1227 平均最小公倍数
题意：求(frac{1}{n} sum_{i=1}^n lcm(n,i))
和的弱化版？
[ans = frac{1}{2}((sum_{i=1}^n sum_{d=1}^{lfloor frac{n}{i}
floor} dcdot varphi(d) ) - sum_{i=1}^n)
]
求(idcdot varphi)的前缀和，卷上(id)就行了
我竟然把整除分块打错了，直接i++，gg
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
typedef long long ll;
const int N=1664512, U=1664510, mo=1e9+7, inv2 = 500000004, inv6 = 166666668;
inline int read(){
char c=getchar(); int 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;}
bool notp[N]; int p[N/10], phi[N], s[N];
void sieve(int n) {
phi[1]=1; s[1]=1;
for(int i=2; i<=n; i++) {
if(!notp[i]) p[++p[0]] = i, phi[i] = i-1;
for(int j=1; j <= p[0] && i*p[j] <= n; j++) {
notp[i*p[j]] = 1;
if(i%p[j] == 0) {phi[i*p[j]] = (ll) phi[i] * p[j] %mo; break;}
phi[i*p[j]] = (ll) phi[i] * (p[j]-1) %mo;
}
mod(s[i] += s[i-1] + (ll) phi[i] * i %mo);
}
}
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;
for(int i=h[u];i;i=e[i].ne) if(e[i].r == x) return;
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;}
inline ll sum2(ll n) {return n * (n+1) %mo * (2*n+1) %mo *inv6 %mo;}
int dj_s(int n) { //printf("dj_s %d
", n);
if(n <= U) return s[n];
if(quer(n) != -1) return quer(n);
int ans = sum2(n), r;
for(int i=2; i<=n; i=r+1) {
r = n/(n/i);
mod(ans -= (ll) (sum(r) - sum(i-1)) * dj_s(n/i) %mo);
}
insert(n, ans);
return ans;
}
int solve(int n) {
int ans=0, r;
for(int i=1; i<=n; i=r+1) {
r = n/(n/i);
mod(ans += (ll) dj_s(n/i) * (r-i+1) %mo);
}
mod(ans += n);
return (ll) ans * inv2 %mo;
}
int l, r;
int main() {
freopen("in", "r", stdin);
sieve(U);
l=read(); r=read();
int ans = solve(r) - solve(l-1); mod(ans);
printf("%d", ans);
}