51nod2026 Gcd and Lcm [杜教筛]-CFANZ编程社区

51nod2026 Gcd and Lcm [杜教筛]

传送门

记住这个性质

$\sum_{i=1}^n\sum_{j=1}^mf(gcd(i,j))f(lcm(i,j))=\sum_{i=1}^n\sum_{j=1}^mf(i)f(j)$

那么原式就是

$(\sum_{i=1}^nf(i))^2$

考虑求

$\sum_{i=1}^nf(i)=\sum_{i=1}^n\sum_{d|i}\mu[d]*d=\sum_{d=1}^n\mu[d]*d*[n/d]$

然后整除分块一波, 求

$\sum_{d=1}^n\mu[d]*d$

的前缀和就可以了, 杜教筛解决

#include<bits/stdc++.h>
#define N 5000050
#define LL long long
#define Mod 1000000007
using namespace std;
int prim[N], tot, isp[N], mu[N];
LL val[N];
map<int,LL> F;
void prework(){
  mu[1] = val[1] = 1;
  for(int i=2;i<=N-50;i++){
    if(!isp[i]) prim[++tot] = i, mu[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) break;
      mu[i * prim[j]] = -mu[i]; 
    }
  }
  for(int i=1;i<=N-50;i++) val[i] = mu[i] * i;
  for(int i=2;i<=N-50;i++) val[i] = (val[i] + val[i-1]) % Mod;
}
LL getf(int x){
  if(x <= N-50) return val[x];
  if(F[x]) return F[x];
  LL ans = 1;
  for(int l=2,r;l<=x;l=r+1){
    int v = x/l; r = x/v;
    ans -= (LL)(l+r) * (LL)(r-l+1) / 2 * getf(v);
    ans = (ans % Mod + Mod) % Mod;
  }
  return F[x] = ans;
}
int main(){
  prework();
  int n; scanf("%d",&n);
  LL ans = 0;
  for(int l=1,r;l<=n;l=r+1){
    int v = n/l; r = n/v;
    ans += (LL)v * (getf(r) - getf(l-1)) % Mod;
    ans = (ans % Mod + Mod) % Mod;
  } ans = (ans * ans) % Mod;
  printf("%lld",ans); return 0;
}

0 条评论