1220 约数之和
题意：求(sum_{i=1}^n sum_{j=1}^n sigma_1(ij))
[sigma_0(ij) = sum_{xmid i}sum_{ymid j}[(x,y)=1]\
sigma_1(ij) = sum_{xmid i}sum_{ymid j}xcdotfrac{j}{y}[(x,y)=1] \
]
怎么证明第二个式子？
[sigma_1(n) = prod_i(1 + p_i + p_i^2 + ... + p_i^{a_i})
]
和第一个式子类似，每个质因子是独立的，x和y枚举了i和j的所有质因子组合，gcd为1保证了对于每个质因子(p_i)，x和y有一个指数为0.这样实际上不重不漏的枚举了一个质因子在ij中的所有指数。
反演后得到
[sum_{d=1}^n mu(d) cdot d cdot g^2(frac{n}{d}) \
g(n) = sum_{i=1}^n sigma_1(i)=sum_{i=1}^n i lfloor frac{n}{i}
floor
]
杜教筛(mu cdot id)，卷上(id)即可；
(g)预处理+分块。
总复杂度(O(n^{frac{2}{3}}))
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <ctime>
using namespace std;
typedef long long ll;
const int N=1664512, mo=1e9+7;
int U=1664510;
inline ll read(){
char c=getchar(); ll x=0,f=1;
while(c<'0' || c>'9') {if(c=='-')f=-1; c=getchar();}
while(c>='0' && c<='9') {x=x*10+c-'0'; c=getchar();}
return x*f;
}
inline void mod(int &x) {if(x>=mo) x-=mo; else if(x<0) x+=mo;}
inline void mod(ll &x) {if(x>=mo) x-=mo; else if(x<0) x+=mo;}
bool notp[N]; int p[N/10], lp[N], si[N], mu[N];
void sieve(int n) {
si[1] = 1; mu[1] = 1;
for(int i=2; i<=n; i++) {
if(!notp[i]) p[++p[0]] = i, lp[i] = i, si[i] = 1+i, mu[i] = -1;
for(int j=1; j <= p[0] && i*p[j] <= n; j++) {
int t = i*p[j]; notp[t] = 1;
if(i % p[j] == 0) {
lp[t] = lp[i] * p[j];
if(lp[t] == t) mod(si[t] = si[i] + lp[t]);
else si[t] = (ll) si[t / lp[t]] * si[lp[t]] %mo;
mu[t] = 0;
break;
}
lp[t] = p[j];
si[t] = (ll) si[i] * (1 + p[j]) %mo;
mu[t] = -mu[i];
}
}
for(int i=1; i<=n; i++) mod(si[i] += si[i-1]), mod(mu[i] = mu[i-1] + mu[i] * i);
}
namespace ha {
const int p = 1001001;
struct meow{int ne, val, r;} e[3000];
int cnt, h[p];
inline void insert(int x, int val) {
int u = x%p;
e[++cnt] = (meow){h[u], val, x}; h[u] = cnt;
}
inline int quer(int x) {
int u = x%p;
for(int i=h[u];i;i=e[i].ne) if(e[i].r == x) return e[i].val;
return -1;
}
} using ha::insert; using ha::quer;
inline ll sum(ll n) {return n * (n + 1) / 2 %mo;}
int dj_u(int n) {
if(n <= U) return mu[n];
if(quer(n) != -1) return quer(n);
int ans=1, r;
for(int i=2; i<=n; i=r+1) {
r = n/(n/i);
mod(ans -= (ll) (sum(r) - sum(i-1)) * dj_u(n/i) %mo);
}
insert(n, ans);
return ans;
}
int g(int n) {
if(n <= U) return si[n];
int ans=0, r;
for(int i=1; i<=n; i=r+1) {
r = n/(n/i);
mod(ans += (ll) (sum(r) - sum(i-1)) * (n/i) %mo);
}
return ans;
}
int solve(int n) {
int ans=0, r, now, last=0;
for(int i=1; i<=n; i=r+1, last=now) {
r = n/(n/i); now = dj_u(r); ll t = g(n/i); //printf("solve [%d, %d] %d %d %d
", i, r, now, last, g(n/i));
mod(ans += (ll) (now - last) * t %mo * t %mo);
}
return ans;
}
int main() {
freopen("in", "r", stdin);
sieve(U);
int n=read();
printf("%d", solve(n));
}