[51nod1238] 最小公倍数之和V3 [杜教筛]-CFANZ编程社区

传送门

先从一道简单题入手 (LCM sum)

求

$\sum_{i=1}^nlcm(i,n)$

$=n\sum_{d|n}\sum_{i=1}^ni/d[gcd(i,n)=d]$

令k=i/d

$=n\sum_{d|n}\sum_{k=1}^{n/d}k[gcd(kd,n)=d]$

gcd里面同除d

$=n\sum_{d|n}\sum_{k=1}^{n/d}k[gcd(k,n/d)=1]$

我们发现枚举到n/d 与 d是等价的

$=n\sum_{d|n}\sum_{k=1}^{d}k[gcd(k,d)=1]$

$=n\sum_{d|n}\varphi(d)d+[d=1]/2$

这道题就可以转化为

$Ans=2\sum_{i=1}^n\sum_{j=1}^ilcm(i,j)-\sum_{i=1}^ni$

$=\sum_{d=1}^nd\varphi(d)\sum_{i=1}^{n/d}id=\sum_{d=1}^nd^2\varphi(d)\sum_{i=1}^{n/d}i$

杜教筛筛第一坨, 然后整除分块

#include<bits/stdc++.h>
#define N 1000050
#define LL long long
#define Mod 1000000007
#define inv2 500000004
#define inv6 166666668
using namespace std;
int prim[N], isp[N], tot, phi[N];
LL val[N],n;
map<LL,LL> F; 
void prework(){
  phi[1] = val[1] = 1;
  for(int i=2;i<=N-50;i++){
    if(!isp[i]) prim[++tot] = i, phi[i] = i-1;
    for(int j=1;j<=tot;j++){
      if(i*prim[j] > N-50) break;
      isp[i*prim[j]] = 1;
      if(i%prim[j] == 0){
        phi[i*prim[j]] = phi[i] * prim[j];
        break;
      }
      phi[i*prim[j]] = phi[i] * (prim[j] - 1);
    }
  }
  for(int i=2;i<=N-50;i++) val[i] = (LL)phi[i] * i % Mod * i % Mod;
  for(int i=2;i<=N-50;i++) val[i] = (val[i] + val[i-1]) % Mod;
} 
LL Sum(LL x){ x %= Mod; return x * (x+1) % Mod * inv2 % Mod;}
LL calc(LL x){ x %= Mod; return x * (x+1) % Mod * ((x*2+1) % Mod) % Mod * inv6 % Mod;}
LL getf(LL x){
  if(x<=N-50) return val[x];
  if(F[x]) return F[x];
  LL ans = Sum(x); ans = (ans * ans) % Mod;
  for(LL l=2,r;l<=x;l=r+1){
    LL v = x/l; r = x/v;
    ans -= (calc(r) - calc(l-1)) * getf(v) % Mod;
    ans = (ans % Mod + Mod) % Mod;
  } return F[x] = ans;
}
int main(){
  prework(); scanf("%lld",&n); LL ans = 0;
  for(LL l=1,r;l<=n;l=r+1){
    LL v = n/l; r = n/v;
    ans += Sum(v) * (getf(r) - getf(l-1)) % Mod;
    ans = (ans % Mod + Mod) % Mod;
  } printf("%lld",ans); return 0;
}