题目链接

https://lydsy.com/JudgeOnline/problem.php?id=4916

题解

∑i=1Nμ(i2)=1∑i=1Nφ(i2)=∑i=1Niφ(i) \sum_{i=1}^N \mu(i^2)=1\\ \sum_{i=1}^N \varphi(i^2)=\sum_{i=1}^N i\varphi(i)

直接杜教筛即可。

代码

#include <map>
#include <cstdio>

int read()
{
  int x=0,f=1;
  char ch=getchar();
  while((ch<'0')||(ch>'9'))
    {
      if(ch=='-')
        {
          f=-f;
        }
      ch=getchar();
    }
  while((ch>='0')&&(ch<='9'))
    {
      x=x*10+ch-'0';
      ch=getchar();
    }
  return x*f;
}

const int maxn=1000000;
const int mod=1000000007;
const int inv6=166666668;

int p[maxn+10],prime[maxn+10],cnt,f[maxn+10];

int getprime()
{
  p[1]=f[1]=1;
  for(int i=2; i<=maxn; ++i)
    {
      if(!p[i])
        {
          prime[++cnt]=i;
          f[i]=i-1;
        }
      for(int j=1; (j<=cnt)&&(i*prime[j]<=maxn); ++j)
        {
          int x=i*prime[j];
          p[x]=1;
          if(i%prime[j]==0)
            {
              f[x]=f[i]*prime[j];
              break;
            }
          f[x]=f[i]*(prime[j]-1);
        }
    }
  for(int i=1; i<=maxn; ++i)
    {
      f[i]=(f[i-1]+1ll*f[i]*i)%mod;
    }
  return 0;
}

std::map<int,int> mp;

int getsum(int n)
{
  if(n<=maxn)
    {
      return f[n];
    }
  if(mp.count(n))
    {
      return mp[n];
    }
  int ans=1ll*n*(n+1)%mod*(2*n+1)%mod*inv6%mod;
  for(int l=2,r; l<=n; l=r+1)
    {
      r=n/(n/l);
      ans=(ans-((1ll*r*(r+1)/2)-(1ll*(l-1)*l)/2)%mod*getsum(n/l))%mod;
      if(ans<0)
        {
          ans+=mod;
        }
    }
  return mp[n]=ans;
}

int n;

int main()
{
  getprime();
  n=read();
  printf("1\n%d\n",getsum(n));
  return 0;
}