv_2%p)
 #define s2(n) ((long long)(n)%p*(((n)+1)%p)%p*((((long long)(n)%p)<<1)%p+1)%p*inv_6%p)
 using namespace std;
 using namespace __gnu_pbds;
 const int table_size=,maxn=table_size+,p=,inv_2=,inv_6=;
 void get_table(int);
 int S(long long);
 bool notp[maxn]={false};
 int prime[maxn]={},phi[maxn]={};
 gp_hash_table<long long,int>hashmap;
 long long n;
 int main(){
     scanf("%lld",&n);
     get_table(min((long long)table_size,n));
     int ans=;
     for(long long i=,last;i<=n;i=last+){
         last=n/(n/i);
         ans+=S(n/i)*((s1(last)-s1(i-))%p)%p;
         ans%=p;
     }
     if(ans<)ans+=p;
     printf("%d",ans);
     return ;
 }
 void get_table(int n){
     phi[]=;
     for(int i=;i<=n;i++){
         if(!notp[i]){
             prime[++prime[]]=i;
             phi[i]=i-;
         }
         for(int j=;j<=prime[]&&i*prime[j]<=n;j++){
             notp[i*prime[j]]=true;
             if(i%prime[j])phi[i*prime[j]]=phi[i]*(prime[j]-);
             else{
                 phi[i*prime[j]]=phi[i]*prime[j];
                 break;
             }
         }
     }
     for(int i=;i<=n;i++){
         phi[i]=(long long)phi[i]*i%p*i%p;
         phi[i]=(phi[i]+phi[i-])%p;
     }
 }
 int S(long long n){
     if(n<=table_size)return phi[n];
     else if(hashmap.find(n)!=hashmap.end())return hashmap[n];
     int ans=s1(n)*s1(n)%p;
     for(long long i=,last;i<=n;i=last+){
         last=n/(n/i);
         ans-=S(n/i)*((s2(last)-s2(i-))%p)%p;
         ans%=p;
     }
     if(ans<)ans+=p;
     return hashmap[n]=ans;
 }
 /*
 h(i)=phi(d)*d^2
 g(i)=sum{h(j)|1<=j<=i}
 g(n)=sum{i^3|1<=i<=n}-sum{i^2*g(n/i)|2<=i<=n}
 ans=sum{i*g(n/i)|1<=i<=n}
 线筛预处理一部分g，大一些的部分直接上杜教筛即可
 s_3(n)=s_1(n)^2,s_2(n)=n(n+1)(2n+1)/6
 */